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ISC Class XII Board Exam 2026 : Mathematics

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Renuka Manthina
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MATHEMATICS DPP DPP No. 3 DAILY PRACTICE PAPER COMPLEX NUMBER Integral power of iota, Algebraic operations and Equality of complex numbers 2 3 = 1. 2. 3. (a) 6 (c) i 6 If is a positive integer, then which of the following relations is false n (b) (d) None of these (a) i4 n = 1 (b) i4n 1 = i (c) i4n+1 = i (d) i 4n = 1 If n is a positive integer, then (a) 1 (c) 4. 5. 6. If (d) i m 1+ i = 1, then 1 i 10. = i the least integral value of (b) 4 (c) 8 (d) None of these If m is (1 i)n = 2n , then n = (a) 1 (b) 0 (c) (d) None of these 1 The value of is (1 + i)5 (1 i)5 (a) 8 (b) (c) 8 (d) 32 1+ i 1 i + 1 i 1+ i 2 8i is equal to (b) (d) 2i 2 The value of (a) (c) 9. 4 n +1 (a) 2 (a) (c) 8. 1+ i 1 i (b) 1 2 7. 6 i 592 +i 590 +i 588 2i 2 + i 586 + i 584 i 582 + i 580 + i 578 + i 576 + i 574 1 = (b) 2 (d) 4 1 3 1 + i 2 + i 4 + i 6 + ..... + i 2n is (a) Positive (b) Negative (c) Zero (d) Cannot be determined i 2 + i 4 + i 6 + ...... (a) i (c) 1 upto (2n + 1) terms = (b) i (d) 1 Page | 1 MATHEMATICS DPP DPP No. 3 DAILY PRACTICE PAPER 11. If i = 1 , (a) then If is equal to 1 + i2 + i3 i6 + i8 (b) 1 2 i (c) 3 12. COMPLEX NUMBER (d) i 2 = 1 , 1 200 i then the value of n is n =1 13. (a) 50 (c) 0 (b) 50 (d) 100 The value of the sum (i 13 n + in +1 ) , where i = 1 , equals n =1 (a) (c) 14. 15. 16. 17. (b) i 1 (d) 0 i i The least positive integer which will reduce i 1 i +1 n to a real number, is (a) 2 (b) 3 (c) 4 (d) 5 1+ 3+ 5 +...+(2n+1) The value of i is (a) i if n is even, i if n is odd (b) 1 if n is even, 1 if n is odd (c) 1 if n is odd, 1 if n is even (d) i if n is even, 1 if n is odd 1 = 2 cos , x If x+ (a) cos + i sin (b) cos i sin (c) cos i sin (d) sin i cos The value of then x is equal to i n + i n+1 + i n+ 2 + i n+ 3 , (n N ) (a) 0 (d) None of these The value of 8 (1 + i) + (1 i)8 (a) 16 , where 19. (1 + i) 20. (a) 32 i (c) 24 i 32 The value of (a) 0 (c) 26 is (b) 16 (c) 32 10 is (b) 1 (c) 2 18. n (d) 32 2 i = 1, is equal to (b) 64 + i (d) None of these (1 + i)6 + (1 i)6 is (b) 27 (d) None of these Page | 2 COMPLEX NUMBER MATHEMATICS DPP DPP No. 3 DAILY PRACTICE PAPER 21. 22. 23. If i 2 = 1 , then sum i + i 2 + i3 + ... to (a) 1 (b) 1 (c) i (d) 0 3 2 If x = 3 + i , then x 3x 8 x + 15 = (a) 6 (b) 10 (c) 18 (d) 15 The smallest positive integer n for which (a) 1 (c) 3 24. 25. 26. 27. 1000 terms is equal to (b) 2 (d) 4 The values of x and y satisfying the equation (a) x = 1, y = 3 (b) x = 3, y = 1 (c) x = 0, y = 1 (d) x = 1, y = 0 If z1 and z2 (1 + i)x 2i (2 3i) y + i + =i 3+i 3 i are be two complex number, then Re (z1z2 ) = (a) Re (z1). Re(z2 ) (b) Re (z1) . (c) Im (z1). Re (z2 ) (d) None of these Im (z2 ) 3 3 + 4i 1 + = 1 2 i 1 + i 2 4i (a) 1 9 + i 2 2 (b) 1 9 i 2 2 (c) 1 9 i 4 4 (d) 1 9 + i 4 4 Additive inverse of (a) 0 + 0i (c) 1 + i (1 + i)2 3 i 1 i is (b) 1 i (d) None of these = 28. Re 29. (a) 1 / 5 (b) 1/5 (c) 1/10 (d) 1/10 If (1 i)x + (1 + i)y = 1 3i, then (x, y) = 30. (1 + i)2n = (1 i)2n is (a) (2, 1) (b) (c) ( 2, 1) (d) (2, 1) 3 + 2i sin 1 2i sin will be real, if ( 2, 1) = (a) 2n (b) (c) n (d) None of these n + 2 Page | 3 MATHEMATICS DPP DPP No. 3 DAILY PRACTICE PAPER 5 + 12i + 5 12i 31. 32. 5 + 12i 5 12i 33. = (a) 3 i 2 (b) 3 i 2 (c) 3 2 (d) 3 2 If z and z are complex numbers such that (a) 0 + i 0 (b) 1 + 0i (c) If (d) 0+i a 2 + b 2 = 1, then 35. 36. then z' = 1+i (b) 2 (d) a + ib 3 + 2i sin 1 2i sin (a) 2n (c) n will be purely imaginary, if = (b) 3 n + 3 (d) None of these 3 The real part of (1 cos + 2i sin ) 1 is (a) 1 3 + 5 cos (b) 1 5 3 cos (c) 1 3 5 cos (d) 1 5 + 3 cos If z.z' = z , 1 + b + ia = 1 + b ia (a) 1 (c) b + ia 34. COMPLEX NUMBER (x + iy)1 / 3 = a + ib, then x y + a b is equal to (a) 4(a 2 + b 2 ) (b) (c) 4(b 2 a 2 ) (d) None of these 4(a 2 b 2 ) 2 37. 38. 2i = 1 + i (a) 1 (b) 2i (c) (d) 1 2i 1 i The real values of x and y for which the equation is is satisfied, are 8 5 ,y = 13 13 (a) x= 5 8 ,y = 13 13 (b) (c) x= 5 14 ,y = 13 13 (d) None of these x= (x + iy) (2 3i) = 4 + i Page | 4 MATHEMATICS DPP DPP No. 3 DAILY PRACTICE PAPER 39. The real values of (a) x and (b) x = 2, y = 3 (c) Both (a) and (b) 40. 41. 42. 44. 45. 46. (1 + i)2 (2 i) is 1 5 (b) (c) 4 5 (d) None of these If z 0 3 5 is a complex number, then (a) Re(z) = 0 Im( z 2 ) = 0 (b) (c) Re(z) = 0 Re(z 2 ) = 0 (d) None of these 5( 8 + 6i) = a + ib , (1 + i)2 If is satisfied, are 1 3 (a) Re(z 2 ) = 0 Im( z 2 ) = 0 then (a, b) equals (b) (20, 15) (d) None of these The true statement is (a) 1 i 1 + i (b) 2i + 1 2i + 1 (c) 2i 1 (d) None of these 1 2i 4 i + = 2 + i 3 + 2i (a) 24 10 + i 13 13 (b) 24 10 i 13 13 (c) 10 24 + i 13 13 (d) 10 24 i 13 13 a + ib c + id can (a) b = 0, c = 0 (c) a = 0, c = 0 If x + iy = (a) (c) 47. x = 2, y = (x 4 + 2xi) (3 x 2 + yi) = (3 5i) + (1 + 2yi) (d) None of these The imaginary part of (a) (15, 20) (c) ( 15, 20) 43. for which the equation y COMPLEX NUMBER If be explained only when (b) b = 0, d = 0 (d) 3 2 + cos + i sin 3x 4 4x + 3 a = 0, d = 0 , then x 2 + y 2 is equal to (b) 4 x 3 (d) None of these ( p + i)2 = + i , then 2 + 2 2p i is equal to (a) ( p 2 + 1)2 4 p2 1 (b) ( p 2 1)2 4 p2 1 (c) ( p 2 1)2 4 p2 + 1 (d) ( p 2 + 1)2 4 p2 + 1 Page | 5 MATHEMATICS DPP DPP No. 3 DAILY PRACTICE PAPER 48. 49. If z = 3 4i , then (a) 5 (c) 5 If z1 = 1 i and z 4 3z 3 + 3z 2 + 99z 95 is z2 = 2 + 4i , 51. then , then (a) x = 1, y = 3 (b) x = 1, y = 3 (c) x = 1, y = 3 (d) x = 1, y = 3 or x = 1, y = 3 If z z Im 1 2 = z1 (b) 2 (d) 4 3 x + 2iy 15 = 5i 2 8 x + 3iy If equal to (b) 6 (d) 4 (a) 1 (c) 3 50. COMPLEX NUMBER 100 i k = x + iy , then the values of x and y are k =0 52. 53. 54. 55. (a) x = 1, y = 0 (b) x = 1, y = 1 (c) x = 1, y = 0 (d) x = 0, y = 1 If z(1 + a) = b + ic and a2 + b2 + c 2 = 1 , then 1 + iz = 1 iz (a) a + ib 1+ c (b) (c) a + ic 1+b (d) None of these Let z1 , z 2 b ic 1+ a be two complex numbers such that (a) z1 = z 2 (b) z1 = z2 (c) z1 = z 2 (d) z1 = z 2 If (x + iy)(p + iq) = (x 2 + y 2 )i , p = x, q = y (b) (c) x = q, y = p (d) None of these form of and z1 z 2 both are real, then then (a) A + iB z1 + z 2 p = x 2, q = y 2 (cos x + i sin x)(cos y + i sin y) (cot u + i)(1 + i tan v) is (a) sin u cos v [cos(x + y u v) + i sin(x + y u v)] (b) sin u cos v [cos(x + y + u + v) + i sin(x + y + u + v)] (c) sin u cos v [cos(x + y + u + v) i sin(x + y + u + v)] (d) None of these Page | 6 COMPLEX NUMBER MATHEMATICS DPP DPP No. 3 DAILY PRACTICE PAPER 56. If x, y R and (x + iy)(3 + 2i) = 1 + i , (a) (c) 57. 58. 59. 1 1, 5 5 1 , 13 13 100 1 i 1+ i If = a + ib , 64. 1 1 , 5 5 (b) a = 1, b = 0 (c) a = 0, b = 1 (d) a = 1, b = 2 If z1 = (4,5) and 23 2 , 12 13 (b) (c) 2 23 , 13 13 (d) If z = 1 + i, If 1 1 i 2 23 , 13 13 2 23 , 13 13 1 ) = x + iy , then (x, y) is (b) (1, 3) (d) (0, 0) then cot 2 Solving equals (b) 1 i (d) i/2 a = cos + i sin , i cot is then the multiplicative inverse of z2 is (where i = 6i 3i 4 3i 20 3 If z1 z2 z 2 = ( 3,2) then (a) (c) 63. (d) a = 2, b = 1 (a) 62. 1 1 , 13 13 (a) (a) (3, 1) (c) (0, 3) 61. (b) (x, y) then (a) 2 i (c) i/2 60. then 3 2yi = 9 x 7i , 1+ a = 1 a (b) cot (d) i tan where 2 2 i 2 = 1, for x and y real, we get (a) x = 0.5 , y = 3.5 (b) x=5 , y=3 (c) x= 1 , y=7 2 (d) x = 0, y = The complex number 1 + 2i 1 i (a) First (c) Third (b) Second (d) Fourth The real part of 3 + 7i 2i lies in which quadrant of the complex plane 1 1 cos + i sin is equal to (a) 1/4 (b) 1/2 (c) tan /2 (d) 1/1 cos Page | 7 COMPLEX NUMBER MATHEMATICS DPP DPP No. 3 DAILY PRACTICE PAPER 65. 66. 67. The statement (a + ib) (c + id) is true for (a) a2 + b2 = 0 (b) b2 + c 2 = 0 (c) a2 + c 2 = 0 (d) b2 + d 2 = 0 The multiplication inverse of a number is the number itself, then its initial value is (a) i (b) 1 (c) 2 (d) i If z = x + iy, z1 / 3 = a ib and x y = k (a 2 b 2 ) a b (a) 2 (b) 4 (c) 6 (d) 1 then value of k equals Conjugate, Modulus and Argument of complex numbers 1. 2. 3. The complex numbers sin x + i cos 2 x and x = n (b) (c) x=0 (d) No value of x is a complex number, then z (z 1 )(z) = (a) 1 (b) 1 (c) 0 (d) None of these If z are conjugate to each other for 1 x = n + 2 (a) If cos x i sin 2 x is a complex number such that z 2 = (z )2 , then (a) z is purely real (b) z is purely imaginary (c) Either z is purely real or purely imaginary (d) None of these 4. If z is a complex number, then z. z = 0 if and only if (a) z=0 (b) (c) Im (z) = 0 (d) None of these Re(z) = 0 5. If 6. (a) A2 + B2 (b) A2 B2 (c) A2 (d) B2 The number of solutions of the equation (a) 1 (b) 2 (c) 3 (d) 4 (a + ib)(c + id)(e + if )(g + ih) = A + iB, then (a 2 + b 2 )(c 2 + d 2 )(e 2 + f 2 )(g 2 + h 2 ) z2 + z = 0 = is Page | 8 COMPLEX NUMBER MATHEMATICS DPP DPP No. 3 DAILY PRACTICE PAPER 7. 8. 9. 10. For the complex number z , one from (a) 7 26i 25 (b) 7 26i 25 (c) 7 + 26i 25 (d) 7 + 26i 25 (z + a)(z + a) , where If is 3 + ix 2y and x 2 + y + 4i are conjugate complex can be is (d) None of these | z + a |2 z i (z i) z+i is a purely imaginary number, then z.z is equal to (b) 1 (d) None of these c+i = a + ib , c i 12. If 13. (a) 1 (c) c 2 If the conjugate of where 1 5 (a) x= (c) x + iy = a, b, c are (b) (d) 1 i 1 2i real, then (d) (2 + i)2 , 3+i a2 + b2 = 1 c2 (x + iy)(1 2i) be 1 + i , (b) The conjugate of y= then 3 5 x iy = 1 i 1 + 2i in the form of a + ib, is (a) 13 15 + i 2 2 (b) 13 15 + i 10 2 (c) 13 9 + i 10 10 (d) 13 9 + i 10 10 If zz is real, is equivalent to (b) z 2 + a 2 a (a) 0 (c) 2 15. 2 + 5i 4 3i The conjugate of the complex number (c) 14. and (a) A real number (b) A imaginary number (c) Both are real numbers (d) Both are imaginary numbers The values of x and y for which the numbers (a) ( 2, 1) or (2, 1) (b) ( 1, 2) or ( 2, 1) (c) (1, 2) or ( 1, 2) (d) None of these (a) | z a | 11. z+z z = 3 + 5i, then z 3 + z + 198 = (a) (c) 3 5i 3 + 5i (b) (d) 3 + 5i 3 5i Page | 9 COMPLEX NUMBER MATHEMATICS DPP DPP No. 3 DAILY PRACTICE PAPER 16. 17. 18. The conjugate of complex number (a) 3i 4 (b) 11 + 10i 17 (c) 11 10i 17 (d) 2 + 3i 4i (a) Re(z) 0 (b) (c) Re(z) 2 (d) None of these 2 z1 3z2 If 20. (a) 3/2 (c) 2/3 If z1 and 22. Re(z) 0 is a purely imaginary number, then z2 (a) 2 | z1 | | z 2 | (b) 2 | z1 | + 2 | z 2 | (c) | z1 |2 +| z 2 |2 (d) 2| z1 | | z2 | 2 z1 z 2 z1 + z 2 = (b) 1 (d) 4/9 are any two complex numbers then | z1 + z2 |2 2 2 If z is a complex number such that (a) | z |= 0 (b) | z |= 1 (c) (d) | z | 1 | z | 1 + | z1 z 2 |2 is equal to 2 z 1 z +1 is purely imaginary, then If z is a complex number, then which of the following is not true (a) | z 2 |=| z |2 (b) | z 2 |=| z |2 (c) 23. is Conjugate of 1 + i is (a) i (b) 1 (c) 1 i (d) 1 + i The inequality | z 4 | |z 2| represents the region given by 19. 21. 2 3i , 4 i (d) z=z z2 = z2 The maximum value of | z | where z satisfies the condition (a) (c) (b) (d) 3 1 3 and is 2+ 3 If 25. which is (a) Positive real (b) Negative real (c) Zero or purely imaginary(d) None of these The solution of the equation | z | z = 1 + 2i is z2 2 =2 z 3 +1 24. z1 z+ are two complex numbers satisfying the equation (a) 2 3 i 2 (b) 3 + 2i 2 (c) 3 2i 2 (d) 2+ z1 + z 2 z1 z 2 =1, then z1 z2 is a number 3 i 2 Page | 10 COMPLEX NUMBER MATHEMATICS DPP DPP No. 3 DAILY PRACTICE PAPER 26. If z1 and z2 be complex numbers such that negative imaginary part, then 27. 28. 29. z1 has positive real part and z2 has be (d) None of these | z | 1 If | z1 |=| z2 |= .......... =| zn |= 1, (a) 1 (c) 31. and | z1 |=| z2 | . If (a) Purely imaginary (b) Real and positive (c) Real and negative (d) None of these The moduli of two complex numbers are less than unity, then the modulus of the sum of these complex numbers (a) Less than unity (b) Greater than unity (c) Equal to unity (d) Any The product of two complex numbers each of unit modulus is also a complex number, of (a) Unit modulus (b) Less than unit modulus (c) Greater than unit modulus (d) None of these 4 Let z be a complex number, then the equation z + z + 2 = 0 cannot have a root, such that (a) | z | 1 (b) | z |= 1 (c) 30. (z1 + z 2 ) may (z1 z 2 ) z1 z2 then the value of | z1 + z2 + z3 + .......... ... + zn |= (b) | z1 | + | z2 | +....... + | zn | 1 1 1 + + ......... + z1 z2 zn (d) None of these For any complex number 1 z, z = if z and only if (a) z is a pure real number (b) | z |= 1 32. 33. (c) z is a pure imaginary number (d) z = 1 If z1 and z 2 are two complex numbers, then (a) | z1 | | z 2 | (b) | z1 | | z 2 | (c) | z1 | + | z 2 | (d) | z 2 | | z1 | The values of z for is which | z + i |=| z i | are (a) Any real number 34. | z1 z 2 | (b) Any complex number (c) Any natural number (d) None of these The value of | z 5 |if z = x + iy , is (a) (x 5) 2 + y 2 (b) x 2 + (y 5) 2 (c) (x y) 2 + 5 2 (d) x 2 + (y 5) 2 Page | 11 COMPLEX NUMBER MATHEMATICS DPP DPP No. 3 DAILY PRACTICE PAPER 35. (1 + i) (a) (2 + i) = (3 + i) 1 2 (c) 1 36. If z1 , z 2 37. 38. 39. 40. z1 z 2 and (d) 1 z1 + z 2 2k tan 1 2 k + 1 (c) 1 2 are two complex numbers such that between (a) (b) 2 tan 1 k and iz1 = kz 2 , where k R, (b) 2k tan 1 1 k2 (d) 2 tan 1 k (a) Im(z) = 0 (b) (c) amp(z) = (d) None of these (a) | z1 | (b) | z 2 | (c) (d) | z1 + z2 | + | z1 z2 | 1 1 (z1 + z 2 ) + z1 z 2 + (z1 + z 2 ) z1 z 2 2 2 (b) | z1 z 2 | (c) (d) | z1 | | z 2 | Modulus of 3 + 2i 3 2i Then z12 z22 | + | z1 z12 z22 | is equal to is (b) 1/2 (d) 2 z 1 | z |= 1, (z 1) and z = x + iy, then z +1 41. If 42. (a) Purely real (b) Purely imaginary (c) Zero (d) Undefined The minimum value of | 2z 1| + | 3z 2| is (a) 0 (b) (c) (d) 2/3 1/ 3 If | z |= 1 and 1 = 1. z = (a) | z1 + z 2 | | z1 | + | z 2 | z+ Re(z) = 0 If z1, z2 are any two complex numbers, then | z1 + | z1 + z2 | then the angle is Let z be a complex number (not lying on X-axis of maximum modulus such that (a) 1 (c) 2 43. z1 z 2 =1 z1 + z 2 = z 1 z +1 1/ 2 (where (a) 0 (b) (c) z 1 . z + 1 | z + 1 |2 (d) is z 1) , then Re( ) is 1 | z + 1|2 2 | z + 1 |2 Page | 12 COMPLEX NUMBER MATHEMATICS DPP DPP No. 3 DAILY PRACTICE PAPER 44. A real value of x will satisfy the equation (a) (c) (b) (d) 2 2 = 1 2 + 2 =1 45. Let 46. (a) 0 (c) 1 If z1 and z1 3 4ix = i ( , real), 3 + 4ix 2 2 =1 2 2 = 2 be a complex number with | z1 |= 1 and z2 if z 2 be any complex number, then (b) 1 (d) 2 are two non-zero complex numbers such that z1 z2 = 1 z1z2 | z1 + z2 |=| z1 | + | z2 |, then arg (z1) arg (z2 ) is equal to 47. 48. (a) (b) (c) (d) 0 2 (a) tan 1 5 (c) tan 1 3 5 3 2 (b) 5 tan 1 3 (d) 3 tan 1 5 Argument and modulus of 2 and 1 (c) 0 and 2 1+ i 1 i (b) (d) 2 2 are respectively and 2 and 1 If z be the conjugate of the complex number z , then which of the following relations is false (a) | z |=| z | (b) z. z =| z |2 (c) 50. arg (5 3i) = (a) 49. If | z |= 4 and (a) (c) arg z = (b) (d) 2 3 + 2i If z = 1 i 52. (a) 60o (c) 240o If arg (z) = , then 3 1+ i 3 arg z = arg z 5 , then 6 2 3 2i 51. (a) (c) (d) z1 + z 2 = z1 + z 2 z= 2 3 + 2i 3 +i , then arg(z) = (b) (d) 120o 300o arg (z) = (b) (d) Page | 13 COMPLEX NUMBER MATHEMATICS DPP DPP No. 3 DAILY PRACTICE PAPER 53. 54. 55. The amplitude of the complex number (a) 2 sin (c) (b) 2 (a) (c) 1 + 3i 3 6 120o 3+i 3 i arg + 2 i 2+i If 2 z1.z2........ zn = z, 60. (b) (d) 3 6 1+ i 3 120o then (b) (d) 2 4 arg z1 + arg z2 + .... + arg zn (b) Multiple of (b) (c) 0 (d) and z (a) (b) (c) 0 (d) z arg z differ by a 2 Im (z) 0 . Then Im(z) 0 . Then (b) (c) 0 (d) is equal to 2 Re(z) 0 , then arg(z) is equal to 2 2 Let z be a complex number. Then the angle between vectors (a) (b) 0 2 arg (z) 2 is a purely real number such that is equal to 2 be a purely imaginary number such that (c) arg(z) 2 Let If is 60o (a) 61. is (c) Greater than (d) Less than Let z be a purely imaginary number such that (a) 59. 2 is equal to (a) Multiple of 58. 3 +1 (b) (d) 60o (c) 0 57. The argument of the complex number (a) is (d) None of these The amplitude of (a) (c) 56. z = sin + i(1 cos ) z and iz is (d) None of these Page | 14 COMPLEX NUMBER MATHEMATICS DPP DPP No. 3 DAILY PRACTICE PAPER 62. 63. 64. 65. 66. 67. 68. 69. 70. For any two complex numbers (a) z Re 1 z2 (c) Re(z1 z 2 ) = 0 =0 z 1 , z 2 we (b) z Im 1 z2 (d) Im(z1 z 2 ) = 0 If for complex numbers and z1 z 2 , arg(z1 / z 2 ) = 0, (b) | z1 | | z 2 | (c) (d) 0 If | z1 + z 2 |=| z1 z 2 | , (a) (c) z arg 1 = z2 , then z1 + z 2 is (d) None of these 0 amp (z) , then amp (z) amp ( z) = (a) 0 (b) 2 amp (z) (c) (d) z = 1 cos + i sin , (c) If 2 2 + 2 z1 , z 2 C , then (a) amp (z1 z 2 ) (c) z amp 2 z1 then amp z = (b) (d) 2 2 2 z amp 1 = z2 (b) amp (z1z 2 ) z amp 1 z2 (d) The argument of the complex number (a) (c) If is equal to (c) Purely real z2 3 (b) Purely imaginary (a) and (a) 0 If z1 (d) 0 2 If | z1 |=| z 2 |and If then then | z1 z 2 | is equal to then the difference in the amplitudes of (b) 4 | z1 |2 + | z 2 |2 =0 (a) | z1 | + | z 2 | || z1 | | z 2 || have | z1 + z 2 |2 = (b) 3 (d) 5 | z1 |=| z2 | and 13 5i 4 9i is 4 6 amp z1 + amp z 2 = 0 , (a) z1 = z2 (b) z1 = z2 (c) z1 + z 2 = 0 (d) z1 = z2 then Page | 15 COMPLEX NUMBER MATHEMATICS DPP DPP No. 3 DAILY PRACTICE PAPER 71. 72. | z1 + z 2 |=| z1 | + | z 2 | is possible if z2 = z1 (b) (c) arg (z1 ) = arg (z 2 ) (d) | z1 |=| z 2 | Amplitude of 1 i 1+ i 74. 75. 76. (b) /2 (c) /4 (d) /6 Which of the following are correct for any two complex numbers (a) | z1z2 |=| z1 || z2 | (b) arg (z1z2) = (arg z1)(arg z2) (c) | z1 + z2 |=| z1 | + | z2 | (d) | z1 z2 | | z1 | | z2 | The amplitude of (a) (c) 1+ 3 i (b) 2 (d) The amplitude of If z= (a) (c) 79. 1+ 3i 3 i 6 If 2 1+ 3 i (b) (d) 2 / 3 6 (c) | z |= + i sin 6 /5 / 10 /6 /2 arg z = arg (z) is /3 /4 then (b) | z |= 1, arg z = 4 6 3 5 , arg z = 2 24 The amplitude of (a) (c) 2 then the value of z = cos is (b) (d) (a) | z |= 1, 80. z2 (d) None of these 3 (a) 0 (c) / 3 78. and is 3 +i 6 z1 The amplitude of 0 is (a) 0 (b) / 2 (c) (d) None of these If arg z 0 then arg ( z) arg (z) is equal to (a) (b) (c) 77. z2 = is (a) /2 73. 1 z1 (a) sin (d) | z |= 5 3 1 , arg z = tan 1 2 2 + i 1 cos 5 (b) (d) 2 / 5 / 15 Page | 16 COMPLEX NUMBER MATHEMATICS DPP DPP No. 3 DAILY PRACTICE PAPER 81. Argument of (a) 2 3 (c) 1 i 3 3 (b) (d) 3 2 3 82. If z and 83. equal to (a) 1 (b) 1 (c) i (d) i The sum of amplitude of z and another complex number is written (a) z (b) z (c) z (d) z 84. The modulus and amplitude of (a) are is 2 and (c) 1 and 85. 86. 87. If two non zero complex numbers such that z1 = 1 + 2i (d) 1 and 3 z 2 = 3 + 5i , 31 17 (b) 17 22 (c) 17 31 (d) 22 17 (3 + i)z = (3 i)z, 2 , then z is . The other complex number can be are z z Re 2 1 z2 is equal to then complex number z is x (3 i), x R (b) x ,x R 3+i (c) x(3 + i), x R (d) x( 3 + i), x R ( 8 + i)50 = 349 (a + ib) (a) 3 (c) 9 (e) 4 4 (a) If arg(z) arg( ) = and then (a) If and (b) 1 and 0 6 and 1 + 2i 1 (1 i) 2 | z |= 1 then a2 + b2 is (b) 8 (d) 8 Page | 17 MATHEMATICS DPP DPP No. 3 DAILY PRACTICE PAPER COMPLEX NUMBER Square root, Representation and Logarithm of complex numbers 1. If x + iy = a + ib , then (x 2 + y 2 )2 = c + id (a) a2 + b2 c2 + d2 (b) a+b c+d (c) c2 + d2 a2 + b2 (d) a2 + b2 c2 + d2 (b) (1 3i) 2 8 6i = 2. (a) 1 3i (c) (1 + 3i) (d) If 4. (a) 15 (b) 25 (c) 25 (d) None of these If x + iy = (a + ib), then x iy is equal to 5. 6. 7. 8. = x iy, 2 x + y2 = (a) (b + ia) (b) (c) (b ia) (d) None of these (a ib) The square root of 3 4i is (a) (2 + i) (b) (2 i) (c) (1 2i) If a + ib = x + iy , (a) x2 + y2 (b) (c) x + iy (d) (d) (1 + 2i) then possible value of a ib is x2 + y2 x iy The number of non-zero integral solutions of the equation |1 i |x = 2x is (a) Infinite (b) 1 (c) 2 (d) None of these 1 + 7i = (2 i)2 (a) (c) 9. ( 7 24i) then (3 i) 3. 1/ 2 If 3 3 2 cos + i sin 4 4 3 3 + i sin cos 4 4 (b) 2 cos + i sin 4 4 (d) None of these z = rei , then | e iz | = (a) er sin (b) e r sin (c) e r cos (d) er cos Page | 18 MATHEMATICS DPP DPP No. 3 DAILY PRACTICE PAPER 10. 11. 12. 13. 14. 15. 16. 17. 18. 1 i 1+ i is equal to (a) cos (c) sin If 2 2 + i sin (b) 2 + i cos (c) 2 3 3 y = cos + i sin (a) 2 cos (c) 2cosec The value of cos 2 i sin (b) (d) 3 2 3 ,then the value of 1/ 3 ( i) (b) 2 sin (d) 2 tan (b) 1 3i 2 (c) 3 i 2 (d) 3 i 2 (1 + i 3 )9 = a + ib, then b (b) 256 (c) 0 (d) 93 i 1 y is (a) e cos [cos(sin )] (b) e cos [cos(cos )] (c) e sin [sin(cos )] (d) e sin [sin(sin )] The amplitude of ee i is equal to (a) sin (b) sin (c) cos (d) e sin If e z= 1+ i 3 3 +i , then (z)100 lies in (a) I quadrant (b) II quadrant (c) III quadrant (d) IV quadrant If x+ 1 = 3, x (a) cos (c) sin 3 6 is is equal to (a) 1 ee y+ is 1 + 3i 2 Real part of 2 is equal to (a) If (d) None of these 2 1 + 3 = re i , then (a) If COMPLEX NUMBER then x = + i sin + i cos 3 6 (b) cos (d) cos 2 6 + i sin + i sin 2 6 Page | 19 COMPLEX NUMBER MATHEMATICS DPP DPP No. 3 DAILY PRACTICE PAPER 19. 20. 21. 22. 23. 24. ( 1 + i 3 )20 is equal to (a) 2 20 ( 1 + i 3 ) 20 (b) (c) 2 20 ( 1 i 3 ) 20 (d) None of these 2 20 (1 i 3 ) 20 The imaginary part of 5i tan 1 3 (a) 0 (b) (c) (d) log 4 log 2 The real part of is (1 i) i is (a) 1 e / 4 cos log 2 2 (b) 1 e / 4 sin log 2 2 (c) 1 e / 4 cos log 2 2 (d) 1 e / 4 sin log 2 2 2 tan 1 x x i i log x+i is equal to (a) + 2 tan 1 x (b) (c) + 2 tan 1 x (d) 2 tan 1 x If ei = cos + i sin , then in ABC value of (a) i (b) 1 (c) 1 (d) None of these If z= 7 i 3 4i then e iA .e iB .e iC is z14 = (a) 27 (b) 27 i (c) 214 i (d) 27 i (e) 214 Geometry of complex numbers 1. 2. Length of the line segment joining the points (a) 5 (b) 15 (c) 5 (d) 25 The points z1 , z 2 , z 3 , z 4 1 i and 2 + 3i is in the complex plane are the vertices of a parallelogram taken in order, if and only if 3. (a) z1 + z 4 = z 2 + z 3 (b) (c) z1 + z2 = z3 + z4 (d) None of these The equation z1 + z3 = z2 + z4 z z + a z + a z + b = 0, b R represents a circle if (a) | a |2 = b (b) | a |2 b (c) (d) None of these | a |2 b Page | 20 COMPLEX NUMBER MATHEMATICS DPP DPP No. 3 DAILY PRACTICE PAPER 4. Let the complex numbers and z1, z2 circumcentre of the triangle, then 5. 6. (a) z02 (b) z02 (c) 3z02 (d) 3z02 The equation bz + b z = c, where b z3 be the vertices of an equilateral triangle. Let z0 be the z12 + z 22 + z32 = is a non-zero complex constant and c is real, represents (a) A circle (b) A straight line (c) A parabola (d) None of these If three complex numbers are in A.P., then they lie on (a) A circle in the complex plane (b) A straight line in the complex plane (c) A parabola in the complex plane 7. (d) None of these If a and b are real numbers between 0 and 1 such that the points an equilateral triangle, then (a) a = b = 2 + 3 (b) a = b = 2 (c) 8. 9. 10. 11. a = 2 3, b = 2 + 3 and z3 = 0 form 3 None of these If | z |= 2 , then the points representing the complex numbers 1 + 5 z will lie on a (a) Circle (b) Straight line (c) Parabola (d) None of these If the vertices of a quadrilateral be A = 1 + 2i, B = 3 + i, C = 2 3i and D = 2 2i , then the quadrilateral is (a) Parallelogram (b) Rectangle (c) Square (d) Rhombus In the Argand plane, the vector z = 4 3i is turned in the clockwise sense through 180o and stretched three times. The complex number represented by the new vector is (a) 12 + 9i (b) 12 9i (c) 12 9i (d) 12 + 9i If is a complex number satisfying (a) 2 + 3 12. (d) z1 = a + i, z2 = 1 + bi + 1 = 2, then maximum distance of from origin is (b) 1 + 2 (c) 1 + 3 (d) None of these The vector z = 3 4i is turned anticlockwise through an angle of complex number corresponding to the newly obtained vector is 180o and stretched 2.5 times. The 15 + 10i 2 (a) 15 10i 2 (b) (c) 15 10i 2 (d) None of these Page | 21 COMPLEX NUMBER MATHEMATICS DPP DPP No. 3 DAILY PRACTICE PAPER 13. is a straight line through the origin POQ respectively and 14. OP = OQ , (b) (c) (d) None of these Let a (a) | z a |= a (c) 15. z2 (b) z z1 , 19. n | a | 1 and be vertices of a polygon such that z1 , z 2 ,...... sides is located at the point then z2 (b) (c) z1 cos i sin 2n 2n (d) None of these and one of its vertex z1 is known. is equal to The vertices B and D of a parallelogram are 1 2i and and AC = 2BD , the complex number representing A is (b) z=0 z1 cos i sin n n 2 2 z1 cos i sin n n 3i 4 + 2i , If the diagonals are at right angles 3 2 (c) 3i 4 (d) 3i + 4 If z1 , z 2 , z 3 , z 4 are the affixes of four points in the Argand plane and | z z1 |=| z z 2 |=| z z 3 |=| z z 4 | , 18. and c + id (d) | z (1 a)|=|1 a | be the vertex adjacent to 5 2 a + ib 1 =| 1 a | 1 a (a) (a) 17. 1 1 = 1 a |1 a | represent the complex numbers Then the vertices of the polygon lie within a circle The centre of a regular polygon of If 16. z Q a+c =b+d be a complex number such that z k = 1 + a + a 2 + ..... + a k 1 . and then (a) | a + ib |=| c + id | arg(a + ib) = arg(c + id) O, P then z1 , z 2 , z 3 , z 4 z is the affix of a point such that are (a) Concyclic (b) Vertices of a parallelogram (c) Vertices of a rhombus (d) In a straight line ABCD is a rhombus. Its diagonals AC and BD intersect at the point M and satisfy BD = 2 AC . If the points D and M represents the complex numbers 1 + i and 2 i respectively, then A represents the complex number 3 1 i or 1 i 2 2 (a) 3 (c) 1 1 i or 1 i 2 2 (b) 1 3 3i i or 2 2 (d) None of these The complex numbers z1 , z 2 , z 3 are the vertices of a triangle. Then the complex numbers z which make the triangle into a parallelogram is (a) z1 + z 2 z 3 (b) (c) z 2 + z 3 z1 (d) All the above z1 z 2 + z 3 Page | 22 COMPLEX NUMBER MATHEMATICS DPP DAILY PRACTICE PAPER 20. 21. 22. The equation DPP No. 3 z z + (2 3i)z + (2 + 3i)z + 4 = 0 (a) 2 (b) 3 (c) 4 (d) 6 represents a circle of radius A rectangle is constructed in the complex plane with its sides parallel to the axes and its centre is situated at the origin. If one of the vertices of the rectangle is a + ib 3 , then the area of the rectangle is (a) ab 3 (b) 2ab 3 (c) 3ab 3 (d) 4ab 3 If the points P1 and P2 represent two complex numbers z1 and z2 , then the point P3 represents the number Y P3 P2 P1 O X (a) z1 + z 2 (b) z1 z 2 (c) z1 z 2 (d) z1 z 2 23. If | z 2| / | z 3 |= 2 represents a circle, then its radius is equal to 24. (a) 1 (b) 1 / 3 (c) 3 / 4 (d) 2 / 3 If complex numbers z1 , z 2 and z 3 represent the vertices A, B and C respectively of an isosceles triangle ABC of which (a) z12 + z22 + z3 2 = z1z2z3 25. is right angle, then correct statement is (b) (z 3 z1 ) 2 = z 3 z 2 (c) (z1 z 2 ) 2 = (z1 z 3 )(z 3 z 2 ) (d) (z1 z 2 ) 2 = 2(z1 z 3 ) (z 3 z 2 ) If centre of a regular hexagon is at origin and one of the vertex on argand diagram is 1 + 2i, then its perimeter is (a) 26. C 2 5 (b) 6 2 (c) 4 5 (d) 6 5 In the argand diagram, if O, P and Q represents respectively the origin, the complex numbers z and z + iz, then the angle OPQ is (a) (c) 4 2 (b) (d) 2 3 3 Page | 23 COMPLEX NUMBER MATHEMATICS DPP DPP No. 3 DAILY PRACTICE PAPER 27. A circle whose radius is r and centre (a) zz zz0 zz0 + z0 z0 = r 2 (b) zz + zz 0 zz 0 + z 0 z 0 = r 2 (c) zz zz 0 + zz 0 z 0 z 0 = r 2 z0 , then the equation of the circle is (d) None of these 28. Let z1 = (a) be three vertices of an equilateral triangle circumscribing the circle z1 , z 2 , z 3 3i 1 + 2 2 and are in anticlockwise sense then (b) 1+ 3i (c) 1 29. z1 , z 2 , z 3 z2 | z |= 1 2 . If is 1 3i (d) 1 For all complex numbers z1, z2 satisfying | z1 |= 12 and | z 2 3 4i |= 5, the minimum value of | z1 z 2 | is 30. (a) 0 (b) 2 (c) 7 (d) 17 If P, Q, R, S are represented by the complex numbers 4 + i, 1 + 6i, 4 + 3i, 1 2i respectively, then PQRS is a (a) Rectangle (c) Rhombus (b) Square (d) Parallelogram z1 1 z2 1 = z3 1 If 32. (a) 0 (b) 1 (c) 1 (d) 2 If z is a complex number in the Argand plane, then the equation | z 2| + | z + 2|= 8 represents 33. (a) Parabola (b) Ellipse (c) Hyperbola (d) Circle The points 1 + 3i, 5 + i and 3 + 2i in the complex plane are z1 , z 2 , z 3 are three collinear points in argand plane, then z1 z2 z3 31. (a) Vertices of a right angled triangle (b) Collinear (c) Vertices of an obtuse angled triangle (d) Vertices of an equilateral triangle Page | 24 COMPLEX NUMBER MATHEMATICS DPP DPP No. 3 DAILY PRACTICE PAPER 34. 35. 36. 37. If z1 and are two complex numbers, then | z1 + z2 | is z2 (a) | z1 | + | z2 | (b) | z1 | | z2 | (c) | z1 | + | z2 | (d) | z1 | + | z2 | If z = x + iy, then area of the triangle whose vertices are points (a) 2 | z |2 (b) 1 | z |2 2 (c) | z |2 (d) 3 | z |2 2 If A, B, C are represented by 3 + 4i, (a) Collinear (b) Vertices of equilateral triangle (c) Vertices of isosceles triangle (d) Vertices of right angled triangle If z1 , z 2 C, then 5 2i , 1 + 16i , z, iz and z + iz is then A, B, C are (a) | z1 + z 2 | | z1 | + | z 2 | (b) | z1 z 2 | | z1 | + | z 2 | (c) 38. If | z1 z 2 | | z1 | | z 2 | z1 , z 2 , z 3 (d) | z1 + z2 | | z1 | | z 2 | are affixes of the vertices and A, B C respectively of a triangle ABC having centroid at G such that z = 0 is the mid point of AG, then (a) z1 + z 2 + z 3 = 0 (b) z1 + 4 z 2 + z 3 = 0 39. (c) z1 + z 2 + 4 z 3 = 0 Let z1 (a) z 1 , z 2 are (b) z 1 , z 2 and the origin form a right angled triangle (c) z 1 , z 2 and the origin form an equilateral triangle and z2 (d) 4 z1 + z 2 + z 3 = 0 be two complex numbers such that z1 z 2 + =1. z 2 z1 Then collinear (d) None of these 40. 41. If the area of the triangle formed by the points value of | z | is (a) 6 (c) 3 2 If z1 = 1 + i, z 2 (a) 1 (c) 4 (b) 9 (d) 2 z, z + iz and iz on the complex plane is 18, then the 3 = 2 + 3i and z 3 = ai / 3 , where i 2 = 1, are collinear then the value of a is (b) 3 (d) 5 Page | 25 COMPLEX NUMBER MATHEMATICS DPP DPP No. 3 DAILY PRACTICE PAPER 42. Which one of the following statement is true (a) | x y|=| x | | y| (b) | x + y| | x | | y| (c) 43. 44. (d) | x + y| | x | | y| | x y | | x | | y | The area of the triangle whose vertices are represented by the complex numbers 0, z, equals (a) 1 | z |2 cos 2 (b) 1 | z |2 sin 2 (c) 1 | z |2 sin cos 2 (d) 1 | z |2 2 If z1 = 1 + 2i, z 2 = 2 + 3i, z 3 = 3 + 4i, (a) Equilateral triangle 45. then z1 , z 2 , z 3 ze i , (0 ) represent the vertices of a/an (b)Isosceles triangle (c) Right angled triangle (d) None of these The complex numbers which satisfy the equation z = x + iy z 5i =1 z + 5i lie on (a) Real axis (b) The line y=5 (c) A circle passing through the origin (d) None of these 46. When z+i z+2 is purely imaginary, the locus described by the point (a) Circle of radius 47. 48. (b)Circle of radius (d) Parabola If | z + 1| = the locus described by the point 2 | z 1|, then (a) Straight line (b) Circle (c) Parabola (d) None of these The region of the complex plane for which x axis (c) The straight line (b) x=a in the Argand diagram is a 5 4 (c) Straight line (a) 49. 5 2 z z a z+a =1 z in the Argand diagram is a [R(a) 0] is y axis (d) None of these The region of Argand plane defined by | z 1| + | z + 1| 4 is (a) Interior of an ellipse (b) Exterior of a circle (c) Interior and boundary of an ellipse (d) None of these Page | 26 COMPLEX NUMBER MATHEMATICS DPP DPP No. 3 DAILY PRACTICE PAPER 50. 51. 52. 53. The locus of the points z which satisfy the condition arg (a) A straight line (b) A circle (c) A parabola (d) None of these If the imaginary part of (b) A straight line (c) A parabola (d) None of these If z = ( + 3) + i 5 2 , (b) Straight line (c) Parabola (d) None of these A point z moves on Argand diagram in such a way that |z 3i| y axis (b) A straight line (c) A circle (d) None of these If z = x + iy The locus of x -axis (b) z lies on z y then its locus will be -axis (d) None of these z 1 =1, z i given by is (a) A circle (b) An ellipse (c) A straight line (d) A parabola R(z 2 ) = 1 is = 2, and | z zi|= 1, then (c) z lies on a circle 56. 3 then the locus of z is a (a) Circle (a) z lies on 55. = is is 2, then the locus of the point representing z in the complex plane is (a) A circle (a) 54. 2z + 1 iz + 1 z 1 z + 1 represented by (a) The parabola x 2 + y2 = 1 (b) The hyperbola x 2 y2 = 1 (c) Parabola or a circle (d) All the above 57. The locus represented by | z 1|=| z + i | is (a) A circle of radius 1 (b) An ellipse with foci at (1, 0) and (0, 1) (c) A straight line through the origin (d) A circle on the line joining 58. If log 3 | z |2 | z | +1 2, 2+ | z | (1, 0), (0, 1) as diameter then the locus of (a) | z |= 5 (b) | z | 5 (c) (d) None of these | z | 5 z is Page | 27 COMPLEX NUMBER MATHEMATICS DPP DPP No. 3 DAILY PRACTICE PAPER 59. 60. 61. 62. If arg (z a) = 4 , where then the locus of (a) Hyperbola (b) Parabola (c) Ellipse (d) Straight line If z = x + iy z C is a and | z 2 + i |=| z 3 i |, then locus of z is (a) 2x + 4 y 5 = 0 (b) 2x 4 y 5 = 0 (c) x + 2y = 0 (d) x 2y + 5 = 0 Locus of the point z satisfying the equation |iz 1| + | z i |= 2 is (a) A straight line (b) A circle (c) An ellipse (d) A pair of straight lines If z = x + iy (a) is a complex number satisfying 2y = x (c) y-axis 63. a R, (b) z+ i 2 2 2 = z i , 2 then the locus of z is y=x (d) x-axis The locus of the point z satisfying z 1 arg = k, z + 1 (where k is non zero) is (a) Circle with centre on y axis (b) Circle with centre on x axis (c) A straight line parallel to x axis 64. 65. (d) A straight line making an angle 60 o If the amplitude of then the locus of z 2 3i is /4, (a) x + y 1 = 0 (b) x y 1 = 0 (c) x + y +1 = 0 (d) x y +1 = 0 with the x axis z = x + iy is If | z 2 1|=| z |2 +1 , then z lies on (a) An ellipse (b) The imaginary axis (c) A circle (d) The real axis If z = x + iy and = 1 iz z i than | |= 1 shows that in complex plane (a) z will be at imaginary axis(b)z will be at real axis (c) z will be at unity circle (d) 66. None of these The equation | z 5i | | z + 5i |= 12, where z = x + iy, (a) Circle (b) Ellipse (c) Parabola (d) No real curve represents a/an Page | 28 COMPLEX NUMBER MATHEMATICS DPP DPP No. 3 DAILY PRACTICE PAPER 67. 68. 69. 70. If z = x + iy and z 2 arg = z + 2 6 (a) A straight line (b) A circle (c) A parabola (d) An ellipse If w= z z 1 i 3 and | w |= 1 , then z lies on (a) A straight line (b) A parabola (c) An ellipse (d) A circle If | 8 + z | + | z 8 |= 16 where z is a complex number, then the point z will lie on (a) A circle (b) An ellipse (c) A straight line (d) None of these PQ and PR are two infinite rays. QAR is an arc. Point lying in the shaded region excluding the boundary satisfies (a) | z 1| 2;| arg(z 1)| ( 1 + 2 , 2i) 4 (b) | z 1| 2;| arg(z 1)| (c) | z + 1| 2;| arg(z + 1)| (d) | z + 1| 2;| arg(z + 1)| 71. 72. 73. , then locus of z is Q 2 P ( 1,0) (1,0) A 4 R ( 1 + 2 , 2i) 2 Which of the following equations can represent a triangle (a) | z 1|=| z 2| (b) | z 1|=| z 2|=| z i | (c) (d) | z 1|2 + | z 2 |2 = 4 | z 1| | z 2|= 2a The number of solutions for the equations | z 1|=| z 2 |= (a) One solution (b) 3 solutions (c) 2 solutions (d) No solution If | z 2 3i | + | z + 2 6i |= 4 , where i = 1 , then locus of P(z) |z i| is is (a) An ellipse (b) (c) Line segment joining of point 2 + 3i and 2 + 6i (d) None of these 74. If z = 2 i 2 is rotated through an angle 45 in the anti-clockwise direction about the origin, then the coordinates of its new position are (a) (2, 0) (b) ( (c) (d) ( 2, 2 ) 2, 2 ) ( 2 ,0) (e) (4, 0) Page | 29 MATHEMATICS DPP DPP No. 3 DAILY PRACTICE PAPER COMPLEX NUMBER De Moivre's theorem and Roots of unity i= 1. 2. (a) 1 i (c) If (a) (c) 3. (b) 2 1+ i xr = cos r + i sin r , 2 2 5. 2 then x1.x2...... is (b) 2 (d) 0 3 1 (cos + i sin )4 (sin + i cos )5 is equal to (a) (b) cos i sin (c) sin i cos (d) sin 9 i cos 9 If 1 i (d) None of these 2 cos 9 i sin 9 5 4. 3 i 3 i z= + + 2 2 2 2 5 , then (a) Re(z) = 0 (b) Im(z) = 0 (c) Re(z) 0, Im(z) 0 (d) Re(z) 0, Im(z) 0 The roots of (2 2i)1 / 3 are (a) 2 cos i sin , 2 sin + i cos , 1 i 12 12 12 12 (b) 2 cos + i sin , 2 sin i cos , 1 + i 12 12 12 12 (c) 1 + 2i, 1 i, 2 2i (d) None of the above 6. 7. The value of 0.4(cos 30 o + i sin 30 o ) is (a) 2 (1 + i) 10 (b) 2 (1 i) 10 (c) 10 (d) 10 2 (1 i) 2 (1 + i) The following in the form of A + iB (cos 2 + i sin 2 ) 5 (a) (cos 25 + i sin 25 ) (b) i(cos 25 + i sin 25 ) (c) 8. 4(cos 75 o + i sin 75 o ) i (cos 25 i sin 25 ) (d) (cos 3 i sin 3 )6 (sin i cos )3 in the form of A + iB is (cos 25 i sin 25 ) If a = 2i then which of the following is correct (a) a = 1 + i (b) a = 1 i (c) a = ( 2 )i (d) None of these Page | 30 COMPLEX NUMBER MATHEMATICS DPP DPP No. 3 DAILY PRACTICE PAPER 9. If (cos + i sin )(cos 2 + i sin 2 )........ (cos n + i sin n ) = 1 , (a) 4m (b) 2m n(n + 1) (c) 4m n(n + 1) (d) m n(n + 1) (b) cos n + i sin n (d) sin n i cos n then the value of is n 10. 11. 1 + cos + i sin = 1 + cos i sin (a) cos n i sin n (c) sin n + i cos n 4 1 + cos + i sin = cos n + i sin n i + sin + i cos If (a) 1 (c) 3 12. 14. 15. 4 equals sin 8 i cos 8 (b) cos 8 i sin 8 (c) sin 8 + i cos 8 (d) cos 8 + i sin 8 sin + sin + sin = 0 = cos + cos + cos , (a) 2/3 (b) 3/2 (c) 1/2 (d) 1 cos + cos + cos = 0 = sin + sin + sin (a) (b) 2 cos( + + ) (c) 0 16. 17. 1 (a) If ( 3 + i)53 is (d) 2 cos + i sin sin + i cos If is equal to cos 2 + i sin 2 ........to 2 2 cos + i sin 2 2 (b) 1 (c) 0 13. n (b) 2 (d) 4 The value of expression (a) , then then the value of then sin 2 + sin 2 + sin 2 cos 2 + cos 2 + cos 2 is equals cos 2( + + ) (d) 1 where i 2 = 1 is equal to (a) 2 53 ( 3 + 2i) (b) 252 ( 3 i) (c) 3 1 2 53 + i 2 2 (d) 2 53 ( 3 i) The value of 1 cos 10 + i sin 10 1 cos i sin 10 10 (a) 0 (b) 1 (c) 1 (d) 2 10 = Page | 31 COMPLEX NUMBER MATHEMATICS DPP DPP No. 3 DAILY PRACTICE PAPER 18. 19. We express (cos 2 i sin 2 ) 4 (cos 4 + i sin 4 ) 5 (cos 3 + i sin 3 ) 2 (cos 3 i sin 3 ) 9 (a) cos 49 i sin 49 (b) cos 23 i sin 23 (c) cos 49 + i sin 49 (d) cos 21 + i sin 21 (sin + i cos )n in the form of x + iy , we get is equal to (a) cos n + i sin n (b) sin n + i cos n (c) cos n + i sin n 2 2 (d) None of these 20. 21. 22. 23. 24. The value of (cos + i sin ) (cos + i sin ) (cos + i sin ) (cos + i sin ) (a) cos( + ) i sin( + ) (b) cos( + ) + i sin( + ) (c) sin( + ) i cos( + ) (d) sin( + ) + i cos( + ) 1 + cos( / 8) + i sin( / 8) 1 + cos( / 8) i sin( / 8) 8 is equal to (a) 1 (b) 0 (c) 1 (d) 2 If xn = cos n + i sin n , 4 4 then x1 . x 2 . x 3 .... = (a) 1+i 3 2 (b) 1+ i 3 2 (c) 1 i 3 2 (d) 1 i 3 2 (cos + i sin ) 4 (sin + i cos )5 is = (a) cos(4 + 5 ) + i sin(4 + 5 ) (b) cos(4 + 5 ) i sin(4 + 5 ) (c) sin(4 + 5 ) i cos(4 + 5 ) (d) None of these The value of i1/3 is (a) 3+i 2 (b) 3 i 2 (c) 1+i 3 2 (d) 1 i 3 2 Page | 32 COMPLEX NUMBER MATHEMATICS DPP DPP No. 3 DAILY PRACTICE PAPER 25. Given then z = (1 + i 3 )100 , (a) 2100 (c) 26. 27. 28. 30. 1 33. n 3 = n n cos n + i sin n 2 2 (b) n n cos + n + i sin + n 2 2 (c) n n sin n + i cos n 2 2 (d) cos n + 2 + i sin n + 2 2 2 If n is a positive integer, then (1 + i)n + (1 i)n (a) n ( 2 )n 2 cos 4 (b) n ( 2 )n 2 sin 4 (c) n ( 2 )n+ 2 cos 4 (d) n ( 2 )n+ 2 sin 4 If 1 + x = 2 cos , x If then cos n iz 4 + 1 = 0 , (a) 1+i (c) 1 4i 2 xn + 1 then z is equal to is equal to xn (b) (d) 2 cos n 2 sin n sin n can take the value (b) cos 8 + i sin 8 (d) i The two numbers such that each one is square of the other, are (a) , 3 (b) i, i If 1, 1 (d) , 2 is a cube root of unity, then (a) 1 (c) 2 32. (d) 3 (a) (c) 31. equals (b) 250 1 + sin + i cos 1 + sin i cos (a) (c) 29. Re(z) Im( z) (1 + 2 ) (1 + 2 ) = (b) 0 (d) 4 (27)1 / 3 = (a) 3 (b) (c) (d) None of these 3, 3 , 3 2 3, 3i, 3i 2 If n is a positive integer not a multiple of 3, then (a) 3 (b) 1 (c) 0 (d) None of these 1 + n + 2n = Page | 33 COMPLEX NUMBER MATHEMATICS DPP DPP No. 3 DAILY PRACTICE PAPER 34. Square of either of the two imaginary cube roots of unity will be (a) Real root of unity (b) Other imaginary cube root of unity (c) Sum of two imaginary roots of unity (d) None of these 35. 36. 37. 38. 39. 40. 41. If is a cube root of unity, then (a) 0 (b) (c) 2 (d) None of these If and are imaginary cube roots of unity, then (a) 3 (b) 0 (c) 1 (d) 2 If is a complex cube root of unity, then (a) 0 (b) 1 (c) 1 (d) 9 If (a) 16 (b) 32 (c) 48 (d) 32 If x = a, y = b , z = c 2 , where (b) 1 (c) 0 (d) None of these is a complex cube root of unity, then (a) x 2 + y2 (b) x 2 y2 (c) x 3 y3 (d) x 3 + y3 If + 1 = (1 + 2 )5 + (1 + 2 )5 = is a complex cube root of unity, then (a) 3 If 4 + 4 (1 )(1 2 ) (1 4 )(1 8 ) = is a cube root of unity, then the value of is a complex cube root of unity, then (a) 0 (c) 1 42. (1 + ) 3 (1 + 2 ) 3 = x y z + + = a b c (x y)(x y) (x 2 y) = (1 + )(1 + 2 ) (1 + 4 )(1 + 8 )... to 2n factors = (b) 1 (d) None of these The product of all the roots of (a) 1 (b) 1 (c) 3 2 (d) cos + i sin 3 3 3/4 is 1 2 Page | 34 COMPLEX NUMBER MATHEMATICS DPP DPP No. 3 DAILY PRACTICE PAPER 43. If x +1 x + 2 2 1 is a cube root of unity, then a root of the equation 44. (a) x = 1 (b) x = (c) x = 2 (d) x = 0 If x = a + b, y = a + b and z = a + b , where 45. (a) a2 + b2 (b) a3 + b3 (c) a3b3 (d) a3 b3 If x = a + b, y = a + b 2, z = a 2 + b , then the value of (a) a 3 + b3 (b) (c) 3(a 2 + b 2 ) 2 1 =0 x + is and are complex cube roots of unity, then x 3 + y3 + z 3 3(a 3 + b 3 ) (d) None of these 2 a + b + c a + b + c 2 + 2 b + c + a c + a + b 2 The value of 47. (a) 1 (b) 1 (c) 2 (d) 2 The cube roots of unity when represented on the Argand plane form the vertices of an (a) Equilateral triangle (b) Isosceles triangle (c) Right angled triangle (d) None of these 48. 1 + 3 i 2 2 50. 1 3 + i 2 2 (c) = (b) 1 3 + i 2 2 1 3 i 2 2 (d) None of these If , , are the cube roots of p(p 0) , then for any (a) 1 ( 1 + i 3 ) 2 (b) (c) 1 (1 i 3 ) 2 (d) None of these z= 3 +i 2 (b) (d) x, y and z, x + y + z = x + y + z 1 (1 + i 3 ) 2 , then the value of (a) i (c) 1 51. will be 1000 (a) If = is equal to 46. 49. xyz z 69 is i 1 The roots of the equation x4 1 = 0 , are (a) 1, 1, i, i (b) 1, 1, i, i (c) 1, 1, , 2 (d) None of these Page | 35 COMPLEX NUMBER MATHEMATICS DPP DPP No. 3 DAILY PRACTICE PAPER 52. If 2 . . 3 is a complex cube root of unity, then for positive integral value of n , the product of ........ n , will be 1 i 3 2 (a) (b) (c) 1 53. 54. 55. 56. 57. One of the cube roots of unity is (a) 1+ i 3 2 (b) 1+ i 3 2 (c) 1 i 3 2 (d) 3 i 2 If ( 1) is a cube root of unity and (a) 0, 1 (b) 1, 0 (c) 1, 1 (d) 60. (1 + )7 = A + B 1 1+i + 2 2 1 i 1 2 1 i i + 1 1 (a) 0 (b) 1 (c) (d) i th The n roots of unity are in (a) A.P. (b) G.P. (c) H.P. (d) None of these 2 If 1, , are the three cube roots of unity, then 2 2 4 3 6 4 Let (a) =0 (b) =1 (c) =2 (d) =3 z and B are respectively, the numbers is equal to (3 + 2 + 4 )6 = factors is (b) 2 2n (d) 1 1 = 2 2 2 3 3 3 where is the cube root of unity, then is a positive integer greater than unity and = (z + 1)n , then n n A (b) 729 (d) 0 (1 + 2 )(1 2 + 4 )(1 4 + 8 )......... .. to 2n If , then 1, 1 If ( 1) is a cube root of unity, then (a) 2n (c) 0 59. 1 i 3 2 (d) (b) and (c) both (a) 64 (c) 2 58. (a) Re(z) 0 (b) (c) Re(z) = 0 (d) None of these z is a complex number satisfying the equation Re(z) 0 Page | 36 COMPLEX NUMBER MATHEMATICS DPP DPP No. 3 DAILY PRACTICE PAPER 61. If is an nth root of unity, other than unity, then the value of (a) 0 62. 63. is (b) 1 (c) 1 (d) None of these th If z1 , z 2 , z 3 ...... nn are n , roots of unity, then for (a) | z k |= k | z k +1 | (b) | z k +1 |= k | z k | (c) (d) | z k |=| z k +1 | If 1 + + 2 + ... + n 1 | zk +1 |=| zk | + | zk +1 | 1, , 2 are three cube roots of unity, then (a) 27 abc (b) 0 (c) 3 abc (d) None of these k = 1, 2,....., n (a + b + c 2 )3 + 64. The common roots of the equations x 12 1 = 0 , (a) (b) 2 (c) , 2 (d) None of these x4 + x2 + 1 = 0 65. If then the value of z1 , z 2 z 3 , z 4 are the roots of the equation z4 = 1, is equal to, if (a + b 2 + c )3 a+b+c = 0 are 4 z 3 i is i =1 66. (a) 0 (b) 1 (c) i (d) 1 + i If is an imaginary cube root of unity, then for (a) 1 (b) 0 (c) 1 (d) 3 67. 1+ i 3 2 (a) 20 1 i 3 + 2 1 1 If 69. (a) 1 (b) 1 (c) 0 (d) None of these If is the cube root of unity, then (3 + 5 + 3 2 )2 + (a) 4 (b) 0 (c) 4 (d) None of these 70. If (a) (c) and are imaginary cube roots of unity, then the value of 71. (3 + 3 + 5 2 )2 is an imaginary cube root of unity, then the value of (b) (d) 3/2 1/ 2 6 3 +i i 3 + 2 2 (a) 2 (c) 2 is = 68. 3n+1 + 3n+ 3 + 3n+5 20 (d) 2 19 the value of (b) 1 20 3i (c) n N , 4 + 28 + 1 ,is = sin ( 10 + 23 ) 4 is 1 / 2 3/2 6 is equal to (b) 0 (d) 1 Page | 37 COMPLEX NUMBER MATHEMATICS DPP DPP No. 3 DAILY PRACTICE PAPER 72. If (a) (c) 73. 74. 75. 76. 77. 78. 79. is an imaginary cube root of unity, (b) 128 (d) 128 2 128 128 2 ( 1 + i 3 )15 (1 i) 20 + ( 1 i 3 )15 is equal to (1 + i) 20 (a) 64 (b) 32 (c) 16 (d) 1 16 If is a complex root of the equation (a) 1 (b) 0 (c) 9 (d) i If cube root of 1 is , (b) 16 (c) (d) 16 The value of (a) 128 (b) 128 2 (c) (d) 128 2 1, , 2 (b) (c) 1 (d) 1 z= is where , 2 are cube roots of unity 3 +i 2 , then z 69 is equal to (a) 1 (b) 1 (c) i (d) i 2 2 2 + i sin , i = 1 , n n Let (3 + + 3 2 )4 is equal to are the cube roots of unity, then their product is (a) 0 If 27 1 3 9 + +... + + 8 32 128 then + 2 16 2 (1 + 2 ) (1 2 + )6 , 128 z3 = 1 , then the value of (a) 0 If (1 + 2 )7 equals n = cos then ( x + y 3 + z 3 2 ) ( x + y 3 2 + z 3 ) is equal to (a) 0 80. (b) x 2 + y2 + z2 (c) x 2 + y 2 + z 2 yz zx xy (d) x 2 + y 2 + z 2 + yz + zx + xy If z + z 1 = 1, then z 100 + z 100 is equal to (a) i (b) i (c) 1 (d) 1 Page | 38 COMPLEX NUMBER MATHEMATICS DPP DPP No. 3 DAILY PRACTICE PAPER 81. 82. If 1+ 3i 2 is a root of equation x4 x3 + x 1 = 0 (a) 1, 1 (b) 1, 1 (c) 1, 1 (d) 1, 2 If 1+ i 3 1 i 3 n is an integer, then n is (a) 1 (b) 2 (c) 3 (d) 4 83. Find the value of 84. (a) 0 (b) 1 (c) (d) 2 If is a non real cube root of unity, then (a) (c) 85. 86. 87. then its real roots are (1 + 2 + 2 )3n (1 + + 2 2 )3n = (b) (d) a3 + b3 a2 + b2 (a + b) (a + b ) (a + b 2 ) is a3 b3 a2 b2 Which of the following is a fourth root of (a) cis 2 (b) cis 12 (c) cis 6 (d) cis 3 1 i 3 + 2 2 The value of (8)1/3 is (a) 1 + i 3 (b) 1 i 3 (c) 2 (d) All of these If is a complex cube root of unity, then 225 + (3 + 8 2 )2 + (3 2 + 8 ) 2 = 88. (a) 72 (b) 192 (c) 200 (d) 248 2 If 1, , are the cube roots of unity, then 1 = n 2n n 2n 2n 1 n 1 (a) 0 (c) 89. If = = (b) 1 (d) 2 1 + 3i 2 (a) 16 (c) 16 then (3 + + 3 2 )4 = (b) 16 (d) 16 2 Page | 39 MATHEMATICS DPP DPP No. 3 DAILY PRACTICE PAPER 90. 91. 92. 93. If 1, , 2 are the roots of unity, then (a) cos 1( sin ) (b) sinh 1( sin ) (c) sin 1( sin ) (d) sin 1( cos ) sinh ix is (a) i sin(ix) (b) i sin x (c) i sin x (d) sin(ix) If 95. (a) cos u + sinh v (b) (c) cos 2 u + cosh 2 v (d) The value of sec h(i ) is cos(u + iv) = + i , 2 then 98. 99. 2 + 2 +1 2 (a) 1 (c) 0 97. is equal to (a) 729 (b) 246 (c) 243 (d) 81 If is a complex cube root of unity, then the value of 99 + 100 + 101 is (a) 1 (b) 1 (c) 3 (d) 0 1 i The real part of sin (e ) is 94. 96. (1 2 + 2 )6 COMPLEX NUMBER equals 2 sin u + cosh 2 v sin 2 u + sinh 2 v (b) i (d) 1 cosh( + i ) cosh( i ) is equal to (a) 2 sinh sinh (b) 2 cosh cosh (c) 2i sinh sin (d) 2 cosh cos The imaginary part of cosh( + i ) is (a) cosh cos (b) sinh sin (c) cos cosh (d) cos cos Which one is correct from the following (a) sin(ix) = i sinh x (b) cos(ix) = i cosh x (c) sin(ix) = i sinh x (d) tan(ix) = i tanh x cos(x + iy) is equal to (a) sin x cosh y + i cos x sinh y (b) cos x cosh y + i sin x sinh y (c) cos x cosh y i sin x sinh y (d) None of these 100. If tan(u + iv) = i , then the value of v is (a) 0 (b) (c) 1 (d) None of these Page | 40 COMPLEX NUMBER MATHEMATICS DPP DPP No. 3 DAILY PRACTICE PAPER 101. If tan 1 ( + i ) = x + iy, then x = (a) 1 2 tan 1 1 2 2 2 (b) 1 2 tan 1 2 2 2 1+ + (c) 2 tan 1 2 2 1 (d) None of these 102. If is a cube root of unity but not equal to 1 then minimum value of | a + b + c 2 | (where a, b, c are integers but not all equal) is (a) 0 (b) (c) 1 103. 104. If 1, 3 2 (d) 2 , 2 are the cube roots of unity then (a) 1 (b) 1 (c) i (d) 0 Let x = + , y = + 2 , z = 2 + , (a) 2 + 2 (b) 2 2 (c) 3 + 3 (d) 3 3 2 (1 + )3 (1 + 2 ) = is an imaginary cube root of unity. Product of xyz is CRITICAL THINKING 1. 2. The number of real values of a satisfying the equation (a) Zero (b) One (c) Two (d) Infinite For positive integers n1 ,n2 the value of the expression a2 2a sin x + 1 = 0 is (1 + i)n1 + (1 + i 3 )n1 + (1 + i 5 )n2 + (1 + i 7 )n2 where i = 1 is a real number if and only if 3. (a) n1 = n2 + 1 (b) n1 = n2 1 (c) n1 = n2 (d) n1 0, n2 0 Given that the equation z 2 + ( p + iq)z + r + i s = 0, where p, q, r, s are real and non-zero has a real root, then (a) pqr = r 2 + p 2 s (b) prs = q 2 + r 2 p (c) qrs = p 2 + s 2q (d) pqs = s 2 + q 2r Page | 41 COMPLEX NUMBER MATHEMATICS DPP DPP No. 3 DAILY PRACTICE PAPER 4. 5. 6. If x = 5 + 2 4 , then the value of the expression (a) 160 (b) 160 (c) 60 (d) 60 If 3 + i = (a + ib)(c + id) , (a) (c) n If 3 + 2n , n I 3 ,n I 69. 70. (d) 2n b c a + + = 1, c a b 6 3 If ,n I then (d) 1 If (1 + i)(1 + 2i)(1 + 3i).....(1 + ni) = a + ib , z (b) a2 b2 then is equal to 2.5.10.... (1 + n2 ) is equal to a2 + b2 (d) a2 + b2 cos( ) + cos( ) + cos( ) a2 b2 is a complex number, then the minimum value of | z | + | z 1| is (a) 1 (b) 0 (c) 1/2 (d) None of these For any two complex numbers z1 and z2 (a) (a 2 + b 2 )(| z1 | + | z 2 |) (b) (c) (a 2 + b 2 )(| z1 |2 | z 2 |2 ) (d) None of these log 1 / 3 | z + 1| log 1 / 3 | z 1| (a) R (z) 0 (b) (c) I (z) 0 (d) None of these z1 = a + ib and z2 = c + id and any real numbers a and b; |(az1 bz2 )|2 + |(bz1 + az2 )|2 = (a 2 + b 2 )(| z1 |2 + | z 2 |2 ) The locus of z satisfying the inequality If has the value ,n I (c) 0 complex numbers 71. n + (b) 3/2 (c) 68. (b) (a) 3/2 (a) 8. b d tan 1 + tan 1 c a is a = cos + i sin , b = cos + i sin , c = cos + i sin and 7. then x 4 + 9 x 3 + 35x 2 x + 4 is R (z) 0 are complex numbers such that w1 = a + ic and w2 = b + id | z1 |=| z2 |= 1 and R(z1 z 2 ) = 0, then the pair of satisfies (a) |w1 |= 1 (b) |w2 |= 1 (c) R(w1 w2 ) = 0, (d) All the above Let z and w be two complex numbers such that | z | 1, | w | 1 and | z + iw |=| z iw |= 2 . Then z is equal to (a) 1 or i (c) 1 or 1 (b) i or i (d) i or 1 Page | 42 COMPLEX NUMBER MATHEMATICS DPP DPP No. 3 DAILY PRACTICE PAPER 72. The maximum distance from the origin of coordinates to the point z+ z satisfying the equation 1 = a is z (a) 1 ( a 2 + 1 + a) 2 (b) 1 ( a 2 + 2 + a) 2 (c) 1 ( a 2 + 4 + a) 2 (d) None of these 73. 74. 75. Find the complex number z satisfying the equations (a) 6 (b) (c) (d) None of these If 6 + 8i, 6 + 17i z1 , z 2 , z 3 6 8i | z 3 |= 1 1 1 + + = 1, z1 z 2 z 3 is a complex number such that z z1 = , amp z z2 4 are complex numbers such that | z1 |=| z 2 |= (a) Equal to 1 (b) Less than 1 (c) Greater than 3 (d) Equal to 3 If z1 = 10 + 6i, z2 = 4 + 6i and z z 12 5 z 4 = , =1 z 8i 3 z 8 then | z1 + z 2 + z 3 | is then the value of | z 7 9i | is equal to 76. (a) 2 (b) 2 2 (c) 3 2 (d) 2 3 If z1 , z 2 , z 3 be a b c b c a = 0, c a b that 77. three non-zero complex number, such that (a) z z1 arg 2 z 3 z1 2 (c) z z1 arg 3 z 2 z1 2 Let z and w then z arg 3 z2 z 2 z1 , a =| z1 |, b =| z 2 | and c =| z 3 | suppose is equal to (b) z z1 arg 2 z 3 z1 (d) z z1 arg 3 z 2 z1 be the two non-zero complex numbers such that | z |=| w | and arg z + arg w = . Then z is equal to (a) w (b) w (c) w (d) w Page | 43 COMPLEX NUMBER MATHEMATICS DPP DPP No. 3 DAILY PRACTICE PAPER 78. 79. If | z 25i | 15 , then | max .amp(z) min.amp(z)|= (a) 3 cos 1 5 (b) 2 cos 1 (c) (d) 3 3 sin 1 cos 1 5 5 If 2 3 + cos 1 5 z1 , z 2 and z3 , z4 are two pairs of conjugate complex numbers, then (a) 0 80. 81. 82. 83. (b) (d) 3 2 Let z, w be (a) 5 / 4 (b) /2 (c) 3 / 4 (d) /4 If 2n (c) 2n sin If complex numbers such that (1 + x)n = C0 + C1 x + C 2 x 2 + ..... + Cn x n , (a) n 2 and x = cos + i sin (a) cos(m + n ) (b) cos(m n ) (c) 2 cos(m + n ) (d) 2 cos(m n ) The value of 8 sin 2n cos (d) 2n / 2 cos equals z + iw = 0 and arg zw = then the value of . Then arg z equals C0 C2 + C4 C6 + ..... is n 4 y = cos + i sin , then x m y n + x m y n is equal to 2r 2r + i cos is 9 9 1 (b) 1 (c) i (d) i and u, v, w are complex numbers representing the vertices of two triangles such that c = (1 r)a + rb and w = (1 r)u + rv , where r is a complex number, then the two triangles a, b, c (a) Have the same area (c) Are congruent 85. z + arg 2 z 3 n 2 (b) (a) If z arg 1 z4 2 (c) r =1 84. 3 5 Suppose z1, z2 , z3 then values of z3 (b) Are similar (d) None of these are the vertices of an equilateral triangle inscribed in the circle and z2 | z |= 2 . If z1 = 1 + i 3 , are respectively (a) 2, 1 i 3 (b) (c) 1 + i 3 , 2 (d) None of these 2, 1 + i 3 Page | 44 COMPLEX NUMBER MATHEMATICS DPP DPP No. 3 DAILY PRACTICE PAPER 86. 87. If the complex number the origin form an equilateral triangle then z1, z2 (a) z1 z2 (b) (c) z 2 z1 (d) | z1 |2 =| z2 |2 z12 + z 22 = z1 z 2 If at least one value of the complex number z = x + iy satisfy the condition | z + 2 |= a 2 3a + 2 inequality | z + i 2 | a 2 , 88. (a) a 2 (c) a 2 If z, iz and (b) a = 2 (d) None of these are the vertices of a triangle whose area is 2 units, then the value of 89. (a) 2 (b) 2 (c) 4 (d) 8 2 2 If z + z | z | + | z | = 0 , then the locus of 90. (a) A circle (b) A straight line (c) A pair of straight lines (d) None of these If cos + cos + cos = sin + sin + sin = 0 then cos 3 + cos 3 + cos 3 equals to 91. then z (a) 0 (b) cos( + + ) (c) (d) 3 sin( + + ) If 3 cos( + + ) z r = cos (a) 92. z + iz r n2 + i sin r n2 , (b) lim z1 z 2 z 3 ... z n n 1, 1 + 2 , 1 + 2 2 (b) 1, 1 2 , 1 2 2 (c) 1, 1, 1 is is equal to cos( /2) i sin( /2) (c) e i / 2 (d) 3 e i If the cube roots of unity be 1, , 2, then the roots of the equation (a) |z| is where r = 1, 2, 3, .,n, then cos + i sin and the (x 1)3 + 8 = 0 are (d) None of these 93. 94. If 1, , 2 , 3 ......., n 1 are the (a) 0 (b) 1 (c) (d) n n, nth roots of unity, then (1 )(1 2 ).....(1 n 1) equals n2 The value of the expression 1.(2 )(2 2 ) + 2.(3 )(3 2 ) + ....... .... + (n 1).(n )(n 2 ), where is an imaginary cube root of unity, is (a) 1 (n 1)n(n 2 + 3n + 4) 2 (b) 1 (n 1)n(n 2 + 3n + 4) 4 (c) 1 (n + 1)n(n 2 + 3n + 4) 2 Page | 45 COMPLEX NUMBER MATHEMATICS DPP DPP No. 3 DAILY PRACTICE PAPER (d) 95. 96. If 1 (n + 1)n(n 2 + 3n + 4) 4 i = 1, 1 i 3 4 + 5 + 2 2 then 334 1 i 3 + 3 + 2 2 (a) 1 i 3 (b) 1+ i 3 (c) i 3 (d) i 3 If a = cos(2 / 7) + i sin(2 / 7), 365 is equal to then the quadratic equation whose roots are = a + a2 + a4 and = a3 + a5 + a6 is (a) x2 x + 2 = 0 (c) x 2 x 2 = 0 97. Let z1 and z2 (b) x2 + x 2 = 0 (d) x2 + x + 2 = 0 be nth roots of unity which are ends of a line segment that subtend a right angle at the origin. Then n must be of the form 98. (a) 4k + 1 (b) 4k + 2 (c) 4k + 3 (d) 4k Let is an imaginary cube roots of unity then the value of 2( + 1)( 2 + 1) + 3(2 + 1)(2 2 + 1) + ..... + (n + 1)(n + 1)(n 2 + 1) 99. 2 (a) n(n + 1) 2 +n (c) n(n + 1) 2 n 2 (b) n(n + 1) 2 is 2 (d) None of these is an imaginary cube root of unity. If (a) 6 (b) 5 (c) 4 (d) 3 (1 + 2 )m = (1 + 4 )m , then least positive integral value of m is Page | 46 COMPLEX NUMBER MATHEMATICS DPP DPP No. 3 DAILY PRACTICE PAPER Integral power of iota, Algebraic operations and Equality of complex numbers 1 6 11 16 21 26 31 36 41 46 51 56 61 66 b d a c d d a b a b c c c b 2 7 12 17 22 27 32 37 42 47 52 57 62 67 b c c a d c b b a d a b a b 3 8 13 18 23 28 33 38 43 48 53 58 63 c b b c b a c c d a b c b 4 9 14 19 24 29 34 39 44 49 54 59 64 b d a a b a c c d d c c b 5 10 15 20 25 30 35 40 45 50 55 60 65 b d c a d c d c b d a d d Conjugate, Modulus and Argument of complex numbers 1 d 2 a 3 c 4 a 5 a 6 11 16 21 26 31 36 41 46 51 56 61 66 71 76 81 86 d b b b a b c b d c c c c c a d a 7 12 17 22 27 32 37 42 47 52 57 62 67 72 77 82 87 c a c c d a b c d b a a d a d d c 8 13 18 23 28 33 38 43 48 53 58 63 68 73 78 83 a c d b a a d a d b b c c a,d c b 9 14 19 24 29 34 39 44 49 54 59 64 69 74 79 84 b c b c a a c c d a d c b a b b 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 c c b c c c a b c c a a b d c d Page | 47 COMPLEX NUMBER MATHEMATICS DPP DPP No. 3 DAILY PRACTICE PAPER Square root, Representation and Logarithm of complex numbers 1 a 2 b 3 b 4 c 5 a 6 d 7 d 8 a 9 b 10 b 11 c 12 a 13 c,d 14 c 15 a 16 b 17 c 18 d 19 d 20 c 21 a 22 b 23 c 24 d Geometry of complex numbers 1 c 2 b 3 b 4 c 5 b 6 b 7 b 8 a 9 c 10 d 11 b 12 b 13 a,b 14 c 15 a 16 b 17 a 18 a 19 d 20 b 21 d 22 a 23 d 24 d 25 d 26 c 27 a 28 d 29 b 30 b 31 a 32 b 33 b 34 a 35 b 36 a 37 d 38 d 39 c 40 a 41 d 42 c 43 b 44 d 45 a 46 a 47 b 48 b 49 c 50 b 51 b 52 a 53 c 54 c 55 c 56 b 57 c 58 b 59 d 60 a 61 a 62 d 63 a 64 d 65 b 66 b 67 a 68 b 69 a 70 b 71 c 72 b 73 a 74 b 75 d De Moivre's theorem and Roots of unity 1 c 2 c 3 d 4 b 5 a 6 d 7 c 8 a 9 c 10 b 11 d 12 a 13 d 14 b 15 c 16 c 17 b 18 a 19 c 20 b 21 a 22 a 23 c 24 a 25 c 26 a 27 c 28 a 29 b 30 d 31 d 32 c 33 c 34 b 35 a Page | 48 COMPLEX NUMBER MATHEMATICS DPP DPP No. 3 DAILY PRACTICE PAPER 36 b 37 d 38 b 39 c 40 c 41 b 42 b 43 d 44 b 45 b 46 b 47 a 48 c 49 a 50 a 51 b 52 d 53 a 54 c 55 a 56 b 57 a 58 b 59 a 60 a 61 a 62 d 63 a 64 c 65 a 66 b 67 d 68 c 69 c 70 c 71 a 72 d 73 a 74 a 75 c 76 c 77 d 78 c 79 c 80 d 81 c 82 c 83 a 84 a 85 b 86 d 87 d 88 a 89 c 90 a 91 d 92 a 93 b 94 c 95 a 96 c 97 b 98 a 99 c 100 b 101 a 102 c 103 d 104 c Critical Thinking Questions 1 c 2 d 3 d 4 b 5 b 6 d 7 b 8 a 9 b 10 a 11 d 12 c 13 c 14 c 15 a 16 c 17 c 18 d 19 b 20 a 21 c 22 d 23 c 24 d 25 b 26 a 27 a 28 a 29 b 30 c 31 c 32 c 33 b 34 c 35 b 36 c 37 d 38 d 39 a 40 d Page | 49 MATHEMATICS DPP DPP No. 3 DAILY PRACTICE PAPER COMPLEX NUMBER Integral power of iota, Algebraic operations and Equality of complex numbers 1. (b) 2. (b) We know that 2 3 = i 2 i 3 = i2 6 = 6 i 2 = 1 (i 2 )2 = ( 1)2 = 1 i 4 n = 1n and therefore 3. i4n 1 = i 1 + i (1 + i)(1 + i) = =i 1 i (1 i)(1 + i) (c) Since Therefore 4. (b) 4 n+1 = i 4 n+1 = ii 4 n = i ( i 4 n = 1) 1 + i 1 + i 1 + i (1 + i) 2 2i = = = =i 1 i 1 i 1+i 2 2 1 + i 1 i 5. 1+ i 1 i m (as given) = im = 1 So the least value of m = 4 { i 4 = 1} (b) If (1 i)n = 2n ......(i) We know that if two complex numbers are equal, their moduli must also be equal, therefore from (i), we have | (1 i)n |=| 2 n | | 1 i |n =| 2 |n , ( 2 n 0) n 1 2 + ( 1) 2 = 2 n ( 2 )n = 2n 2n / 2 = 2n n = n n = 0 2 Trick : By inspection, (1 i)0 = 20 1 = 1 6. (d) (1 + i)5 (1 i)5 = (1 i 2 )5 = 25 = 32 . 7. (c) 2i 2i 1+ i 1 i + + = = 2 2i 2i 1 i 1+ i 8. (b) i 584 (i 8 + i 6 + i 4 + i 2 + 1) i 584 1 = 1 i 574 (i 8 + i 6 + i 4 + i 2 + 1) i 574 2 2 = i10 1 = 1 1 = 2 9. (d) S = 1 + i 2 + i 4 + ..... + i 2n = 1 1 + 1 1 + ...... + ( 1)n Obviously it depends on n . Hence cannot be determined unless n is known. 10. (d) Given expression is 1 + 1 1 + 1 + ..... upto (2n + 1) terms Obviously number of terms is odd, so expression has the value 1. 11. (a) 1 + i 2 + i 3 i6 + i8 = 1 1 i + 1 + 1 = 2 i . 12. (c) 200 i n = i + i 2 + i 3 + .... + i 200 = n =1 = 13. (b) 13 (i n i(1 i 200 ) 1 i (since G.P.) i(1 1) = 0. 1 i + i n + 1 ) = (i + i 2 + i 3 + .... + i13 ) + (i 2 + i 3 + .... + i14 ) n =1 Page | 50 MATHEMATICS DPP DPP No. 3 DAILY PRACTICE PAPER = COMPLEX NUMBER i(1 i13 ) i 2 (1 i13 ) + 1 i 1 i 1 i i 2 (1 i) = i = i + i2 = i 1 . + 1 i (1 i) 14. (a) n n i 1 i 1 2i = = in i +1 i 1 2 Hence, to make the real number the least positive integar is 2. 15. (c) Let z = i[1+3+5+....+(2n+1)] Clearly series is A.P. with common difference = 2 Tn = 2n 1 and Tn+1 = 2n + 1 So, number of terms in A. P. Now, Sn+1 = Sn+1 = n+1 [2.1 + (n + 1 1)2] 2 n+1 [2 + 2n] = (n + 1)2 2 Now put = n+1 i.e. 2 i(n+1) n = 1, 2, 3, 4, 5, ..... n = 1, z = i = 1 , n = 2, z = i 6 = 1 , 4 n = 3, z = i 8 = 1 , n = 4, z = i10 = 1 , n = 5, z = i12 = 1,........ 16. (c) x+ 1 = 2 cos x 2 2 x cos + 1 = 0 x x= 2 cos 4 cos 2 4 2 x = cos i sin 17. (a) i n + i n+1 + i n+ 2 + i n+3 = i n [1 + i + i 2 + i 3 ] = i n [1 + i 1 i] = 0 . 18. (c) (1 + i)2 = 1 + i 2 + 2i = 2i and . (1 i)2 = 1 + i 2 2i = 2i (1 + i)8 + (1 i)8 = (2i)4 + ( 2i)4 = 2 4 (i 4 + i 4 ) = 2 5 = 32. 19. (a) 10 (1 + i) = [(1 + i) ] = (2i)5 = 32 i . 2 5 20. (a) (1 + i)6 + (1 i)6 = [(1 + i)2 ]3 + [(1 i)2 ]3 = (2i)3 + ( 2i)3 = (8 8)i3 = 0 . 21. (d) i + i 2 + i 3 + ....... up to 1000 terms = 22. (d) i(1 i1000 ) i(1 (i 4 )250 ) i(1 1) = = =0. 1 i 1 i 1 i x = 3 + i x 3 = i x 2 6 x + 10 = 0 x 3 3x 2 8 x + 15 = x(x 2 6 x + 10) + 3 (x 2 6 x + 10) 15 = x(0) + 3(0) 15 = 15 . 23. (b) We have 1+ i 1 i 2n (1 + i) 2n = (1 i) 2n = 1 (i) 2n = 1 (i) 2n = ( 1) 2 (i) 2n = (i 2 ) 2 (i) 2n = (i)4 2n = 4 n = 2 . Page | 51 COMPLEX NUMBER MATHEMATICS DPP DPP No. 3 DAILY PRACTICE PAPER 24. (b) (1 + i)x 2i (2 3i)y + i + =i 3+i 3 i (4 + 2i)x + (9 7i)y 3i 3 = 10i Equating real and imaginary parts, we get 2x 7y = 13 and 4 x + 9y = 3 . Hence x = 3 and Trick : After finding the equations, no need to solve them, put the values of x and the options and get the appropriate option. 25. (d) If z1 and z 2 be two complex number then Re 26. (d) y = 1 . y given in (z1z 2 ) = Re(z1 ) Re(z 2 ) Im(z1 ) Im(z 2 ) 3 3 + 4i 1 + 1 2i 1 + i 2 4i 3 3i 6 16 + 12i + 8i 1 + 2i = 2 + 2 2 1 + 2 1 + 12 22 + 4 2 2 + 4i + 15 15i 1 + 2i = 10 2 = 27. (c) If (17 11i)( 1 + 2i) 5 + 45i 1 9 = = + i. 20 20 4 4 is the additive inverse of 1 i then x = 1 , y = 1 The additive inverse of 1 i is z = 1 + i Trick : Since (1 i) + ( 1 + i) = 0 . 28. (a) z = x + iy (x + iy) + (1 i) = 0 x +1 = 0 , y 1 = 0 (1 + i)2 2i 3 + i Re = Re 3 i 3 i 3 + i (1 + i)2 2i 3 + i Re = Re 3 i 3 i 3 + i 6 1 6i 2 2 = Re = Re 10 + 10 i = 5 9 + 1 29. (a) . (1 i)x + (1 + i)y = 1 3i (x + y) + i ( x + y) = 1 3 i Equating real and imaginary parts, we get (2, 1) . 30. (c) (3 + 2i sin )(1 + 2i sin ) = (1 2i sin )(1 + 2i sin ) 3 4 sin 2 1 + 4 sin 2 8 sin 1 + 4 sin2 =0 x = 2, y = 1 . Thus point is 8 sin + i 1 + 4 sin 2 Now, since it is real, therefore Im x + y = 1 and x + y = 3 ; (z) = 0 sin = 0 , = n where n = 0 , 1, 2, 3, ...... Trick : Check for (a), if n = 0, = 0 the given number is absolutely real but (c) also satisfies this condition and in (a) and (c), (c) is most general value of . 31. (a) ( 5 + 12i + 5 12i )( 5 + 12i + 5 12i ) ( 5 + 12i 5 12i )( 5 + 12i + 5 12i ) 5 + 12i + 5 12i + 2 5 + 12i 5 12i 5 + 12i 5 + 12i 10 + 2 ( 13) 3 2i = = i or 24i 2 3 = . Page | 52 COMPLEX NUMBER MATHEMATICS DPP DPP No. 3 DAILY PRACTICE PAPER 32. (b) According to condition z is multiplicative identity therefore 33. (c) Given that a 2 + b 2 = 1 , therefore z = 1 + 0i . 1 + b + ia (1 + b + ia)(1 + b + ia) = 1 + b ia (1 + b ia)(1 + b + ia) = = (1 + b)2 a 2 + 2ia(1 + b) (1 a 2 ) + 2b + b 2 + 2ia(1 + b) = 2(1 + b) 1 + b 2 + 2b + a 2 2b 2 + 2b + 2ia(1 + b) = b + ia 2 (1 + b) Trick : Put a = 0, b = 1 , 1 + b + ia 1 + 1 + 0 = =1 1 + b ia 1 + 1 0 But options (a) and (c) give 1. So again put a = 1, b = 0, 1 + b + ia 1 + i = =i. 1 + b ia 1 i Which gives (c) only. 34. (c) 35. (d) 3 + 2i sin 1 2i sin will be purely imaginary, if the real part vanishes, i.e., 3 4 sin 2 = 0 sin = = n + ( 1)n (only if 1 + 4 sin 2 =0 be real) 3 = sin 2 3 = n 3 3 {(1 cos ) + i.2 sin } 1 = 2 sin 2 + i.4 sin cos 2 2 2 = 3 4 sin 2 1 2 sin sin + i.2 cos 2 2 2 = 2 sin 2 sin = 2 sin 1 sin 2 2 1 + i.2 cos i.2 cos 1 sin sin 2 2 2 i.2 cos i.2 cos 2 2 2 + 4 cos 2 sin 2 2 2 2 1 . it s real part sin = 2 sin 36. (b) 2 2 1 + 3 cos 2 2 = 1 2 1 + 3 cos 2 2 = 1 5 + 3 cos (x + iy)1 / 3 = a + ib (x + iy) = (a + ib)3 = a 3 + 3a 2 .ib + 3a.(ib) 2 + (ib) 3 = a 3 3ab2 + i(3a 2b b3 ) Equating real and imaginary parts, we get y x = 3a 2 b 2 = a 2 3b 2 and b a x y + = 4(a 2 b 2 ) a b Page | 53 MATHEMATICS DPP DPP No. 3 DAILY PRACTICE PAPER 37. (b) COMPLEX NUMBER 2 2 2i 1 i 2i 2 2 1 + i = 1 + i 1 i = (i + 1) = i + 1 + 2i = 2i . 38. (c) Equation (x + iy)(2 3i) = 4 + i (2x + 3y) + i( 3x + 2y) = 4 + i Equating real and imaginary parts, we get ......(i) 2x + 3y = 4 ......(ii) 3 x + 2y = 1 5 14 ,y = 13 13 4 + i (4 + i)(2 + 3i) 5 14 x + iy = = = + i 2 3i 13 13 13 From (i) and (ii), we get Aliter : x= . 39. (c) Given equation (x 4 + 2 xi) (3 x 2 + yi) = (3 5i) + (1 + 2yi) (x 4 3 x 2 ) + i(2 x 3y) = 4 5i Equating real and imaginary parts, we get ......(i) x 4 3x 2 = 4 and 2x 3y = 5 .....(ii) 1 3 1 x = 2, y = , 3 From (i) and (ii), we get x = 2 and y = 3, Trick : Put then 40. (c) We have Thus Im x = 2, y = 3 and 2 4 (1 + i)2 (2i)(2 + i) = +i = 5 5 2 i (2 i)(2 + i) (z ) = 4 5 we see that they both satisfy the given equation. . . 41. (a) If z 0 . Let z = x + iy z 2 = x 2 y 2 + i(2xy) Re(z)= 0 x = 0 . Therefore Im(z 2 ) = 2xy = 0 Thus Re(z) = 0 Im(z 2 ) = 0 . 42. (a) 5( 8 + 6i) 40 + 30i = a + ib = 15 + 20i = a + ib 2i (1 + i)2 Equating real and imaginary parts, we get a = 15 and b = 20 . 43. (d) The two complex numbers can be compare only when their real and imaginary parts are equal. In other words, there is no meaning of >, < in complex numbers. 44. (d) 1 2i 4 i (1 2i)(3 + 2i) + (4 i)(2 + i) + = 2 + i 3 + 2i (2 + i)(3 + 2i) = 45. (b) 50 120i 10 24 = i. 65 13 13 a + ib c + id , it is defined if and only if imaginary parts must be equal to zero. Therefore ib = id = 0 b = d = 0 ( i 0) 46. (b) If = x + iy = 3 2 + cos + i sin 3(2 + cos i sin ) 2 2 (2 + cos ) + sin = 6 + 3 cos 3i sin 4 + cos 2 + 4 cos + sin 2 Page | 54 MATHEMATICS DPP DPP No. 3 DAILY PRACTICE PAPER COMPLEX NUMBER 6 + 3 cos 3 sin = +i 5 + 4 cos 5 + 4 cos 3(2 + cos ) 3 sin ,y = 5 + 4 cos 5 + 4 cos 9 x 2 + y2 = (5 + 4 cos )2 x= [4 + cos 2 + 4 cos + sin 2 ] 9 6 + 3 cos = 4 3 = 4x 3 5 + 4 cos 5 + 4 cos 3(2 + cos i sin ) Trick : x + iy = (2 + cos + i sin )(2 + cos i sin ) = Let y = 0 , then sin = 0 i.e., = 0 . Now put x = 1 then x 2 + y 2 = 12 + 0 = 1 . Also option (b) gives 4(1) 3=1. 47. (d) + i = ( p + i)2 ( p 2 1 + 2 pi)(2 p + i) = 2p i (2 p i)(2 p + i) = 2 p( p 2 2) + i(5 p 2 1) 4 p2 + 1 2 + 2 = 4 p 2 ( p 2 2) 2 + (5 p 2 1) 2 (4 p 2 + 1) 2 = = 4 p6 + 6p2 + 9p4 + 1 (4 p 2 + 1) 2 p 4 (4 p 2 + 1) + 2 p 2 (4 p 2 + 1) + (4 p 2 + 1) (4 p 2 + 1) 2 = 48. (a) Given that p 4 + 2p 2 + 1 4p2 + 1 = ( p 2 + 1) 2 4p2 + 1 . z = 3 4i z 2 = 7 24i , z 4 = 117 44i and z 4 = 527 + 336i z 4 3z 3 + 3z 2 + 99z 95 = 5 Aliter : z = 3 4i (z 3)2 = 16 z 2 6z + 25 = 0 z 4 3z 3 + 3z 2 + 99z 95 2 = (z 2 + 3z 4)(z 2 6 z + 25) + 5 = (z + 3z 4)(0) + 5 = 5 49. (d) If z1 = 1 i and z 2 = 2 + 4i Then 50. (d) z1 z 2 (1 i)( 2 + 4i) = = 2 + 4i z1 1 i Given that z z Im 1 2 = 4 . z1 3 x + 2iy 15 = 5i 2 8 x + 3iy 24 x 2 + 9ixy 6y 2 + 16ixy = 75i 30 24 x 2 6y 2 + 25ixy = 75i 30 Equating real and imaginary parts, we get 24 x 2 6y 2 = 30 or 4 x 2 y 2 = 5 and xy = 3 On solving we get x = 1, y = 3 Page | 55 MATHEMATICS DPP DPP No. 3 DAILY PRACTICE PAPER 51. (c) 100 i k COMPLEX NUMBER = x + iy, 1 + i + i 2 + ...... + i100 = x + iy k =0 Given series is G.P. 1.(1 i101 ) = x + iy 1 i 1 i = x + iy 1 i 1 + 0i = x + iy Equating real and imaginary parts, we get the required result. 52. (a) 1 + iz 1 + i(b + ic) /(1 + a) 1 + a c + ib = = 1 iz 1 i(b + ic) /(1 + a) 1 + a + c ib = (1 + a c + ib)(1 + a + c + ib) (1 + a + c)2 + b 2 = 1 + 2a + a 2 b 2 c 2 + 2ib + 2iab) 1 + a 2 + c 2 + b 2 + 2ac + 2(a + c) a 2 + b 2 + c 2 + 2a + a 2 b 2 c 2 + 2ib(1 + a) 1 + 1 + 2ac + 2(a + c) 2a(a + 1) + 2ib(1 + a) a + ib = = . 2(1 + a)(1 + c) 1+ c = 53. (b) Let z1 = a + ib, z 2 = c + id , then (a + c) + i(b + d) is real b+d =0 d = b .....(i) (ad bd) + i(ac + bc) is real z1 z 2 is real ad + bc = 0 a( b) + bc = 0 a = c z1 = a + ib = c id = z2 ( a = c and b = d) z1 + z 2 54. (c) is real (x + iy)(p + iq) = (x 2 + y 2 )i (xp yq) + i(xq + yp) = (x 2 + y 2 )i x y = q p Let x y = = . q p xp yq = 0, xq + yp = x 2 + y 2 and xq + yp = x 2 + y 2 then xq + yp = x 2 + y 2 x = q, y = p . 55. (a) L.H.S. = x = q, y = p = 2 = 1 (cos x + i sin x)(cos y + i sin y) sin u cos v (cos u + i sin u)(cos v + i sin v) = sin u cos v[cos(x + y u v) + i sin(x + y u v)] 56. (c) 1 + i (3 2i) 5 + i x + iy = = 13 3 + 2i (3 2i) Hence 57. (b) Given, x= 5 1 ,y = 13 13 1 i 1+ i 100 . 1 i 1 i = a + ib ; = a + ib 1 + i 1 i (1 i)2 a + ib = 2 100 2i = 2 100 = ( i)100 Page | 56 COMPLEX NUMBER MATHEMATICS DPP DPP No. 3 DAILY PRACTICE PAPER 58. (c) a + ib = (i)4 25 = 1 + 0i, Hence, a = 1, b = 0 . z1 4 + 5i 3 2i 12 8i 15i + 10 = = z 2 3 + 2i 3 2i 9 (2i)2 z1 2 23 2 23 = i = , z2 13 13 13 13 59. (c) Given z = 1+ i and i = 1. Squaring both sides, we get z 2 = (1 + i)2 = 1 + 2i + i 2 = 1 + 2i 1 Since it is multiplicative identity, therefore multiplicative inverse of 60. (d) 6i 4 20 61. (c) 3i 3i 3 1 1 i 6i + 4 4 20 0 3i 3 1 i i i = = . 2i i 2i 2 2 0 1 = x + iy [R1 R1 + R2 ] i (6i + 4)(3i 2 + 3) = x + iy (6i + 4)( 3 + 3) = x + iy x + iy = 0 = 0 + i.0 (x, y) = (0, 0) . a = cos + i sin . 1 + a (1 + cos ) + i sin = . 1 a (1 cos ) i sin Rationalization of denominator, we get 1 + a (1 + cos ) + i sin (1 cos ) + i sin = 1 a (1 cos ) i sin (1 cos ) + i sin (1 + cos )(1 cos ) + (1 + cos ) i sin + (1 cos )i sin + i 2 sin 2 (1 cos )2 (i sin )2 = 1 cos 2 + i sin + i sin cos + i sin i sin cos sin 2 1 + cos 2 2 cos + sin 2 = 2i sin 1 (cos 2 + sin 2 ) + 2i sin = 2 2 2(1 cos ) 1 + (cos + sin ) 2 cos = z 2 = 2i. = x + iy = z2 = or i.2 sin 2 2 sin 62. (a) cos 2 2 2 =i cos 2 = i cot . 2 sin 2 3 2yi = 9 x 7i Equating real and imaginary parts both sides 9 x = 3 3 2 x = 31 2 x = 1 x = 0.5 2y = 7 y = 3.5 . 63. (b) z= 1 + 2i 3 1 + 2i 1 + i 1 z = = +i 1 i 2 1 i 1+ i 2 This complex number will lie in the II quadrant. 64. (b) Ans. (b) (1 cos ) i sin 1 1 = 1 cos + i sin (1 cos ) + i sin (1 cos ) i sin = (1 cos ) i sin (1 cos )2 + sin 2 = (1 cos ) i sin 2(1 cos ) Page | 57 COMPLEX NUMBER MATHEMATICS DPP DPP No. 3 DAILY PRACTICE PAPER (1 cos ) sin i . 2(1 cos ) 2(1 cos ) = Therefore its real part = 65. (d) a + ib c + id, b = d = 0. So, b 2 1 cos 1 = 2 (1 cos ) 2 defined if and + d2 = 0 . only if its imaginary parts must be equal to zero, i.e. 66. (b) We know that, multiplicative inverse of x + iy is 1 x + iy According to question x + iy = 1 x + iy and only (b) is satisfying such condition. 67. (b) (x + iy)1 / 3 = a ib x + iy = (a ib)3 = (a 3 3ab 2 ) + i(b 3 3a 2b) x = a 3 3ab 2 , y = b 3 3a 2 b x y = a 2 3b 2 , = b 2 3a 2 a b x y = a 2 3b 2 b 2 + 3a 2 a b x y = 4(a 2 b 2 ) = k(a 2 b 2 ) a b k = 4. Conjugate, Modulus and Argument of complex numbers 1. (d) sin x + i cos 2 x and cos x i sin 2 x are tan x = 1 x = or and or tan 2 x = 1 x= 5 9 8 , 8 , 8 conjugate to each other if 5 9 , , ,...... 4 5 9 2x = , , ,........ 4 4 4 4 4 ....... sin x = cos x and cos 2x = sin 2x (i) ..(ii) There exists no value of x common in (i) and (ii). Therefore there is no value of given complex numbers are conjugate. 2. (a) Let 3. 4. z = x + iy, z = x iy and z 1 = (z 1 ) = x + iy x 2 + y2 ; (z 1 ) z = x for which the 1 x + iy x + iy ( x iy) = 1 x 2 + y2 (c) Let z = x + iy , then its conjugate z = x iy Given that z 2 = (z)2 x 2 y 2 + 2ixy = x 2 y 2 2ixy 4ixy = 0 If x 0 then y = 0 and if y 0 then x = 0 (a) Let z = x + iy, z = x iy z z = 0 (x + iy)(x iy) = 0 x 2 + y 2 = 0 It is possible only when x and y both simultaneously zero i.e., z = 0 + 0i = 0 Page | 58 MATHEMATICS DPP DPP No. 3 DAILY PRACTICE PAPER 5. (a) COMPLEX NUMBER .....(i) ......(ii) (a + ib)(c + id)(e + if )(g + ih) = A + iB (a ib)(c id)(e if )(g ih) = A iB Multiplying (i) and (ii), we get (a 2 + b 2 )(c 2 + d 2 )(e 2 + f 2 )(g 2 + h2 ) = A 2 + B 2 6. (d) Let z = x + iy, 2 so that 2 therefore z = x iy, 2 z + z = 0 ( x y + x) + i (2 xy y) = 0 Equating real and imaginary parts, we get x 2 y2 + x = 0 .....(i) and 2xy y = 0 If y = 0, If x= Then 7. 8. 9. y = 0 or x = then (i) gives 1 2 x 2 + x = 0 x = 0 or x = 1 1 , 2 x 2 y2 + x = 0 y2 = 1 1 3 + = 4 2 4 y = 3 2 Hence, there are four solutions in all. (c) Here z + z = (x + iy) + (x iy) = 2x (Real) and z z = (x + iy)(x iy) = x 2 + y 2 (Real). (a) According to condition, 3 ix 2 y = x 2 + y + 4i (b) x 2 + y = 3 and x 2 y = 4 (x, y) = (2, 1) or ( 2, 1) 2 + 5i (2 + 5i)(4 + 3i) 7 + 26i = = 4 3i 25 25 x = 2, y = 1 . Therefore conjugate of the complex number is 10. (c) As we know if a is real, then a = a (z + a)(z + a) = (z + a)(z + a) = (z + a)(z + a) 11. (b) Here . =| z + a |2 z i x + i(y 1) x i(y + 1) = . z + i x + i(y + 1) x i(y + 1) = As 7 26i 25 z i z+i (x 2 + y 2 1) + i( 2 x) x 2 + (y + 1)2 is purely imaginary, we get x 2 + y 2 1 = 0 x 2 + y 2 = 1 zz = 1 . 12. (a) c+i = a + ib c i c i = a ib c+i .....(i) .....(ii) Multiplying (i) and (ii), we get c2 + 1 = a2 + b2 a2 + b2 = 1 . c2 + 1 13. (c) Given that x iy = (x + iy)(1 2i) = 1 + i 1+ i 1 i x + iy = 1 + 2i 1 2i . Page | 59 MATHEMATICS DPP DPP No. 3 DAILY PRACTICE PAPER 14. (c) z= (2 + i) 2 9 3 + 4i 3 i 13 = = +i 3+i 3 + i 3 i 10 10 Conjugate 15. (c) COMPLEX NUMBER = 13 9 . i 10 10 z = 3 + 5i , z = 3 5i z 3 = (3 + 5i)3 = 3 3 + (5i)3 + 3.3.5i (3 + 5i) = 198 + 10i Hence, 16. (b) z 3 + z + 198 = 10 i 198 + 3 5 i + 198 = 3 + 5 i . 2 3i (2 3i) (4 + i) 8 + 3 12i + 2i 11 10i = = = 4 i (4 + i) (4 i) 16 + 1 17 Conjugate = 11 + 10i 17 . 17. (c) Let z = 1 + i z = 1 i 18. (d) Given inequality | z 4 | | z 2| | z 4 |2 | z 2 |2 (x 4)2 + y 2 (x 2)2 + y 2 19. (b) As 4 x 12 Re(z) 3 . given, let 2z1 = iy or z1 = 3 iy , z2 2 3z 2 z1 3 1 iy 1 1 z1 z 2 z2 = = 2 = z1 3 z1 + z 2 iy + 1 1 + +1 2 z2 so that 3 iy 2 = 1 | z |=| z | 3 iy 2 20. (b) | z1 + z 2 |2 + | z1 z 2 |2 = (x1 + x 2 )2 + (y1 + y2 )2 + (x1 x 2 )2 + (y1 y2 )2 = 2(x12 ) + 2(y12 ) + 2(x 22 ) + 2(y 22 ) = 2 | z1 |2 +2 | z 2 |2 z 1 = iy where y R z +1 21. (b) Let This gives | z |= z= 1 + iy 1 + iy 1 + iy (1 y 2 ) + 2iy = = 1 iy 1 iy 1 + iy 1 + y2 1 1 + y2 2 2 2 ( 1 y ) + 4 y = =1. 1 + y2 1 + y2 22. (c) L.H.S.= | z 2 |=| (x + iy)2 | = | x 2 y 2 + 2ixy |= (x 2 y 2 )2 + (2xy)2 = (x 2 + y2 R.H.S. ) 2 ..(i) =| z |2 =| x + iy |2 = (x 2 + y 2 )2 (ii) = x 2 + y2 Therefore | z |=| z | (b) True (c) False (since 2 23. (b) z+ 2 z z ). 2 2 = 2 | z | 2 | z |2 2 | z | 2 0 z |z| | z | 2 4 +8 1 3 2 . Hence max. value of | z |is 1 + 3 Page | 60 COMPLEX NUMBER MATHEMATICS DPP DPP No. 3 DAILY PRACTICE PAPER 24. (c) Given z + z2 z1 + z 2 = cos + i sin = 1 1 z1 z 2 z1 z 2 (say) z1 1 + cos + i sin = = i cot z 2 1 + cos + i sin 2 which is zero, if = n (n I), and is otherwise purely imaginary. 25. (c) | z | z = 1 + 2i Let z = x + iy , therefore | x + iy | (x + iy) = 1 + 2i Equating real and imaginary parts, we get x 2 + y 2 x = 1 and y = 2 x = Hence complex number 3 2i . 2 3 3 2i 2i 2 2 Trick : Since = z= 3 2 9 3 5 3 + 4 + 2i = + 2i = 1 + 2i 4 2 2 2 26. (a) Let z1 = a + ib = (a, b) and z 2 = c id = (c, d) Where a 0 and d 0 Then | z1 |=| z 2 | a 2 + b 2 = c 2 + d 2 Now ......(i) z1 + z 2 (a + ib) + (c id) = z1 z 2 (a + ib) (c id) = [(a + c) + i(b d)][( a c) i(b + d)] [(a c) + i(b + d)][( a c) i(b + d)] = (a 2 + b 2 ) (c 2 + d 2 ) 2(ad + bc)i a 2 + c 2 2ac + b 2 + d 2 + 2bd (ad + bc)i a 2 + b 2 ac + bd (z + z 2 ) is purely 1 (z 1 z 2 ) However if [using (i)] imaginary. ad + bc = 0 , then (z 1 + z 2 ) (z 1 z 2 ) will be equal to zero. According to the conditions of the equation, we can have ad + bc = 0 Trick : Assume any two complex numbers satisfying both conditions i.e., Let z1 = 2 + i, z 2 = 1 2i, z1 z 2 and | z1 |=| z 2 | z1 + z 2 3 i = = i z1 z 2 1 + 3i Hence the result. 27. (d) It may be greater than, less than or equal to zero. 28. (a) If | z1 |= 1 and | z 2 |= 1, then | z1z 2 |=| z1 || z 2 |= 1.1 = 1 29. (a) Suppose there exists a complex number | z | 1 . Then z4 + z + 2 = 0 2 2 which satisfies the given equation and is such that 2 = z 4 + z | 2 |=| z 4 + z | 2 | z 4 | + | z | 2 2, But z because| z | 1 is not possible. Hence given equation cannot have a root z such that | z | 1 Page | 61 MATHEMATICS DPP DPP No. 3 DAILY PRACTICE PAPER COMPLEX NUMBER 30. (c) We have | zk |= 1, k = 1, 2,.... n | zk |2 = 1 zk z k = 1 zk = 1 zk Therefore | z1 + z 2 + .... + zn |=| z1 + z 2 + .... + zn | ( | z |=|z |) =| z 1 + z 2 + ..... + z n |= Aliter : Let 1 1 1 + + .... + z1 z 2 zn z k = cos k + i sin k , k = 1, 2,.... n So that | zk |= cos 2 k + sin 2 k = 1 Then 1 = (cos k + i sin k ) 1 = (cos k i sin k ) zk Now, z1 + z 2 + ..... + z n = (cos 1 + ..... + cos n ) i(sin 1 + ..... + sin n ) 1 1 1 + + ..... + z z z 1 2 n and = (cos 1 + ..... + cos n ) i(sin 1 + ..... + sin n ) Hence | z1 + z 2 + ..... + z n |= 1 1 1 + + ..... + z1 z 2 zn Since each side is equal to (cos 1 + ..... + cos n ) 2 + (sin 1 + .... + sin n ) 2 31. (b) Given that z= 1 zz = 1 z | z |2 = 1 | z |= 1 32. (a) It is a fundamental property. 33. (a) Let z = x + iy ......(i) Given | z + i |=| z i | or | x + iy + i |=| x + iy i | or | x + i(y + 1)|=| x + i(y 1)| or x 2 + (y + 1)2 = x 2 + (y 1)2 or x 2 + (y + 1)2 = x 2 + (y 1)2 or y 2 + 2y + 1 = y 2 2y + 1 or 4 y = 0 or y = 0 Hence from (i), we get z = x , where x is any real number. 34. (a) z = x + iy , then | z 5 |=| x + iy 5 |=| x 5 + iy | = (x 5)2 + y 2 . (2 + i) 2+i =|1 + i | = (3 + i) 3+i 2 5 35. (c) (1 + i) 36. (c) z z2 z1 z 2 = cos + i sin = 1 1 z1 + z 2 z1 + z 2 10 =1 2 z1 cos + i sin + 1 = 2z 2 cos 1 + i sin Page | 62 MATHEMATICS DPP DPP No. 3 DAILY PRACTICE PAPER COMPLEX NUMBER (Applying componendo and dividendo) z1 = i cot z2 2 But k = cot tan = 2 iz1 = kz 2 k = cot Now iz1 = cot 2 2k 1 k2 cot 2 2 = k tan = 2k 1 = 2 tan k 2 1 k z1 z 2 = cos + i sin z1 + z 2 is the angle between 37. (b) Let +2k k2 1 = tan 1 Now z2 z1 z 2 and z1 + z 2 . z = r(cos + i sin ) . Then z+ 1 1 =1 z+ z z 2 =1 2 1 r(cos + i sin ) + (cos i sin ) = 1 . r 1 1 2 2 r + cos + r sin = 1 r r r2 + 2 2 1 + 2 cos 2 = 1 r2 Since | z |= r is maximum, therefore dr =0 d Differentiating (i) w.r.t. , we get 2r dr 2 dr 4 sin 2 = 0 d r 3 d Putting dr = 0 ,we d get sin 2 = 0 = 0 or 2 z is purely imaginary or purely real. ( 38. (d) Trick : Check by putting =0 is not given) z1 = 1 + 0i and z 2 = 0 + i 1 1 |( z1 + z 2 )2 | + |( z1 z 2 )2 | 2 2 1 1 2 { | z 2 |=| z |2 } = | z1 + z 2 | + | z1 z 2 |2 2 2 1 = 2 [| z1 |2 + | z 2 |2 ] =| z1 | + | z 2 | 2 39. (c) R.H.S = 40. (a) 3 + 2i 3 + 2i 3 + 2i = 9 4 + 12i = 5 + i 12 = 13 13 13 3 2i 3 2i 3 + 2i Modulus = 41. (b) 2 z = x + iy | z |2 = x 2 + y 2 = 1 Now, 2 5 12 + =1 . 13 13 .....(i) z 1 (x 1) + iy ( x + 1) iy = z + 1 ( x + 1) + iy ( x + 1) iy Page | 63 COMPLEX NUMBER MATHEMATICS DPP DPP No. 3 DAILY PRACTICE PAPER = 2iy (x 2 + y 2 1) + 2iy = (x + 1)2 + y 2 (x + 1)2 + y 2 Hence, z 1 is z +1 purely imaginary. 42. (c) Given expression, expression = 1 1 +0= 3 3 =0+ [by equation (i)] | 2z 1| + | 3z 2| , 1 1 = , 2 2 minimum value of minimum value of | 3z 2 | | 2z 1| is is 0, at . So minimum value of given expression is 1 3 z= 2 . 3 0 at z= 1 2 . So value of given So value of given expression . 43. (a) | z |= 1 | x + i y |= 1 x 2 + y 2 = 1 = = z 1 ( x 1) + i y ( x + 1) i y = z + 1 ( x + 1) + i y ( x + 1) i y 2i y 2i y (x 2 + y 2 1) + = (x + 1)2 + y 2 (x + 1)2 + y 2 (x + 1)2 + y 2 ( x 2 + y 2 = 1) 44. (c) Re ( ) = 0 . i = 3 4 xi . 3 + 4 xi Taking modulus and squaring on both sides, 45. (b) We have | z1 | = 1 and z 2 be z1 z 2 |z z | = 1 2 ; 1 z1z 2 z 1 2 z1 = | z1 z 2 | | z1 |; | z1 z 2 | = | z1 z 2 | | z1 z 2 | = =1. | z1 z 2 | | z1 z 2 | 46. (d) Let 2 + 2 =1. any complex number. z 1 z1 = | z 1 |2 Given that | z1 | = 1 z1 = r1 (cos 1 + i sin 1 ) , z 2 = r2 (cos 2 + i sin 2 ) | z1 + z 2 |= [(r1 cos 1 + r2 cos 2 )2 + (r2 sin 1 + r2 sin 2 )2 ]1 / 2 = [r12 + r22 + 2r1r2 cos( 1 2 )]1 / 2 = [(r1 + r2 )2 ]1 / 2 ( | z1 + z2 |=| z1 | + | z2 |) Therefore cos( 1 2 ) = 1 1 2 = 0 1 = 2 Thus arg (z1 ) arg(z 2 ) = 0 . Trick : | z1 + z 2 |=| z1 | + | z 2 | z1, z 2 lies on same straight line. arg z1 = arg z 2 arg z1 arg z 2 = 0 47. (d) Let 48. (d) z = 5 3i r(cos + i sin ) = 5 3 i r cos = 5 and r sin = 3 tan = 3 3 = tan 1 5 5 1 + i 1 + i 1 + i (1 + i)2 = = 1 i 1 i 1+ i 2 Now 1 + i = r(cos + i sin ) r cos = 1, r sin = 1 Page | 64 MATHEMATICS DPP DPP No. 3 DAILY PRACTICE PAPER COMPLEX NUMBER r = 2 , = / 4 1 + i = 2 cos + i sin 4 4 1 1 (1 + i)2 = . 2 cos + i sin 2 2 4 4 By De Moivre's Theorem, Hence the amplitude is Trick : 2 cos + i sin 2 2 and modulus is 1. 2 1 + i arg = arg(1 + i) arg(1 i) 1 i = 45o ( 45o ) = 90o 1+i 1+i = = 1 i 1 i 49. (d) Let 2 = 1. 2 z = x + iy, z = x iy Since arg(z) = = tan 1 y x y arg(z) = = tan 1 x Thus 50. (c) arg(z) arg(z) . | z |= 4 and arg z = Let z = x + iy , and = 5 = 150 o 6 then | z |= r = x 2 + y2 = 4 5 = 150 o 6 x = r cos = 4 cos 150 o = 2 3 and . 1 y = r sin = 4 sin 150o = 4 = 2 2 z = x + iy = 2 3 + 2i Trick : Since arg z = (b) rejected. Also 51. (c) If z= 1 i 3 1+ i 3 = 5 = 150 o , 6 | z |= 4 here the complex number must lie in second quadrant, so (a) and which satisfies (c) only. (1 i 3 )(1 i 3 ) (1 + i 3 )(1 i 3 ) 1 3 2i 3 2 2i 3 1 3 = = i 1+ 3 4 2 2 Thus arg(z) = tan 1 y = tan 1 3 = = 60o. x 3 = Since the complex number lies in III quadrant, therefore arg (z) is 180o + 60 o = 240o Aliter : 1 i 3 = arg(1 i 3 ) arg(1 + i 3 ) arg 1+ i 3 = 60o 60o = 120o or 240o . 52. (b) It is a fundamental concept. 53. (b) z = sin + i(1 cos ) Page | 65 COMPLEX NUMBER MATHEMATICS DPP DPP No. 3 DAILY PRACTICE PAPER 1 1 cos amp(z) = tan = tan 1 tan = 2 2 54. (a) Let z= 1+i 3 1+ 3 sin 2 2 sin 1 2 = tan 2 sin cos 2 2 . amp(z) or arg(z) 3 /(1 + 3 ) 1 = tan 1 3= = tan 3 1 /(1 + 3 ) 55. (c) 3 = 120o arg( 1 + i 3 ) = tan 1 1 because it lies in second quadrant. 56. (c) 6 + 5i + i 2 + 6 5i + i 2 3+i 3 i arg + = arg 5 2 i 2+i 10 = arg = 0 . 5 57. (a) We know that the principal value of 58. (b) Let z = 0 + ib , where b 0. lies between and Then z lies on +ve y-axis and so . arg(z) = 2 . 59. (d) Let z = 0 + ib , where b 0 . Then y axis), therefore arg(z) = . z is represented by a point on 60. (a) Let is represented by a point on negative side of x-axis, therefore OY (negative direction of 2 z = a + i0 , arg(z) = where a 0. Then z 61. (c) Since the multiplication of a complex number by negative (clockwise) direction. 62. (a) We have | z1 + z 2 |2 =| z1 |2 + | z 2 |2 i rotates through it by a right angle in | z1 |2 + | z 2 |2 +2 | z1 || z 2 | cos( 1 2 ) =| z1 |2 + | z 2 |2 Where 1 = arg(z1 ), 2 = arg(z 2 ) cos( 1 2 ) = 0 1 2 = 2 z1 z1 | z1 | arg = Re = cos = 0 z2 2 z2 | z2 | 2 Note : Also z Re 1 z2 = 0 Re(z1 z 2 ) = 0 z1 z 2 is purely imaginary. 63. (c) We have | z1 z 2 |2 =| z1 |2 + | z 2 |2 2 | z1 || z 2 | cos( 1 2 ) where 1 = arg(z1 ) and Since arg z1 arg z 2 = 0 2 = arg(z 2 ) | z1 z 2 |2 =| z1 |2 + | z 2 |2 2| z1 || z 2 | Page | 66 MATHEMATICS DPP DPP No. 3 DAILY PRACTICE PAPER COMPLEX NUMBER = (| z1 | | z 2 |)2 | z1 z 2 |=|| z1 | | z 2 || 64. (c) Squaring the given relations implies that x1 x 2 + y1y2 = 0 Now = tan amp z1 amp z 2 = tan 1 1 y1 y 2 x1 x 2 y x y 2 x1 = tan 1 1 2 = tan 1 = . y1 y 2 x1 x 2 + y1 y 2 2 1+ x1 x 2 65. (a) We have z arg 1 = z2 arg(z1 ) arg(z 2 ) = Let arg (z 2 ) = y1 y tan 1 2 x1 x2 , then arg (z1 ) = arg (z 2 ) + arg (z1 ) = + z1 =| z1 |[cos( + ) + i sin( + )] =| z1 |( cos i sin ) and z 2 =| z 2 |(cos + i sin ) =| z1 |(cos + i sin ) ( | z1 |=| z 2 |) Hence 66. (c) amp(z) z1 + z 2 = 0 . amp( z) = tan 1 67. (d) amp (z) = tan 1 y y tan 1 = x x sin 1 cos = tan 1 cot = tan 1 tan = 2 2 2 2 2 68. (c) . z arg 1 = arg z1 arg (z 2 ) = arg z1 + arg z 2 = arg(z1 .z 2 ) z2 Option (c) gives the same result. 69. (b) 13 5i arg = arg(13 5i) arg(4 9i) 4 9i 9 5 = tan 1 + tan 1 = 13 4 4 70. (b) Let z1 = r1(cos 1 + i sin 1) Then | z1 |=| z2 | | z2 |= r1 and arg(z1 ) + arg(z2 ) = 0 arg(z2 ) = arg(z1) = 1 z2 = r1[cos( 1) i sin( 1 )] = r1(cos 1 i sin 1) = z1 z1 = z 2 . 71. (c) | z1 + z2 |=| z1 | + | z2 | | z1 + z 2 |2 =| z1 |2 + | z 2 |2 +2 | z1 || z 2 | | z1 |2 + | z 2 |2 +2 Re | z1z 2 | =| z1 |2 + | z 2 |2 +2 | z1 || z 2 | Page | 67 MATHEMATICS DPP DPP No. 3 DAILY PRACTICE PAPER 72. (a) 2 Re | z1z2 |= 2 | z1 || z2 | 2 | z1 || z2 | cos( 1 2 ) = 2 | z1 || z2 | arg(z1) = arg(z2 ) COMPLEX NUMBER cos( 1 2 ) = 1 or 1 2 = 0 1 i 1 i 1 i (1 i)2 2i = = = i = 2 2 1+ i 1+ i 1 i Im(z) 0 , Hence amplitude = / 2 . 73. (a,d) These are the properties of moduli. 74. (a) 1+ 3 i = amp(1 + 3 i ) amp( 3 + i) amp 3 +i = 3 = 6 . 6 75. (d) Amplitude of 0 is not defined. 76. (a) Since arg where arg i.e. ve, we choose arg z = z 0 is +ve ( z) = [+ ( )] = = 2 + ( ) = + arg(z) arg ( z) arg(z) = . . 77. (d) 1 + 3i z= 3 i = = 1 + 3i 3 i 3 + i + 3i 3 3 +1 amp (z) = / 2 78. (c) z= 2 1 + 3i z = cos | z |= and 3 +i = 4i 4 =i [ tan = b / a] 2 1 + 3i 1 3i 1 3i 6 + i sin 6 = sin 2 + 2 3i 1+ 3 . 3 i + 2 2 3 1 + =1 4 4 1/ 2 y = tan 1 1 arg (z) = tan 1 = tan 1 x 3 / 2 3 arg(z) = tan 1 tan = 6 6 80. (c) = 1 3 3 / 2 2 + i arg (z) = tan 1 = 2 2 3 1/ 2 z= 79. (b) = 3 +i . + i 2 sin 2 + i 1 cos = 2 sin cos 10 10 10 5 5 = 2 sin + i sin cos 10 10 10 Page | 68 COMPLEX NUMBER MATHEMATICS DPP DPP No. 3 DAILY PRACTICE PAPER For amplitude, 81. (d) Let tan = sin 10 = tan 10 cos 10 = 10 . z = 1 i 3 then b 3 = tan 1 = a 1 3 = tan 1 Clearly, z is in III quadrant. Therefore argument = ( ) = ( / 3) = 82. (d) | z || |= 1 and 2 3 . ......(i) z z arg = =i 2 z =1 .....(ii) From equation (i) and (ii) and | z |=| |= 1 z = z = 83. (b) We have z z + z = 0; z + z = 0 ; z = i| |2 = i. . z = x + iy and let their complex z2 and given that arg (z) + z 2 = y z 2 = + tan 1 x z 2 = arg(z) ; z 2 = + [arg (z)] which lies in second quadrant, i.e. 84. (b) z . 1 + 2i 1 + 2i 1 + 2i = = = 1 + 0i 1 (1 i) 2 1 (1 1 2i) 1 + 2i Modulus =1 Amplitude 85. (d) Given = tan 1 0 1 = 0. z1 = 1 + 2i , z2 = 3 + 5i and z2 = 3 5i z 2 z1 (3 5i)(1 + 2i) 13 + i = = z2 (3 + 5i) 3 + 5i = 13 + i 3 5i 44 62i = 3 + 5i 3 5i 34 Then z z 44 22 Re 2 1 = = z 2 34 17 86. (a) Given : Let . (3 + i)z = (3 i)z z = x(3 i) , x R L.H.S. = = R.H.S. = (3 + i)z = (3 + i) x (3 i) x (3 + i) (3 i) = x [(3)2 + 12 ] = 10 x (3 i)z = (3 i) x (3 + i) = x [3 2 + 12 ] = 10 x Hence, L.H.S. = R.H.S. Page | 69 MATHEMATICS DPP DPP No. 3 DAILY PRACTICE PAPER z = x(3 i) 87. (c) satisfies the equation, then z = x(3 i) , COMPLEX NUMBER where x is a real number. ( 8 + i) 50 = 3 49 (a + ib) Taking modulus and squaring on both sides, we get (8 + 1)50 = 3 98 (a 2 + b 2 ) 9 50 = 3 98 (a 2 + b 2 ) 3100 = 3 98 (a 2 + b 2 ) (a 2 + b 2 ) = 9 . Square root, Representation and Logarithm of complex numbers 1. (a) x + iy = Also 2. 3. a + ib c + id x iy = a ib c id x 2 + y 2 = (x + iy)(x iy) = ( x 2 + y 2 )2 = a2 + b2 c2 + d2 a2 + b2 c2 + d2 (b) Given that 8 6i = x + iy = z 8 6i = (x + iy)2 x 2 y 2 = 8 .....(i) and 2xy = 6 .....(ii) 2 2 Now x + y = 64 + 36 = 10 .....(iii) From (i) and (iii), we get x = 1 and y = 3 Hence z = (1 3i) Trick : Since { (1 3i)}2 = 8 6i (b) 7 24i = x iy Squaring both sides, 7 24i = x 2 y 2 i(2xy) Equating real and imaginary parts, we get x 2 y 2 = 7 and 2xy = 24 x 2 + y 2 = 49 + 576 = 625 = 25 4. (c) x + iy = (a + bi) x + iy = a 2 b 2 + 2iab x iy = (a 2 b 2 ) 2iab = b 2 a 2 2iab x = a 2 b 2 , y = 2ab = (b ia)2 = (b ia) . 5. (a) Let 3 4i = x + iy 3 4i = x 2 y 2 + 2ixy x 2 y 2 = 3, 2xy = 4 ......(i) (x 2 + y 2 )2 = (x 2 y 2 )2 + 4 x 2 y 2 = (3)2 + ( 4)2 2 2 x +y = 5 = 25 ..(ii) Page | 70 MATHEMATICS DPP DPP No. 3 DAILY PRACTICE PAPER From equation (i) and (ii) y 2 = 1 y = 1. Hence 6. (d) a + ib = x + yi 2 a = x 2 y 2 , b = 2 xy a ib = x2 = 4 x = 2, the square root of ( a + ib) (3 4i) is (2 i) . = (x + yi) 2 and hence x 2 y 2 2xyi = (x yi) 2 = x iy Note: In the question, it should have been given that 7. a, b, x, y R. 1 i = 2 cos i sin ,|1 i |= 2 4 4 (d) Since |1 i |x = 2 x x =x 2 COMPLEX NUMBER ( 2)x = 2 x 2x / 2 = 2 x x=0 Therefore, the number of non-zero integral solutions is nil or zero. 8. (a) 1 + 7i (2 i) 2 Let = (1 + 7i) (3 + 4i) 25 + 25i = = 1 + i (3 4i) (3 + 4i) 25 z = x + iy = 1 + i r cos = 1 and r sin Thus = 3 4 9. r= 2 1 + 7i 1 + 7i = = 2 3 4i (2 i)2 1 + 7i 4 arg = tan 1 7 tan 1 3 4i 3 = tan 1 7 + tan 1 and 1 + 7i 3 3 = 2 cos + i sin 4 4 (2 i)2 Aliter : and =1 4 3 = 3 4 1 + 7i 3 3 = 2 cos + i sin 4 4 (2 i)2 (b) If z = re i = r(cos + i sin ) iz = ir(cos + i sin ) = r sin + ir cos or e iz = e ( r sin +ir cos ) = e sin e ri cos or | e iz |=| e r sin || e ri cos | = e r sin | e ir cos | = e r sin [{cos 2 (r cos ) + sin 2 (r cos )}]1 / 2 = e r sin 10. (b) 1 i (1 i)(1 i) 1 + (i)2 2i = = = i 1 + i (1 + i)(1 i) 1+1 which can be written as 11. (c) Here 1 + 3 = re i cos 2 i sin 1 + i 3 = re 2 i = r cos + ir sin Equating real and imaginary parts, we get r cos = 1 and r sin = 3 Page | 71 MATHEMATICS DPP DPP No. 3 DAILY PRACTICE PAPER Hence tan 12. (a) = 3 tan = tan 1 = 2 cos y y + .Hence = 2 3 . 1 = e i = cos i sin y then y = cos + i sin = e i , 2 3 COMPLEX NUMBER . 3 i = cos + i sin 2 6 6 13. (c,d) Since 3 3 3 i = cos + i sin = i 2 6 6 3 i = cos i sin 2 6 6 and 3 3 i = cos i sin = i . 2 2 2 and Hence the result. 14. (c) 1 3 1 + i 3 = 2 + i = 2 cos + i sin = 2e i / 3 2 2 3 3 (1 + i 3 )9 = (2e i / 3 )9 = 29.e i(3 ) = 29 (cos 3 + i sin 3 ) = 29 15. (a) ee a + ib = (1 + i 3 )9 = 29 ; i = e cos +i sin = ecos [ei sin ] Real part of 16. (b) Let z = ee i ee b=0. = e cos [cos(sin ) + i sin(sin )] i is e cos [cos(sin )] = e cos i sin = e cos e i sin z = e cos [cos(sin ) i sin(sin )] z = e cos cos(sin ) ie cos sin(sin ) e cos sin(sin ) amp(z) = tan 1 cos cos(sin ) e = tan 1[tan( sin )] = sin 17. (c) z= 1+ i 3 3 +i z= z= Now z= 1+ i 3 3 +i 3 i 3 i 3 + 3i i + 3 2( 3 + i) = 3 +1 4 3 +i = cos + i sin 2 6 6 z = cos 6 i sin 6 (z )100 = cos i sin 6 6 (z )100 = cos (z )100 . 100 50 50 2 2 i sin = cos i sin 3 3 3 3 lies in III quadrant. Page | 72 MATHEMATICS DPP DPP No. 3 DAILY PRACTICE PAPER 18. (d) x 2 3x + 1 = 0 x = cos + i sin 6 6 19. (d) Let 3 3 4 2 x= 3 i 3 i = 2 2 2 x= COMPLEX NUMBER z = 1 + i 3 , [Taking +ve sign] r = 1+ 3 = 2 3 2 = 1 3 = tan 1 2 2 z = 2 cos + i sin 3 3 2 2 (z) 20 = 2 cos + i sin 3 3 2 2 = 2 20 cos + i sin 3 3 20. (c) 20 20 1 3 = 2 20 + i 2 2 5i i 1 5 tan = i tan = log 3 3 2 1 20 . 5 +1 3 5 1 3 1 5i 1 Im tan 1 = log 4 = . 2 log 2 = log 2 . 2 3 2 21. (a) Let z = (1 i) i . Taking log on both sides, log z = i log(1 i) = i log 2 cos i sin 4 4 = i log ( 2e i / 4 ) = i 12 log 2 + log e i / 4 i i 1 = i log 2 = log 2 2 4 4 2 z = e / 4 e i / 2 log 2 . Taking real part only, 1 Re(z) = e / 4 cos log 2 . 2 22. (b) Let z x i x i z = i log = log i x + i x +i x 2 1 2ix z x i x i = log = log 2 i x + 1 x + i x i x2 1 z 2x = log 2 i 2 i x + 1 x + 1 ......(i) log(a + ib) = log(re i ) = log r + i = log a 2 + b 2 + i tan 1 (b / a) Hence, z = log i 2 x 2 1 2x 2 1 2 x x 2 + 1 + x 2 + 1 + i tan x 2 1 [by eqn. (i)] Page | 73 COMPLEX NUMBER MATHEMATICS DPP DPP No. 3 DAILY PRACTICE PAPER x 4 + 1 2x 2 + 4 x 2 z = log i 2 (x + 1) 2 2x + i tan 1 2 1 x = log 1 + i (2 tan 1 x) = 0 + i (2 tan 1 x) z = i 2 2 tan 1 x = 2 tan 1 x = 2 tan 1 x . 23. (c) e iA .e iB .e iC = e iA+iB+iC = e i( A+ B+C) = e i [ A + B + C = ] = 24. (d) cos + i sin = ( 1) + i(0) = 1 . z= 7 i 3 + 4i 3 4i 3 + 4i = 21 + 25i + 4 16 + 9 = 25 (1 + i) 25 = (1 + i) z14 = (1 + i)14 = [(1 + i)2 ]7 = (2i)7 = 27 i7 = 27 i . Geometry of complex numbers 1. (c) Since the coordinates in complex plane are (2, 3) and Trick : We know that the distance between | 2 + 3i + 1 + i |= 5 . 2. (b) Diagonals of a parallelogram i.e. 3. z1 + z 3 z 2 + z 4 = 2 2 ABCD are and z2 is | z1 z2 | therefore, the required length bisected each other at a point z z + az + az = b | z + a |2 =| a |2 b, { z z =| z |2 } This equation will represent a circle with centre 4. z = a, if | a |2 b 0, i.e.| a |2 b represents point circle only. (c) Let r be the circum radius of the equilateral triangle and A (Z1 Z0) r since | a |2 = b the cube root of unity. A(Z1) Y 2 3 Hence the required distance is 5. z1 + z 3 = z 2 + z 4 (b) By adding aa on both the sides of we get, (z + a)(z + a) = aa b z1 ( 1, 1) O 2 3 2 3 X O (Z0) C(Z3 Z0) B(Z2) C(Z3) Let ABC be the equilateral triangle with z1, z 2 and z3 as its vertices circumcentre O (z0 ) . The vectors O A, O B, O C are equal and parallel to Then the vectors C respectively with OA , OB , OC respectively. OA = z1 z 0 = re i 2 + 3 OB = z 2 z0 = re 4 i + 3 OC = z3 z0 = re A, B and = r e i = r 2e i Page | 74 COMPLEX NUMBER MATHEMATICS DPP DPP No. 3 DAILY PRACTICE PAPER z1 = z0 + re i , z2 = z0 + r ei , z3 = z0 + r 2ei Squaring and adding z12 + z 22 + z 32 = 3z 02 + 2(1 + + 2 )z 0 re i + (1 + 2 + 4 )r 2e i 2 z1 and = 3z 20 , since 1 + + 2 = 0 = 1 + 2 + 4 5. Note : Students should remember this question as a formula. (b) Equation bz + bz = c Putting z = x + iy, b = b1 + ib2 where x, y, b1 , b 2 are real, then the given equation become (b1 ib2 )(x + iy) + (b1 + ib2 )(x iy) = c or 6. 2b1 x + 2b2 y = c Which is equation of a straight line. Note : It is a fundamental concept. (b) Let z1 , z 2 , z 3 be three complex numbers in A.P. Then 2z 2 = z1 + z 3 . Thus the complex number z 2 is the mid-point of the line joining the points three points z1, z 2 and z3 are in a straight line. 7. (b) Since the triangle with vertices z12 + z 22 + z 32 z1 = a + i, z 2 = 1 + bi and z3 = 0 z3 So the is equilateral, we have = z1 z 2 + z 2 z 3 + z 3 z1 (a + i)2 + (1 + ib)2 + 0 = (a + i)(1 + ib) + 0 + 0 a 2 b 2 + 2i(a + b) = a b + i(1 + ab) Equating real and imaginary parts, a 2 b 2 = a b (i) and 2(a + b) = 1 + ab ..... (ii) From (i), (a b)[(a + b) 1] = 0 either a = b or a + b = 1 Taking a = b , we get from (ii) 4a = 1 + a 2 or a 2 4a + 1 = 0 a= Since 4 16 4 = 2 3 2 0 a 1 and 0 b 1, we Taking a + b = 1 or b = 1 a, we 2 = 1 + a(1 a) or a a + 1 = 0 , 2 value of 8. (a) Let a and = 1 + 5 z , 9. a = b = 2 3 get from (ii) which gives imaginary values of a . Hence a = b = 2 3 is the required b. then + 1 = 5z | + 1|= 5 | z |= 5 2 = 10 Thus have ( | z |= 2, given value) lies on a circle. (c) Given the vertices of quadrilateral Page | 75 COMPLEX NUMBER MATHEMATICS DPP DPP No. 3 DAILY PRACTICE PAPER A(1 + 2i), B( 3 + i), C( 2 3i) and D(2 2i) Now, AB = 16 + 1 = 17 , BC = 16 + 1 = 17 CD = 16 + 1 = 17, DA = 16 + 1 = 17 AC = 9 + 25 = 34 , BD = 25 + 9 = 34 Hence it is a square. 10. (d) | z |= 4 2 + ( 3)2 = 5 Let z1 be the new vector obtained by rotating z in the clockwise sense through 180o , therefore z1 = e i z = (cos i sin ), i.e., z = 4 + 3i The unit vector in the direction of z1 is 4 3 + i 5 5 . Therefore required vector 4 3 4 3 = 3 | z | + i = 15 + i = 12 + 9i 5 5 5 5 11. (b) Since maximum distance of any complex number | |= + 1 1 + 1 + 1 | |2 2 | | 1 0 | | Hence max | |is 12. (b) 1+ 2 = 2+ from origin is given by | | therefore, 1 | | 2 2 2 2 . i.e., (3, 4) lie in fourth quadrant in complex plane, after turned anticlockwise through 180o this will lie in II quadrant, therefore, the number will be 3 + 4i , now after stretching it 2.5 times i.e., multiplying by 2.5, the required complex number will be 15 + 10i . 3 4i 2 13. (a,b) It is given that OP = OQ | OP |=| OQ | | a + ib |=| c + id | Also OP = OQ, OP + OQ = 0 (a + c) + i(b + d) = 0 a + c = 0 = b + d 14. (c) We have z k = 1 + a + a 2 + ..... + a k 1 = 1 ak 1 a zk 1 ak = 1 a 1 a zk 1 | ak | | a |k 1 = = 1 a |1 a | |1 a | |1 a | zk lies within z 1 1 . = 1 a |1 a | Page | 76 COMPLEX NUMBER MATHEMATICS DPP DPP No. 3 DAILY PRACTICE PAPER 15. (a) Let A be the vertex with affix rotating z1 through 2 n B (Z2) z 2 , i.e., z 2 can be obtained by either in clockwise or in anticlockwise direction. 2 n B(Z2) A(Z1) There are two possibilities of O 2 n z1 . z2 z = 2 e z1 z1 i 2 n z 2 = z1e i 2 n , ( | z 2 |=| z1 |) 2 2 z 2 = z1 cos i sin n n 16. (b) We have | BD |=|(4 + 2i) (1 2i)|= 9 + 16 = 5 88. Let the affix of A be 89. The affix of the mid point of BD is z = x + iy 5 ,0 . 2 Since the diagonals of a parallelogram bisect each other, therefore, the affix of the point of intersection of the diagonals is D(4+2i) 5 ,0 . 2 C E(5/2+0i) A(x+iy) B(1 2i) 1 BD = AC = AE 2 90. We have | AE |= 5 91. Which is satisfied by option (b). 17. (a) We have | z z1 |=| z z 2 |=| z z3 |=| z z4 | Therefore the point having affix z is equidistant from the four points having affixes z1 , z 2 , z 3 , z 4 . Thus z is the affix of either the centre of a circle or the point of intersection of diagonals of a square or rectangle. Therefore z1, z 2 , z3 , z4 are either concyclic or vertices of a square. Hence z1 , z 2 , z 3 , z 4 18. (a) are concyclic. BD = 2 AC 2DM = 2(2 AM) or DM = 2 AM or DM 2 = 4 AM 2 or 5 = 4[(x 2)2 + (y + 1)2 ] Again slope of DM = 2 .....(i) and slope of AM is y +1 x 2 Page | 77 MATHEMATICS DPP DPP No. 3 DAILY PRACTICE PAPER COMPLEX NUMBER AM is perpendicular to DM C ( 1 , 1 ) D (2, 1) M (x,y)A B y +1 2 = 1 x 2 = 2(y + 1) x 2 .....(ii) Hence from (i) and (ii), we get y= 19. (d) Let 1 3 , and x = 3,1 2 2 A, B, C be the points represented by the numbers C(z3) P(z) A(z1) z1 , z 2 , z 3 and P be the point represented by z P(z) B(z2) P(z) 92. Now the four points A, B, C, P form a parallelogram in the following three orders. (i) A, B, P, C (ii) B, C, P, A and (iii) C, A, P, B In case (i), the condition for A, B, P, C to form a parallelogram is AB = CP i.e., z 2 z1 = z z3 or z = z 2 + z 3 z1 Similarly in case (ii) and (iii), the required points and 20. (b) Let BC = AP or z 3 z 2 = z z1 i.e., z = z 3 + z1 z 2 CA = BP or z1 z 3 = z z 2 i.e., z = z1 + z 2 z 3 z = x + iy , therefore given equation becomes (x + iy)(x iy) + (2 3i)(x + iy) + (2 + 3i)(x iy) + 4 = 0 x 2 + y 2 + 2x + 3y 3ix + 2iy + 2x 2iy + 3ix + 3y + 4 = 0 x 2 + y 2 + 4 x + 6y + 4 = 0 Therefore given equation represents a circle with radius Radius = 2 2 + 3 2 4 = 4 + 9 4 = 9 = 3 . 21. (d) Area of required rectangle = 4 area of OABC Y C O B(a, b 3 ) A X Page | 78 COMPLEX NUMBER MATHEMATICS DPP DPP No. 3 DAILY PRACTICE PAPER = 4 a b 3 = 4ab 3 . 22. (a) This is a parallelogram of OP1 P2 P3 . Then the mid point of P1 P2 and OP3 are the same. But midpoint x + x 2 y1 + y2 , P1 P2 is 1 2 2 P3 are (x1 + x 2 , y1 + y2 ) So that the coordinates of Thus the point P3 corresponds to sum of the complex number and z1 z2 . OP 3 = OP 1 + P1 P3 = OP 1 + OP 2 = z1 + z 2 23. (d) Given, | z 2| =2 | z 3| (x 2)2 + y 2 = 2 (x 3)2 + y 2 (x 2)2 + y 2 = 4[(x 3)2 + y 2 ] x 2 + y 2 + 4 4 x = 4 x 2 + 4 y 2 + 36 24 x or x 2 + y2 3 x 2 + 3y 2 20 x + 32 = 0 20 32 x+ =0 3 3 .....(i) We know that, standard equation of circle, x 2 + y 2 + 2gx + 2 fy + c = 0 .....(ii) Comparison of (i) from(ii) 2g = 20 10 32 g = , f = 0, c = 3 3 3 Hence, Radius = 24. (d) g2 + f 2 c = 100 32 = 9 3 4 2 = 9 3 and C = / 2 By rotation about C in anti clockwise sense BC = AC A(z1) 90 B(z2) CB = CA e i / 2 C(z3) (z 2 z 3 ) = (z1 z 3 ) e i / 2 = i (z1 z 3 ) (z 2 z 3 ) 2 = (z 1 z 3 ) 2 z 22 + z 32 2z 2 z 3 = z12 z 32 + 2z1 z 3 z12 + z 22 2z1 z 2 = 2z1 z 3 + 2z 2 z 3 2z 32 2z1 z 2 (z1 z 2 )2 = 2[(z1 z 3 z 32 ) (z1 z 2 z 2 z 3 )] (z1 z 2 )2 = 2(z1 z 3 )(z 3 z 2 ) . 25. (d) Let the vertices be z 0 , z1 ,....., z 5 w.r.t centre O at origin and | z 0 |= 5 . Y A4 (1+2i) A3 O (0,0) A5 A0 A2 A1 X Page | 79 MATHEMATICS DPP DPP No. 3 DAILY PRACTICE PAPER COMPLEX NUMBER A0 A1 = | z1 z 0 |=| z 0 e i z o | = | z 0 || cos + i sin 1| = = A0 A1 = 5 (cos 1) 2 + sin 2 5 2 (1 cos ) = 5 2 sin( / 2) 2 5 . 2 sin = 5 = = ....(i) 6 6 3 Similarly, A1 A2 = A2 A3 = A3 A4 = A4 A5 = A5 A0 = 5 . Hence the perimeter of, regular polygon is = Ao A1 + A1 A2 + A2 A3 + A3 A4 + A4 A5 + A5 A0 = 6 5 . 26. (c) It is a fundamental concept. 27. (a) Equation of circle | z z 0 |2 = r 2 (z z 0 )(z z 0 ) = r 2 (z z 0 )(z z 0 ) = r 2 zz zz 0 zz 0 + z 0 z 0 = r 2 . 28. (d) 1 3 2 2 z 2 = z1 e 2i / 3 = + i cos + i sin 2 2 3 3 1 3 = + i 2 2 1 3 1 3 + i = = 1 . 2 4 4 2 29. (b) The two circles are C1 (0, 0), r1 = 12 , C2(3, 4), r2 = 5 C1C2 = 5 r1 r2 = 7 Hence circle A C2 lies inside circle and it passes through origin, the centre of C1 . B 5 5 C 2 (3,4) C1 C1 . Therefore minimum distance between them is (0, 0) AB = C1 B C1 A = r1 2r2 = 12 10 = 2. 30. (b) | PQ |=| QR |=| RS |=| SP | and PQR = 90 o . 31. (a) It is a fundamental concept. 32. (b) | z 2 | + | z + 2 | = 8 (x 2) 2 + y 2 + (x + 2) 2 + y 2 = 8 x 2 + y 2 + 4 4 x = 64 + x 2 + y 2 + 4 + 4 x 16 (x + 2) 2 + y 2 8 x 64 = 16 (x + 2) 2 + y 2 (x + 8) = 2 (x + 2) 2 + y 2 x 2 + 64 + 16 x = 4[ x 2 + y 2 + 4 + 4 x] 3 x 2 + 4 y 2 48 = 0 x2 y2 + = 1, 16 12 which is an ellipse. 33. (b) Let z1 = 1 + 3i, z 2 = 5 + i and Then area of triangle z 3 = 3 + 2i Page | 80 COMPLEX NUMBER MATHEMATICS DPP DPP No. 3 DAILY PRACTICE PAPER x1 1 x2 2 x3 A= Hence y1 1 1 3 1 1 y2 1 = 5 1 1 = 0 2 y3 1 3 2 1 z1 , z 2 and z3 are collinear. 34. (a) | z1 + z2 | | z1 | + | z2 | 35. (b) Let z = x + iy ; z + iz = (x y) + i(x + y) and iz = y + ix If A denotes the area of the triangle formed by Applying transformation R2 R2 R1 R3 , z, z + iz and iz , then A= x y 1 1 x y x+y 1 2 y x 1 we get x y 1 1 1 1 0 0 1 = (x 2 + y 2 ) = | z |2 2 2 2 y x 0 A= 36. (a) Given that points are and C( 1 + 16i) Area of triangle x1 1 x2 2 x3 = A(3 + 4i), B(5 2i) y1 1 3 4 1 1 y2 1 = 5 2 1 =0 2 y3 1 1 16 1 Hence points are collinear. 37. (d) It is a fundamental concept. 38. (d) The affix of G is is 0. AG 39. (c) z1 + z 2 + z 3 3 z = 0 is the mid point of AG . Therefore affix of the mid-point of z1 + z 2 + z 3 + z1 3 = 0 4 z1 + z 2 + z 3 = 0 1+1 We have z1 + z 2 = 1 z12 + z 22 = z1 z 2 z 2 z1 z12 + z 22 + z 32 = z1 z 2 + z1 z 3 + z 2 z 3 , where z 3 = 0 z1 , z 2 and the origin 40. (a) Area of the triangle 41. (d) . Since ( z 3 = 0) form an equilateral triangle. 1 |z |2 = 18 | z |= 6 . 2 z1 = 1 + i z1 = (1,1) z 2 = 2 + 3i z 2 = ( 2, 3) z3 = ai z 3 = (0, a / 3) 3 z1 , z 2 1 2 0 and 1 3 a/3 z3 are collinear 1 1 =0 1 a (1 + 2) + 1(3 + 2) = 0 3 a + 5 = 0 a = 5 . 42. (c) It is based on the properties of moduli. 43. (b) Vertices are 0 = 0 + i0, z = x + iy and ze i = (x + iy) (cos + i sin ) = (x cos y sin ) + i(y cos + x sin ) Page | 81 MATHEMATICS DPP DPP No. 3 DAILY PRACTICE PAPER Area = 44. (d) = 1 2 0 x (x cos y sin ) 0 y (y cos + x sin ) COMPLEX NUMBER 1 1 1 1 [ xy cos + x 2 sin xy cos + y 2 sin ] 2 1 1 = sin (x 2 + y 2 ) = | z |2 sin [ | z |= x 2 + y 2 ] . 2 2 z1 = 1 + 2i, z 2 = 2 + 3i, z 3 = 3 + 4i | z 1 z 2 | = | 1 i | = 2 and | z 2 z 3 | = 1 i = 2 . and | z1 z 3 | = | 2 2i | = 2 2 . Hence vertices are collinear. 45. (a) z 5i x + i(y 5) =1 =1 z + 5i x + i(y + 5) | x + i(y 5) |=| x + i(y + 5) | , z1 | z1 | = z 2 | z 2 | x 2 + 25 10y + y 2 = y 2 + x 2 + 25 + 10y 20y = 0 y = 0. z+i z + 2 46. (a) Given that Im Let = x + iy + i x + iy + 2 z = x + iy = x + i (y + 1) (x + 2) + iy [ x + i(y + 1)][( x + 2) iy] [( x + 2) + iy][( x + 2) iy] x 2 + 2x + y 2 + y (y + 1)( x + 2) xy = +i 2 2 (x + 2) 2 + y 2 ( x + 2) + y If it is purely imaginary then real part must be equal to zero. x 2 + y 2 + 2x + y = 0 x 2 + y 2 + 2x + y = 0 (x + 2)2 + y 2 Which is a circle and its radius is given by g2 + f 2 c = 1+ 1 5 0 = 4 2 Therefore Argand diagram is circle of radius 47. (b) | z + 1|= Putting 5 2 . 2 | z 1| z = x + iy |(x + 1) + iy |= | x + iy + 1|= 2 | x + iy 1| 2 | ( x 1) + iy | (x + 1)2 + y 2 = 2[(x 1)2 + y 2 ] x 2 + y2 6x + 1 = 0 . Which is the equation of a circle. Page | 82 MATHEMATICS DPP DPP No. 3 DAILY PRACTICE PAPER COMPLEX NUMBER z a =1 z+a 48. (b) We have | z a |=| z + a | | z a |2 =| z + a |2 (z a)(z a) = (z + a)(z + a) (z a)(z a) = (z + a)(z + a) z z za a z + aa = z z + za + a z + aa za + z a + a z + a z = 0 (a + a)(z + z) = 0 z + z = 0 ( a + a = 2 Re(a) 0) 2 Re(z) = 0 2 x = 0 x = 0 Which is the equation of y-axis. 49. (c) We have | z 1| + | z + 1| 4 | z 1|2 + | z + 1|2 +2 | z 1|| z + 1| 16 (z 1)(z 1) + (z + 1)(z + 1) + 2 |(z 1)(z + 1)| 16 | z |2 + | z 2 1| 7 2 | z |2 +2 + 2 | z 2 1| 16 | x + iy |2 + |(x + iy)2 1| 7 x 2 4 Therefore the points are on the boundary or in the interior of the ellipse. z 1 x + iy 1 (x 2 + y 2 1) + 2iy = = z + 1 x + iy + 1 (x + 1)2 + y 2 50. (b) Wehave Therefore Hence z y2 1 (ellipse) 3 + arg tan 1 z 1 2y = tan 1 2 z +1 x + y2 1 2y = x 2 + y2 1 3 2y = tan = 3 2 3 x + y 1 x 2 + y2 1 = 2 2 3 y x 2 + y2 2 3 y 1 = 0 Which is obviously a circle. 51. (b) We have = 2z + 1 2(x + iy) + 1 (2 x + 1) + 2iy = = iz + 1 i(x + iy) + 1 (1 y) + ix [(2 x + 1)(1 y) + 2 xy] + i[2y(1 y) x(2 x + 1)] (1 y)2 + x 2 But it is given that imaginary part of (2z + 1) (iz + 1) is 2 x + 2y 2 = 0 . Which is a straight line. 52. (a) Let z = x + iy .Then x = + 3 and y = 5 2 (x 3)2 = 2 ......(i) and y 2 = 5 2 ......(ii) From (i) and (ii), (x 3)2 = 5 y 2 (x 3)2 + y 2 = 5 . Obviously it is a circle. 53. (c) | z 3i |= 2, let z = x + iy | x + i(y 3)|= 2 Squaring both sides, we get [ x 2 + (y 3)2 ] = 4 Page | 83 MATHEMATICS DPP DPP No. 3 DAILY PRACTICE PAPER COMPLEX NUMBER x 2 + y 2 6y + 5 = 0 54. (c) Here | z zi|= 1 | x + iy i(x + iy)|= 1 |(x + y) + i(y x)|= 1 55. (c) 2(x 2 + y 2 ) = 1 . (x + y) 2 + (y x) 2 = 1 Hence z lies on a circle. z 1 = 1 | z 1|=| z i | z i |(x 1) + iy |=| x + i(y 1)| (x 1)2 + y 2 = x 2 + (y 1)2 2x = 2y or x y = 0 Which is the equation of a straight line. 56. (b) z = (x + iy) z = x y + 2ixy Re(z 2 ) = 1 x 2 y 2 = 1 , which is a hyperbola. 57. (c) | z 1|=| z + i | | x 1 + iy |2 =| x + i(y + 1)|2 2 2 (x 1)2 + y 2 = x 2 + (y + 1)2 x+y=0 58. (b) We have i.e., a straight line through the origin. | z |2 | z | +1 2 2+ | z | log 3 | z |2 | z | +1 ( 3 )2 2+ | z | 1 | z | 5 Locus of 59. (d) 2 | z |2 4 | z | 5 0 | z | 5 as | z | 0 is |z|<5 z arg (x a) + iy = 4 y = x a 4 tan 1 y = tan = 1 x a = y x a 4 60. (a) | z 2 + i |=| z 3 i | |(x 2) + i(y + 1)| =|(x 3) + i (y 1)| (x 2)2 + (y + 1)2 = (x 3)2 + (y 1)2 x 2 + 4 4 x + y 2 + 1 + 2y = x 2 + 9 6 x + y 2 + 1 2y 2x + 4 y 5 = 0 . 61. (a) | iz 1| + | z i |= 2 | i(x + iy) 1|+ | x + iy i |= 2 | (y + 1) + ix |+ | x + i(y 1)|= 2 ( (y + 1))2 + x 2 + x 2 + (y 1)2 = 2 (y + 1)2 + x 2 = 2 x 2 + (y 1)2 y 2 + 1 + 2y + x 2 = 4 + x 2 + y 2 + 1 2y 4 x 2 + (y 1)2 4 y = 4 4 x 2 + (y 1) 2 y = 1 x 2 + (y 1)2 x 2 + (y 1)2 = (1 y)2 x 2 + y 2 + 1 2y = 1 + y 2 2y x 2 = 0 x = 0 Page | 84 COMPLEX NUMBER MATHEMATICS DPP DPP No. 3 DAILY PRACTICE PAPER i.e. equation of straight line. 62. (d) z+ i 2 2 = z 2 i 2 1 x + i y + 2 2 x + iy + = x + iy 1 = x + i y 2 2 1 1 x2 + y + = x2 + y 2 2 63. (a) 2 i 2 2 i 2 2 2 2y = 0 i.e. x-axis. (x 1) + iy z 1 = k arg = k arg z +1 (x + 1) + iy arg (x 1) + iy arg (x + 1) + iy = k y 1 y tan 1 tan =k x 1 x +1 y y ( x 1 ) ( x + 1) =k tan 1 y2 1 + x2 1 tan k = y( x + 1) y( x 1) 2y = 2 2 2 x + y 1 x + y2 1 2y 2y = x 2 + y2 1 x 2 + y2 1 = 0 tan k tan k It is an equation of circle whose centre is 64. (d) ( g, f ) = (0, cot k) on y-axis. z 2 3i = x + iy 2 3i = (x 2) + i(y 3) y 3 y 3 tan 1 = tan = 1 = x 2 4 x 2 4 x y + 1 = 0. 65. (b) We have | z 2 1| = | z |2 +1 | (x + iy) 2 1 | = | x + iy |2 +1 | (x 2 y 2 1) + 2 xyi | = 2 x 2 + y 2 + 1 (x 2 y 2 1) 2 + (2xy) 2 = x 2 + y 2 + 1 x 4 + y 4 + 1 2x 2y 2 + 2y 2 2x 2 + 4 x 2y 2 = x 4 + y 4 + 1 + 2 x 2 y 2 + 2y 2 + 2 x 2 2x 2y 2 = 2x 2y 2 + 4 x 2 x = 0 then, z = x + iy = 0 + iy = iy Hence z lies on imaginary axis. 66. (b) w = 1 iz , then | w | = 1 z i 1 iz =1 z i |1 iz | = | z i | |1 i(x + iy)| = | x + iy i | | (1 + y) ix | = | x + i(y 1)| x 2 + 1 + y 2 + 2y = x 2 + y 2 + 1 2y y = 0 Hence z = x + iy = x . So z lies on real axis. Page | 85 COMPLEX NUMBER MATHEMATICS DPP DPP No. 3 DAILY PRACTICE PAPER 67. (a) | z 5i | x 2 + (y 5) 2 = 12 2 = 12 | z + 5i | x + (y + 5) 2 x 2 + y 2 + 25 10y = 12[ x 2 + y 2 + 25 + 10y] 11x 2 + 11y 2 + 130y + 275 = 0. Which represents the equation of a circle. 68. (b) Put z = x + iy in arg z 2 = z + 2 6 (x 2) + iy = arg (x + 2) + iy 6 arg((x 2) + iy) arg ((x + 2) + iy) = tan 1 y y tan 1 = x 2 x+2 6 6 . y y x 2 x + 2 = xy + 2y xy + 2y = tan = 1 tan 6 6 x 2 + y2 4 y2 3 1+ 2 x 4 1 = x 2 + y 2 4 3y 4 = 0 Which is equation of a circle. 69. (a) z i z 3 = 1 | z |= z i 3 Clearly locus of z is perpendicular bisector of line joining points having complex number and 0+ i 3 0 + i0 . Hence z lies on a straight line. 70. (b) Given | 8 + z | + | z 8 |= 16 . Clearly locus of z is an ellipse. 71. (c) Equation of ray PQ Equation of ray PR Shaded region is | arg(z + 1)| 4 ; arg(z + 1) = 4 arg(z + 1) = 4 arg(z + 1) 4 4 | PQ |= ( 2 ) 2 + ( 2 ) 2 = 2 |PA| =2 ; |PR| = 2 so, arc QAR is of a circle of radius 2 unit with centre at region are exterior to this circle All the points in the shaded | z + 1|= 2 . 72. (b) | z 1|=| z 2 |=| z i | (i) | z 1|=| z i | represents a straight line through origin i.e., (ii) | z 1|=| z 2 | x = P( 1,0) . y=x 3 2 Page | 86 MATHEMATICS DPP DPP No. 3 DAILY PRACTICE PAPER COMPLEX NUMBER which is a straight line (iii) | z 2 |=| z i | 4 x 2y = 3 which is a straight line | z 1|=| z 2 |=| z i | can represent a triangle. 73. (a) From | z 1|=| z 2 | | x + iy 1|=| x + iy 2 | (x 1) 2 + y 2 = (x 2) 2 + y 2 x = Similarly from | z 1|=| z i |; Hence, x= 3 2 x=y 3 3 ,y = 2 2 Therefore only one solution for the equation. 74. (b) | z z1 | + | z z 2 |= 2a when | z1 z 2 | 2a , then it is an ellipse z1 = 2 + 3i and z 2 = 2 + 6i z1 z 2 = (2 + 3i) ( 2 + 6i) = 4 3i | z1 z 2 |=| 4 3i | = But 4 2 + ( 3) 2 = 5 P(z) 75. (d) is false, because in any triangle sum of two sides is not smaller than third side. is not represent locus of any point. 5 4 z= 2 i 2 Here, 2 2 = tan 1 = tan 1 ( 1) = 135 o Now, rotate z in opposite direction with 45 angle = 180 0 = tan 1 (0) = tan 1 2 Hence x= 2 and y=0. Page | 87 MATHEMATICS DPP DPP No. 3 DAILY PRACTICE PAPER COMPLEX NUMBER De Moivre's theorem and Roots of unity 1. i = (i)1 / 2 = cos + i sin 2 2 (c) 1/ 2 = cos 2n + + i sin 2n + 2 2 1/ 2 (where n I ) 1 1 = cos 2n + + i sin 2n + 2 2 2 2 (Using De Moivre's theorem) = [cos 4n + + i sin 4n + ] 4 4 Putting n = 0 , 1 we get cos + i sin = 1 + i 1 = 1 + i 4 4 2 2 2 1+ i 5 5 1 1 + i sin = i = 4 4 2 2 2 Therefore i = 1 + i 2 and cos Trick : Check by squaring the options, here (c) is the square root of squaring 1 + i ,we 2 2. (c) x1, x2, x3 ..... upto = i because on get i . cos + i sin cos 2 + i sin 2 2 2 2 2 upto ..... = cos + 2 + ..... + i sin + 2 + ..... 2 2 2 2 = cos 2 + i sin 2 = cos + i sin = 1 . 1 1 1 1 2 2 3. (d) (cos + i sin )4 = (sin + i cos )5 = = (cos + i sin )4 5 i (cos i sin ) (cos+ i sin )4 1 i 5 sin + cos i = 5 (cos + i sin )4 (By property) i (cos + i sin ) 5 1 (cos + i sin )9 = sin 9 i cos 9 i . 5 4. (b) Given that 3 3 1 1 z= +i + i 2 2 2 2 5 5 = cos + i sin + cos i sin 6 6 6 6 = cos 5. 5 5 5 5 + i sin + cos i sin 6 6 6 6 5 . Hence Im (z) = 0. (a) Using De Moivre's theorem (cos + i sin )n = (cos n + i sin n ) and putting n = 0, 1, 2 then we get required roots. Page | 88 MATHEMATICS DPP DPP No. 3 DAILY PRACTICE PAPER 6. (d) COMPLEX NUMBER 4(cos 75 o + i sin 75 o ) 0.4(cos 30 o + i sin 30 o ) = 10(cos 75o + i sin 75o )(cos 30o i sin 30o ) = 10(cos 45 o + i sin 45 o ) = 7. (c) 10 2 (1 + i) sin i cos = i 2 sin i cos = i(cos + i sin ) Given expression is ( i)3 [cos( 10 18 + 3 ) + i sin( 25 )] = 8. i (cos 25 i sin 25 ] . (a) We have a = 2i = 2 i 1 / 2 = 2 cos + i sin 2 2 1/ 2 1 1 = 2 cos + i sin = 2 + i = 1 + i 4 4 2 2 9. Trick : Check with options. (c) We have (cos + i sin )(cos 2 + i sin 2 ) ...... (cos n + i sin n ) = 1 cos( + 2 + 3 + ... + n ) + i sin( + 2 + . + n ) = 1 n(n + 1) n(n + 1) cos + i sin =1 2 2 n (n + 1) n(n + 1) cos = 1 and sin = 0 2 2 n(n + 1) 4m = 2m = , where m I . 2 n(n + 1) 10. (b) L.H.S. 2 cos 2 ( / 2) + 2i sin( / 2) cos( / 2) = 2 2 cos ( / 2) 2i sin( / 2) cos( / 2) n n n e i( / 2) cos ( / 2) + i sin( / 2) i n = = i( / 2) = (e ) e cos( / 2) i sin( / 2) = cos n + i sin n 11. (d) D r = i(1 + cos ) + sin = 2i cos 2 L.H.S + 2 sin cos 2 2 2 cos( / 2) + i sin( / 2) = i cos( / 2) + sin( / 2) = 12. (a) . 4 1 (cos + i sin )4 = cos 4 + i sin 4 i4 cos + i sin cos + i sin 2 ..... 2 2 2 2 to . = cos + 2 + ..... + i sin + 2 + .... 2 2 2 2 = cos 1 1 1 1 1 + + 2 + ..... + i sin 1 + + 2 + ..... 2 2 2 2 2 2 Page | 89 MATHEMATICS DPP DPP No. 3 DAILY PRACTICE PAPER COMPLEX NUMBER 1 1 = cos + i sin = cos+ i sin = 1 2 1 1 2 13. (d) 2 1 1 2 4 cos + i sin (cos + i sin )4 cos 4 + i sin 4 sin + i cos = i 4 (cos i sin )4 = cos 4 i sin 4 = (cos 4 + i sin 4 )(cos 4 + i sin 4 ) (cos 4 i sin 4 )(cos 4 + i sin 4 ) (cos 4 + i sin 4 )2 = cos 8 + i sin 8 cos 2 4 + sin 2 4 14. (b) (i) ....(ii) cos + cos + cos = 0 sin + sin + sin = 0 Let a = cos + i sin , b = cos + i sin c = cos + i sin [by (i) and (ii)] .....(iii) a+b+c = 0 1 1 1 + + a b c = [cos + i sin ] 1 + [cos + i sin ] 1 + [cos + i sin ] 1 = cos i sin + cos i sin + cos i sin ab + bc + ca = 0 ....(iv) [by (i) and (ii)] Squaring both sides of equation (iii), we get a 2 + b 2 + c 2 + 2ab + 2bc + 2ca = 0 or a 2 + b 2 + c 2 = 0 [by (iv)] [cos + i sin ] 2 + [cos + i sin ] 2 + [cos + i sin ] 2 = 0 (cos 2 + cos 2 + cos 2 ) +i (sin 2 + sin 2 + sin 2 ) = 0 separation of real and imaginary part , .....(v) cos 2 + cos 2 + cos 2 = 0 and sin 2 + sin 2 + sin 2 = 0 .....(vi) 1 2 sin + 1 2 sin + 1 2 sin 2 = 0 [by 2 2 sin 2 + sin 2 + sin 2 = 3 2 eq. (v)] . 15. (c) See solution of above question. 16. (c) ( 3 + i) = 53 3 i =2 + 2 2 53 53 253 (cos150 o + i sin150 o )53 = 2 53 [cos(150 o 53) + i sin(150 o 53)] = 2 53 [cos(22 + 30 o ) + i sin(22 + 30 o )] 3 1 = 2 53 [cos 30 o + i sin 30 o ] = 2 53 +i . 2 2 Page | 90 MATHEMATICS DPP DPP No. 3 DAILY PRACTICE PAPER 17. (b) Let cos 10 i sin Therefore, = z and cos 10 1 z 1 1 z 10 + i sin 10 = 1 z 10 (z 1)z = (z 1) = z 10 = cos i sin 10 10 18. (a) COMPLEX NUMBER 10 10 = ( z)10 = cos i sin = 1 . (cos 2 i sin 2 )4 (cos 4 + i sin 4 ) 5 (cos 3 + i sin 3 ) 2 (cos 3 i sin 3 ) 9 = [(cos + i sin ) 2 ]4 [(cos + i sin )4 ] 5 [(cos + i sin )3 ] 2 [(cos + i sin ) 3 ] 9 = (cos + i sin ) 8 (cos + i sin ) 20 (cos + i sin ) 6 (cos + i sin )27 = (cos + i sin ) 8 20 +6 27 = (cos + i sin ) 49 = cos 49 i sin 49 . 19. (c) (sin + i cos )n = cos + i sin 2 2 n cos n + i sin n . 2 2 (cos + i sin ) (cos + i sin ) (cos + i sin ) (cos + i sin ) = 20. (b) = cos( + ) + i sin( + ) [By de-movire's theorem]. 21. (a) 1 + cos( / 8) + i sin( / 8) 1 + cos( / 8) i sin( / 8) 8 2 cos 2 ( / 16) + 2i sin( / 16) cos( / 16) = 2 2 cos ( / 16) 2i sin( / 16) cos( / 16) = 8 [cos( / 16) + i sin( / 16)]8 [cos( / 16) i sin( / 16)]8 8 = cos + i sin cos + i sin 16 16 16 16 8 = [cos( / 16) + i sin( / 16)]16 = cos 16 + i sin 16 16 16 22. (a) = cos = 1 . x1 . x 2 . x 3 ...... = cos + i sin 4 4 cos 2 + i sin 2 cos 3 + i sin 3 ..... 4 4 4 4 = cos + 2 + 3 + .... + i sin + 2 + 3 + ..... 4 4 4 4 4 4 = /4 /4 cos + i sin 1 1/ 4 1 1/ 4 Page | 91 COMPLEX NUMBER MATHEMATICS DPP DPP No. 3 DAILY PRACTICE PAPER = 23. (c) cos( / 3) + i sin ( / 3) = (cos + i sin ) 4 (sin + i cos ) = = = = = 5 1 + 3i 2 . cos 4 + i sin 4 5 i (cos i sin )5 i (cos 4 + i sin 4 ) (cos i sin ) 5 i [cos 4 + i sin 4 ] [cos 5 + i sin 5 ] i [cos(4 + 5 ) + i sin(4 + 5 )] sin(4 + 5 ) i cos(4 + 5 ) . i 1 / 3 = cos + i sin 2 2 24. (a) 25. (c) Let 1/ 3 = cos 6 + i sin 6 = 3 i + . 2 2 z = (1 + i 3 ) r = 3 + 1 = 2 and r cos = 1, r sin = 3 tan = 3 = tan 3 = . 3 z = 2 cos + i sin 3 3 z100 = 2 cos + i sin 3 3 100 100 = 2100 cos + i sin 3 3 100 1 i 3 = 2100 cos i sin = 2100 2 2 3 3 Re(z) 1 / 2 1 = = . Im (z) 3 / 2 3 26. (a) n 1 + sin + i cos 1 + cos + i sin = 1 + sin i cos 1 + cos i sin n ( where = 2 cos 2 = 2 2 cos n cos + i sin 2 2 2 = 2 2 2i sin cos cos i sin 2 2 2 2 2 + 2i sin cos 2 ) n n cis n 2 = cis + = cis(n ) = 2 2 cis 2 = n cis n = cis n 2 2 n n = cos n + i sin n . 2 2 27. (c) (1 + i)n + (1 i)n n n n n = (2)n / 2 cos + i sin + cos i sin 4 4 4 4 Page | 92 MATHEMATICS DPP DPP No. 3 DAILY PRACTICE PAPER n n = 2 2 . 2 cos 28. (a) . x+ 1 = 2 cos x x 2 2x cos + 1 = 0 x = cos i sin x n = cos n i sin n 1 1 = x cos i sin 1 = cos i sin x Thus, 29. (b) +1 n n n = 2 2 cos = ( 2 )n+ 2 cos 4 4 4 xn + 1 = 2 cos n xn 1 xn COMPLEX NUMBER = cos n i sin n . iz 4 = 1 z4 = 1 z 4 = i z = (i)1 / 4 i z = (0 + i)1 / 4 1/ 4 z = cos + i sin 2 2 z = cos 8 + i sin (using De Moivre s theorem) 8 30. (d) Since ( )2 = 2 and ( 2 )2 = 4 = 3 = 31. (d) If is a complex cube root of unity then 3 = 1 and 1+ + 2 = 0 , therefore (1 + 2 )(1 + 2 ) = ( 2 2 )( 2 ) = 4 32. (c) Let x = 271 / 3 = x 3 27 = 0 (x 3)(x 2 + 3 x + 9) = 0 1 i 3 . x = 3, x = 3 2 Trick : As we know 33. (c) Let n = 3k + 1 Hence, roots are (27)1 / 3 3, 3 ,3 2 . must have 3 roots, so (a) option cannot be the best. Here (c) satisfies. n + 2n = 3k +1 + 2(3k +1) = 3k + 6k 2 = ( 3 )k . + ( 3 )2k . 2 = + 2 = 1 Hence 1 + n + 2n = 1 1 = 0 34. (b) Since imaginary cube root of unity are square of each other. 35. (a) (1 + )3 (1 + 2 )3 = ( 2 )3 ( )3 = 6 + 3 = 3 3 + 3 = 1 + 1 = 0 36. (b) Complex cube root of unity are Let = , = 2 ; Then 1, , 2 4 + 4 + 1 1 = 4 + ( 2 )4 + ( 1 ) ( 2 ) 1 = + 2 + 1 = 0 . 37. (d) (1 )(1 2 )(1 4 )(1 8 ) = (1 )(1 2 )(1 )(1 2 ) = (1 )2 (1 2 )2 = 38. (b) ( 3 )( 3 2 ) = 9 3 = 9 . (1 + 2 )5 + (1 + 2 )5 = ( 2 )5 + ( 2 2 )5 = 32 3 2 32( 3 )3 Page | 93 MATHEMATICS DPP DPP No. 3 DAILY PRACTICE PAPER COMPLEX NUMBER = 32( 2 + ) = 32( 1) = 32 39. (c) Given that Then 40. (c) x = a, y = b , z = c 2 x y z a b c 2 + + = + + = 1+ + 2 = 0 a b c a b c (x y)(x y)(x 2 y) = (x 2 xy xy + y 2 )(x 2 y) = x 3 x 2 y(1 + + 2 ) + xy 2 (1 + + 2 ) y 3 = x 3 y3 41. (b) ( 1 + + 2 = 0) upto 2n factors = ( )( )(1 + )(1 + )..... upto 2n factors = 1.1.1..... upto n factors = 1 (1 + )(1 + 2 )(1 + 4 )(1 + 8 )...... 2 2 42. (b) Given that cos 3 + i sin 3 Since the 3/4 = [cos + i sin ]1 / 4 . expression has only 4 different roots, therefore on putting 2n + 2n + cos + i sin 4 4 n = 0, 1, 2, 3 in and multiplying them, 3 3 we get = cos + i sin cos + i sin 4 4 4 4 5 5 7 7 cos 4 + i sin 4 cos 4 + i sin 4 1 1 1 1 1 1 1 1 = +i +i +i i 2 2 2 2 2 2 2 2 1 1 1 1 = = ( 1)( 1) = 1 . 2 2 2 2 43. (d) Given that x +1 2 x + 2 1 2 1 =0 x + Applying transformation 1 x 2 1 x + 2 1 R1 R1 + R2 + R3 , we get 1 1 =0 x + (x + 2 )(x + ) 1 + 2 (x + ) + 2 (x + 2 ) = 0 x2 = 0 x = 0 Trick : Putting x = 0, we get 1 2 2 2 1 = 0 1 44. (b) If x = a + b, y = a + b and z = + b Then xyz = (a + b)(a + b 2 )(a 2 + b ), where = and = 2 = (a + b)(a 2 + ab 2 + ab + b 2 ) = (a + b)(a 2 ab + b 2 ) = a 3 + b 3 Trick : Put a=b=2 Page | 94 COMPLEX NUMBER MATHEMATICS DPP DPP No. 3 DAILY PRACTICE PAPER then x = 4, y = 2( + 2 ) = 2 and z = 2( 2 + ) = 2 xyz = 4( 2)( 2) = 16 and (b) i.e. a 3 + b 3 = 16 . 45. (b) We have x 3 + y 3 + z 3 = (a + b)3 + (a + b 2 )3 + (a 2 + b )3 = 3a 3 + 3b 3 + 3(a 2b + ab 2 )(1 + 2 2 + 4 ) = 3a 3 + 3b 3 + 3(a 2 b + ab 2 )(1 + + 2 ) = 3(a 3 + b 3 ) Trick : As in the previous question x 3 + y 3 + z 3 = (4)3 + ( 2)3 + ( 2)3 = 48 and (b) i.e. 3(a 3 + b 3 ) = 48 46. (b) Multiplying the numerator and denominator by = = a + b + c 2 b + c + a 2 + a + b + c and 2 respectively I and II expressions 2 c + a + b 2 (a + b + c 2 ) 2 (a + b + c 2 ) + = + 2 = 1 (b + c 2 + a) (c 2 + a + a ) 47. (a) The cube roots of unity are 1, , 2 . We know that if z1 , z 2 , z 3 are the vertices of an equilateral triangle in the Argand plane. Then z12 + z 22 + z 32 = z1 z 2 + z 2 z 3 + z 3 z1 If we take Then and z1 = 1, z 2 = , z 3 = 2 z 12 + z 22 + z 32 = 1 + 2 + 4 = 0 z1 z 2 + z 2 z 3 + z 3 z1 = 1. + . 2 + 2 .1 = + 3 + 2 = 1+ + 2 = 0 Thus z12 + z 22 + z 32 = z1 z 2 + z 2 z 3 + z 3 z1 Therefore triangle is equilateral. Note : Students should remember this question as a fact. 48. (c) Here Then 1 1 + i 3 2 2 is one of the two imaginary cube root of unity. If we denote it by 1000 = 999 = ( 3 )333 = = . 1 3 + i. 2 2 49. (a) Since p 0 . Let p = q , where q is positive. Therefore p1 / 3 = q1 / 3 (1)1 / 3 . Hence = q1 / 3 , = q1 / 3 and = q1 / 3 2 The given expression x + y + z 2 x + y 2 + z 50. (a) Given that iz = Now = = 1 1 z + y 2 + z = = 2. . x + y 2 + z z= 3 +i 3 1 = + i 2 2 2 1 3 +i = 2 2 z 69 = z 4(17) z = (iz) 4(17) z = ( )68 z, ( i 4 n = 1) 69 i = ( 3 )23 1 = = i i i Page | 95 MATHEMATICS DPP DPP No. 3 DAILY PRACTICE PAPER Aliter : z= COMPLEX NUMBER 3 1 + i = cos + i sin 2 2 6 6 z 69 = cos + i sin 6 6 69 = cos 69 69 + i sin 6 6 = cos 11 + + i sin 11 + = 0 + i( 1) = i . 2 2 51. (b) Given equation x 4 1 = 0 (x 2 1)(x 2 + 1) = 0 x 2 = 1 and x 2 = 1 x = 1, i 52. (d) The product is given by . 2 . 3 ..... n = 1+ 2 + 3 + ......+ n = n(n +1) / 2 On putting = 1(1+1) / 2 n = 1,2,3,....., = , 2( 2+1) / 2 we get 3 = = 1,..... 4(5) / 2 = 10 = Hence it gives the values 1 and only. 53. (a) It is obvious because the cube roots of unity are 54. (c) 55. (a) 1, 1+ i 3 1 i 3 , 2 2 (1 + )7 = A + B ( 2 )7 = A + B 14 = A B 2 . 12 = A B A + B + 2 = 0 ( 1 + + 2 = 0) A = 1, B = 1 1 1+ i + 2 2 = 1 i 1 2 1 i i + 1 1 1 i 2 = 1 i 1 2 1 = i i + 1 1 0 ( Two rows are identical) 56. (b) x n = 1 = (cos 0 + i sin 0 o ) = cos 2r + i sin 2r = e i 2r x = e i(2r / n) , r = 0, 1, 2 .. Obviously the roots are 1, e 2 i / n , e 4 i / n ...... which are obviously in G.P. with common ratio 57. (a) 58. (b) e 2 i / n . (3 + 2 + 4 )6 = (3 + 2 + )6 = (3 1)6 = 64 4 = , 8 = 2 etc. 2nd. L.H.S 3rd, 5th, 7th factors are each equal to 1st and 4th, 6th, 8th factors are each equal to to 2n factors = (22 3 )(22 3 )...... to n factors = (22 )n = 2 2n = ( 2 )( 2 2 )( 2 )( 2 2 )..... 59. (a) We have 1 = 2 2 2 3 3 3 2 2 1 3 4 = 2 2 2 2 6 4 3 3 3 2 2 = 0 3 3 Page | 96 COMPLEX NUMBER MATHEMATICS DPP DPP No. 3 DAILY PRACTICE PAPER 60. (a) We have 61. (a) n z z n = (1 + z)n =1 z +1 z z is = 11 / n z +1 z +1 |z| z = 1 | z |=| z + 1| =1 | z + 1| z +1 x+ 1 1 = 0 x = 2 2 a n th root of unity Re(z) 0 . n 1 = 0 , ( n = 1) 1 + + 2 + ..... + n 1 = 1 62. (d) The zk = n th roots i 2 (k 1) e n , | zk |= e of unity are given by (k = 1,2...., n) i 2 (k 1) n = 1 for | zk |=| zk +1 | for all 63. (a) Trick : Put all k = 1,2,....., n k = 1,2....., n a = 1, b = 1, c = 2 , a + b + c = 0 2 3 (1 + 2 ) + (1 + 2 2 )3 = ( 3 2 )3 + ( 3 )3 = 27 27 = 54 Also option (a) gives the value 64. (c) x 12 6 6 6 2 1 = (x + 1)(x 1) = (x + 1)(x 1)(x + x 2 + 1) Common roots are given by 65. (a) i.e., 27 1 1( 2) = 54 4 x4 + x2 + 1 = 0 1 i 3 = , 2 or 4 , 2 2 x2 = or x = 2 , ( 3 = 1) z1 = 1, z 2 = i, z 3 = 1, z 4 = i 13 + (i)3 + ( 1)3 + ( i)3 = 0 . 66. (b) Since is an imaginary cube root of unity, let it be = + 1 + 5 , { 3n = 1 and 3 = 1} = + 1 + 2 = 0 67. (d) As then ( )3n +1 + ( )3n + 3 + 3n + 5 1+ i 3 1 i 3 = and = 2 2 2 ( )20 + ( 2 )20 = 18 . 2 + 39 . = 2 + = 1 68. (c) Since Hence, and are complex roots of unity, we may write 4 + 28 + 1 = 4 + ( 2 )28 + = , = 2 1 . 2 = + 56 + 1 = + 2 + 1 = 0 69. (c) (3 + 5 + 3 2 )2 + (3 + 3 + 5 2 )2 = (3 + 3 + 3 2 + 2 )2 + (3 + 3 + 3 2 + 2 2 )2 (1 + + 2 = 0, 3 = 1) = (2 )2 + (2 2 )2 = 4 2 + 4 4 = 4( 1) = 4 . Page | 97 COMPLEX NUMBER MATHEMATICS DPP DPP No. 3 DAILY PRACTICE PAPER 70. (c) Given, sin ( 10 + 23 ) = sin ( + 2 ) 4 4 1 = sin = sin + = sin = 4 4 4 2 . 6 6 6 3 + i i 3 1 + 3i 1 3i + = + 2 2 2i 2i 71. (a) = 6 1 [( )6 + ( 2 )6 ] = [( 3 )2 + ( 3 )4 ] i6 = 1 + 3i , 2 = 1 3i 2 2 = (1 + 1) = 2 . 72. (d) 73. (a) (1 + w 2 )7 = (1 + + 2 2 2 )7 = ( 2 2 )7 = 128 14 = 128 12 2 = 128 2 15 15 1 i 3 1 + i 3 2 2 2 2 215 + 20 (1 i) (1 + i) 20 = 15 30 215 + = 215 1 20 + 1 20 20 20 (1 + i) (1 + i) (1 i) (1 i) (1 + i)20 + (1 i)20 215 [(1 + i)20 + (1 i)20 ] 215 = 2 20 (1 i 2 )20 = 15 [(i i 2 )20 + (1 i)20 ] = 15 (i 20 + 1)(1 i)20 2 2 1 1 2 20 20 = 5 (1 i) = 4 (1 i) = 4 [(1 i)2 ]10 2 2 2 1 1 10 2 10 = 4 [1 + i 2i] = 4 ( 2i) 2 2 ( 2)10 i 10 = 2 6 = 64 . = 24 = 74. (a) + 1 3 9 27 + ..... + + + 2 8 32 128 + 2 = 1 75. (c) 1/ 2 + 1 3 / 4 [ 1 + + 2 = 0] (3 + + 3 2 ) 4 = ( 3 + )4 [ + 2 = 1] = ( 2 )4 = 16 4 = 16 . 76. (c) 77. (d) (1 + )(1 + ) = ( 2 )( 2 ) = 128 . 78. (c) 1 + 3i 3 +i iz = = 2 2 = i z 69 = i 69 . 69 = i ( 3n = i 4 n = 1) z= i 2 2 n = cos + i sin n n 79. (c) 2 2 6 2 6 1. . 2 = 3 = 1. z= 3 = cos 2 2 1 i 3 + i sin = + = 3 3 2 2 Page | 98 MATHEMATICS DPP DPP No. 3 DAILY PRACTICE PAPER and = COMPLEX NUMBER 2 2 2 4 4 32 = cos + i sin + i sin = cos 3 3 3 3 1 i 3 = 2. 2 2 (x + y 3 + z 32 )(x + y 32 + z 3 ) = (x + y + z 2 )(x + y 2 + z ) = x 2 + y 2 + z 2 xy yz zx . 80. (d) For 2 z = , z 100 + z 100 = ( )100 + ( ) 100 = + 1 = + 2 = 1 For z = 2, = 200 + 81. (c) or z + z 1 = 1 z 2 z + 1 = 0 z = 1 z 100 + z 100 = ( 2 )100 + ( 2 ) 100 = 2 + 200 x4 x3 + x 1 = 0 x 1 = 0 or 1 2 = 2 + = 1. x 3 (x 1) + 1(x 1) = 0 x3 +1 = 0 x = 1, 1, 1 + 3i 1 3i , 2 2 so its real roots are 1 and 1. 82. (c) 1 + i 3 1 + i 3 2 + i2 3 = = 4 1 i 3 1 i 3 1 + i 3 1+ i 3 = 1+ i 3 = 2 n 83. (a) 1+ i 3 = n = 3 =1 n = 3 . 1 i 3 [(1 + + 2 ) + ]3n [(1 + + 2 ) + 2 ]3n = 3n ( 2 )3n = ( 3 )n ( 3 ) 2n = 1n 12n = 0 . 84. (a) (a + b) (a + b ) (a + b 2 ) = (a + b) (a 2 + ab ( + 2 ) + b 2 3 ) = (a + b) (a 2 ab + b 2 ) = a 3 + b 3 . 85. (b) 1 3 +i = cos + i sin 2 2 3 3 1/ 4 Now 1 +i 3 2 2 1/ 4 = cos + i sin 3 3 = cos + i sin = cis . 12 12 12 86. (d) (8)1 / 3 = x x 3 8 = 0 (x 2) (x 2 + 2 x + 4) = 0 . x = 2, 2 , 2 2 or x = 2, 1 + i 3 , 1 i 3 . Page | 99 MATHEMATICS DPP DPP No. 3 DAILY PRACTICE PAPER 87. (d) COMPLEX NUMBER 225 + (3 + 8 2 ) 2 + (3 2 + 8 ) 2 = 225 + (5 2 3) 2 + (5 3) 2 = 225 + 18 5( + 2 ) = 225 + 18 5( 1) = 225 + 18 + 5 = 248. 88. (a) = ( 3n 1) + n ( 2n 2n ) + 2n ( n 4 n ) = (1 1) + 0 + 2n [ n ( 3 )n n ] =0 + 0 + 0 =0. 89. (c) (3 + + 3 2 ) 4 = [(3 + 3 2 + )4 ] = [ 3 (1 + 2 ) + ]4 = [3( ) + ]4 = [ 2 ]4 = 16 4 = 16 . 90. (a) (1 2 + 2 )6 = = (1 + 2 2 )6 ( 2 )6 = ( 3 )6 = ( 3)6 ( 3 )2 [Since 1 + + 2 = 0, 3 = 1 ] = 729. 91. (d) 99 + 1 + 101 = 99 [1 + + 2 ] [Since 1 + + 2 = 0, 3 = 1 ] = 0. 92. (a) sin 1 (e i ) = x + iy sin(x + iy) = e i sin x cosh y + i cos x sinh y = cos + i sin On comparison, sinh y = sin cos x cosh y = cos sin x , cosh2 y sinh2 y = 1 1 sin 2 sin 2 =1 2 1 cos x cos 2 x cos 2 x sin2 cos 2 x sin2 + cos 2 x sin2 = cos 4 x = sin 2 x = cos 1 sin 93. (b) sinh ix = i sin x . 94. (c) cos (u + iv) = + i cos 2 x cos 4 x . cos u cos(iv) sin u sin(iv) = + i = cos ucosh v and = sin u sinhv ( cos(ix) = cosh x, sin(ix) = i sinh x) Page | 100 MATHEMATICS DPP DPP No. 3 DAILY PRACTICE PAPER 2 + 2 + 1 = COMPLEX NUMBER cos 2 u cosh2 v + sin 2 u sin h2v + 1 = cos 2 u cosh2 v + sin2 u sinh2 v + cos 2 u + sin2 u = cos 2 u cos h2v + sin 2 u (1 + sin h2v) + cos 2 u = cos 2 u cos h 2v + sin 2 u cos h 2v + cos 2 u [ cosh 2 v sinh 2 v = 1] = cosh 2 v (cos 2 u + sin 2 u) + cos 2 u = cos 2 u + cosh 2 v [ cos 2 u + sin 2 u = 1] . 95. (a) sec h (i ) = sec (i 2 ) = sec( ) = sec = 1 . 96. (c) cosh( + i ) cosh( i ) = cosh cosh(i ) + sinh sinh(i ) cosh cosh(i ) + sinh sinh(i ) = 2 sinh sinh i = 2i sinh sin 97. (b) . cosh( + i ) = cosh cosh (i ) + sinh sinh(i ) Imaginary part = sinh sin . 98. (a) It is obvious. 99. (c) It is obvious. 100. (b) tan(u + iv) = i tan u + tan(iv) =i 1 tan u tan(iv) tan u + i tan hv = i[1 i tan u tan hv] [ tan ix = i tan hx] tan u(1 tan hv) = i(1 tan hv) (tan u i)(1 tanh v) = 0 1 tan hv = 0 tan hv = 1 e v e v e v + e v = 1 e v e v = e v + e v 2e v = 0 e v = e v = . 101. (a) tan 1 ( + i ) = x + iy + i = tan(x + iy) ..(i) Taking conjugate, ( i ) = tan(x iy) ..(ii) tan 2x = tan[(x + iy) + (x iy)] Page | 101 COMPLEX NUMBER MATHEMATICS DPP DPP No. 3 DAILY PRACTICE PAPER tan 2 x = x = 102. (c) Let ( + i ) + ( i ) 2 = 1 ( + i ) ( i ) 1 ( 2 + 2 ) 1 2 tan 1 1 2 2 2 . y =| a + b + c 2 | for y to be minimum must be minimum. y2 y 2 =| a + b + c 2 |2 y 2 = (a + b + c 2 ) (a + b + c 2 ) 1 [(a b)2 + (b c)2 + (c a)2 ] 2 = Since a, b and c are not equal at a time so minimum value of y2 occurs when any two are same and third is differ by 1. Minimum of y =1 (as a, b, c are integers) 103. (d) 2 (1 + )3 (1 + 2 ) = 2 ( 2 )3 ( ) = 104. (c) 2 + 2 = 0. x = + , y = + 2 , z = 2 + xyz = ( + ) ( + 2 )( 2 + ) = ( + )[ 2 + ( + 2 ) + 2 ] = ( + ) ( 2 + 2 ) = 3 + 3 . Critical Thinking Questions 1. (c) Given equation a 2 2a sin x + 1 = 0 2 sin x 4 sin 2 x 4 2 a= = sin x (1 sin 2 x) a = sin x i cos x If 2. x= (d) Using 2 a = 1, x = 270o a = 1 . i 3 = i, i 5 = i and i7 = i , we can write the given expression as (1 + i)n1 + (1 i)n1 + (1 + i)n2 + (1 i)n2 = 2[n1 C0 + n1 C2i 2 + n1 C4 i 4 + .....] + 2[n2 C0 + n2 C2i 2 + n2 C4 i 4 + .....] = 2[n1 C0 n1 C2 + n1 C4 + ....] + 2[n 2 C0 n 2 C2 + n 2 C4 + ....] This is a real number irrespective of the values of n1 and n2 . Page | 102 COMPLEX NUMBER MATHEMATICS DPP DPP No. 3 DAILY PRACTICE PAPER 3. (d) Given that Let (where z = is real) be a root of (i), then 2 + ( p + iq) + r + is or ......(i) z 2 + ( p + iq)z + r + is = 0 =0 2 + p + r + i(q + s) =0 Equating real and imaginary parts, we have Eliminating or 4. (b) , we s 2 pqs + q 2r = 0 get or 2 + p + r = 0 and q + s = 0 2 s s + p + r = 0 q q pqs = s 2 + q 2r x + 5 = 4i x 2 + 10 x + 25 = 16 Now, x 4 + 9 x 3 + 35 x 2 x + 4 = (x 2 + 10 x + 41)(x 2 x + 4) 160 = 160 5. (b) We have 3 + i = (a + ib)(c + id) ac bd = 3 and ad + bc = 1 Now tan-1 b + tan 1 d a b d + a c = tan b d 1 . a c 1 = n + 6. (d) 6 c = tan 1 bc + ad = tan 1 1 ac bd 3 ,n I b cos + i sin cos i sin = c cos + i sin cos i sin b = cos( ) + i sin( ) c Similarly, and ......(i) c = cos( ) + i sin ( ) a a = cos( ) + i sin( ) b ......(ii) .....(iii) from (i) + (ii) + (iii) cos( ) + cos( ) + cos( ) +i[sin( ) + sin( ) + sin( )] = 1 Equating real and imaginary parts, cos( ) + cos( ) + cos( ) = 1 . 7. (b) We have (1 + i)(1 + 2i)(1 + 3i).....(1 + ni) = a + ib .....(i) .....(ii) (1 i)(1 2i)(1 3i).....(1 ni) = a ib Page | 103 MATHEMATICS DPP DPP No. 3 DAILY PRACTICE PAPER Multiplying (i) and (ii), we get 8. COMPLEX NUMBER 2.5.....( 1 + n2 ) = a 2 + b 2 (a) First note that | z |=| z | and | z1 + z 2 | | z1 | + | z 2 | Now | z | + | z 1|=| z | + |1 z | | z + (1 z)|=|1|= 1 Hence, minimum value of 9. (b) | (az1 bz2 ) |2 | z | + | z 1| is 1. + | (bz1 + az 2 ) |2 = a 2 | z1 |2 +b 2 | z 2 |2 2 Re(ab) | z1 z 2 | +b 2 | z1 |2 + a 2 | z 2 |2 +2 Re(ab)| z 1 z 2 | = (a 2 + b 2 )(| z1 |2 + | z 2 |2 ) log a m log a n m n or m n , 10. (a) We know that according as a 1 or 0 a 1 . Hence for z = x + iy log (1 / 3) | z + 1| log (1 / 3) | z 1| | z + 1| | z 1| 1 0 1 3 | x + iy + 1| | x + iy 1| (x + 1)2 + y 2 (x 1)2 + y 2 4 x 0 x 0 Re(z) 0 11. (d) Since | z1 |=| z 2 |= 1 , we have z1 = cos 1 + i sin 1 , z 2 = cos 2 + i sin 2 where Also, 1 = arg(z1 ) z1 = a + ib and and 2 = arg(z 2 ) z 2 = c + id. Therefore a = cos 1 , b = sin 1, c = cos 2 and d = sin 2 Also, R(z1 z 2 ) = 0 R[(cos 1 + i sin 1 )(cos 2 i sin 2 )] = 0 R[(cos( 1 2 ) + i sin( 1 2 )] = 0 cos( 1 2 ) = 0 1 2 = Now, 2 1 = 2 + 2 w1 = a + ic = cos 1 + i cos 2 = cos 1 + i sin 1 |w1 |= 1 Similarly, |w2 |= 1 Next w1 w 2 = (cos 1 + i sin 1 )(cos 2 i sin 2 ) = cos( 1 2 ) + i sin( 1 2 ) | w1 w 2 |= 1 Finally, R(w1w2 ) = R(w2 w1 ) Page | 104 MATHEMATICS DPP DPP No. 3 DAILY PRACTICE PAPER COMPLEX NUMBER = R[(cos 2 + i sin 2 )(cos 1 i sin 1 )] = R[cos( 2 1 ) + i sin( 2 1 )] = cos( 2 1 ) = cos =0 2 z = a + ib,| z | 1 a 2 + b 2 1 12. (c) Let and w = c + id,|w | 1 c 2 + d 2 1 | z + iw|=| a + ib + i(c + id)|= 2 ......(i) (a d)2 + (b + c)2 = 4 | z iw |=| a + ib i(c id) | ......(ii) (a d)2 + (b c)2 = 4 From (i) and (ii), we get Either If b=0, 13. (c) Let b = 0 or c = 0 then a2 1 . Then, only possibility is 1 1 =a z+ z z z+ 1 + 2 cos 2 = a 2 r2 r2 + Differentiating w.r.t. dr 2 dr 3 4 d r d Putting r a = 1 or 1 . z = r (cos + i sin ) . Then 2r bc = 0 sin dr = 0, we d is maximum for r2 + = a2 (i) we get 2 = 0 get = = 0, 2 , 2 therefore from (i) 1 1 a + a2 + 4 2 = a2 r = a r = 2 r 2 r 14. (c) We have Let 2 z 12 5 z 4 = and =1 z 8i 3 z 8 z = x + iy , then z 12 5 = 3 | z 12 |= 5 | z 8i | z 8i 3 3 |(x 12) + iy |= 5 | x + (y 8)i | 9(x 12)2 + 9y 2 = 25 x 2 + 25(y 8)2 and ....(i) z 4 = 1 | z 4 |=| z 8 | z 8 | x 4 + iy |=| x 8 + iy | (x 4)2 + y 2 = (x 8)2 + y 2 x = 6 Page | 105 COMPLEX NUMBER MATHEMATICS DPP DPP No. 3 DAILY PRACTICE PAPER Putting x = 6 in (i), we get y 2 25y + 136 = 0 y = 17,8 Hence z = 6 + 17i or z = 6 + 8i Trick : Check it with options. 15. (a) 1= z z z z z z 1 1 1 = 1 1 + 2 2 + 3 3 + + z1 z2 z3 z1 z 2 z 3 ( | z 1 |2 = 1 = z 1 z 1 , etc ) =| z1 + z 2 + z 3 |=| z1 + z 2 + z 3 |=| z1 + z 2 + z 3 | ( | z1 |=| z1 |) 16. (c) Given numbers are z1 = 10 + 6i, z 2 = 4 + 6i and z = x + iy z z1 amp z z2 ( x 4)(y 6) (y 6)( x 10) =1 ( x 4)( x 10) + (y 6)2 12y y 2 72 + 6y = x 2 14 x + 40 Now = 4 (x 10) + i (y 6) amp = (x 4) + i (y 6) 4 .....(i) | z 7 9i |=|(x 7) + i(y 9)| ....(ii) (x 7)2 + (y 9)2 From (i), (x 2 14 x + 49) + (y 2 18y + 81) = 18 (x 7)2 + (y 9)2 = 18 or [( x 7) 2 + (y 9) 2 ]1 / 2 = [18]1 / 2 = 3 2 |(x 7) + i(y 9)|= 3 17. (c) First deduce that 2 or | z 7 9i |= 3 2 a=b=c, . then it will be equal to z z1 arg 3 z 2 z1 2 . 18. (d) Trick : On checking the options, (d) satisfies both the conditions. 19. (b) We have max amp(z)=amp (z 2 ), min amp (z)=amp (z1 ) Y 15 15 Z2 Z1 1 25 2 1 X O Now 15 1 3 amp(z1 ) = 1 = cos 1 = cos 25 5 amp(z 2 ) = 2 + 2 = 2 15 1 3 + sin 1 = + sin 25 2 5 Page | 106 MATHEMATICS DPP DPP No. 3 DAILY PRACTICE PAPER COMPLEX NUMBER | max amp(z) min amp(z)| = = 2 2 + sin 1 + 2 3 3 cos 1 5 5 20. (a) We have z 2 = z 1 and z 4 = z 3 Therefore Now 3 3 3 cos 1 = 2 cos 1 5 5 5 cos 1 z1 z 2 =| z1 |2 and z 3 z 4 =| z 3 |2 z arg 1 z4 | z |2 = arg 1 2 | z 3 | z + arg 2 z 3 = arg z1 z 3 z z = arg 1 2 z z 4 3 2 = 0 ( Argument of positive real number is zero). 21. (c) Given that arg zw = .....(i) z + i = 0 z = i z = i = iz From (i), arg ( iz 2 ) = arg ( i) + 2arg(z) = 2 arg (z) = 22. (d) Since Put 3 2 ; ; + 2 arg(z) = 2 a rg(z) = 3 4 (1 + x)n = C0 + C1 x + C2 x 2 + ..... + Cn x n x =i, on both the sides, we get n (1 + i) = (C0 C2 + C4 .....) + i(C1 C3 + C5 .....) .....(i) Also, 1 + i = 2 cos + i sin 4 4 (1 + i)n = 2n / 2 cos + i sin 4 4 in amplitude modulus form n n n = 2n / 2 cos + i sin 4 4 ....(ii) Equating the real parts in (i) and (ii) we get, C0 C2 + C4 C6 + ..... = 2n / 2 cos 23. (c) n 4 x = cos + i sin = e i , y = cos + i sin = e i x m y n + x m y n = e im e in + e im e in = e i(m +n ) + e i (m +n ) = cos(m + n ) + i sin(m + n ) + cos(m + n ) i sin(m + n ) = 2 cos(m + n ) Page | 107 COMPLEX NUMBER MATHEMATICS DPP DPP No. 3 DAILY PRACTICE PAPER 24. (d) We have 8 2r 2r + i cos sin = 9 9 r =1 8 =i 2r i e 9 8 =i r =1 , r 8 i cos r =1 when 2r 2r i sin 9 9 = e (2 i / 9) r =1 ( 9 ) 1 (1 ) =i = i = i 1 (1 ) 1 8 = i ( 9 = e i 2 = cos 2 i sin 2 = 1 ) 25. (b) Let the complex number a, b, c and triangle ABC and DEF respectively. Put b a = r1e i 1 represent the vertices u, v, w A, B, C and D, E, F of the two c a = r2 e i 2 v u = 1 e i 1 , w u = 2 e i 2 and r = e i Substituting these values in the given relations c a = r(b a) and w u = (v u)r, we have r2e i 2 = e i r1e i 1 = r1e i( + 1 ) .......(i) and .......(ii) 2e i 2 = 1e i 1 e i = ( 1 )e i( 1 + ) Equating moduli and arguments of the complex numbers on both sides (i), we get r2 = r1 , 2 = + 1 i.e., AC = AB and CAB = 2 1 = Similarly from (ii), we shall get AC AB = DF DE Thus we get Hence the triangle and ABC DF = DE and FDE = 2 1 = CAB = FDE and DEF are similar. 26. (a) One of the number must be a conjugate of z1 = 1 + i 3 i.e. z 2 = 1 i 3 or z 3 = z1e i 2 / 3 and z 2 = z1e i 2 / 3 2 2 z 3 = (1 + i 3 ) cos + i sin = 2 3 3 Aliter : Obviously | z |= 2 is a circle with centre O(0, 0) and radius 2. Therefore, OA = OB = OC and this is satisfied by (a) because two vertices of any triangle cannot be same. 27. (a) Let OA, OB vectors be the sides of an equilateral z1 , z 2 OAB and let OA, OB represent the complex numbers or respectively. Y B(Z2) Z2 Z1 Z2 3 O Z1 A(Z1) X Page | 108 COMPLEX NUMBER MATHEMATICS DPP DPP No. 3 DAILY PRACTICE PAPER From the equilateral z z1 arg 2 z2 and z arg 2 z1 = arg (z 2 z 1 ) arg z 2 = / 3 = arg(z 2 ) arg(z 1 ) = 3 z 2 z1 z =1= 2 z2 z1 Also OAB, AB = z 2 z1 Thus the vectors , since triangle is equilateral. z 2 z1 z2 and z2 z1 have same modulus and same argument, which implies that the vectors are equal, that is z 2 z1 z 2 = z2 z1 z1 z 2 z12 = z 22 z12 + z 22 = z1 z 2 Note : Students should remember this question as a formula. 28. (a) If is a complex number satisfying the given conditions, then z = x + iy a 2 3a + 2 =| z + 2 |=| z + i 2 + 2 i 2 | | z + i 2 | + 2 |1 i | a 2 + 2 .....(i) 3a 0 a 0 Since | z + 2 |= a 2 3a + 2 | z + 2i | a 2 represents a circle with centre at A( 2 ,0) represents the interior of the circle with centre at and radius B(0, 2 ) a 2 3a + 2 and radius a, , and therefore there will be a complex number satisfying the given condition and the given inequality if the distance AB is less than the sum or difference of the radii of the two circles, i.e., if ( 2 0)2 + (0 + 2 )2 a 2 3a + 2 a 2 a a 2 3a + 2 a 2 or But a 0 4 + a 2 4a a 2 3a + 2 7a 2 a 2 or a from (i), therefore 29. (b) Area of the triangle 1 | z |2 = 2 2 30. (c) = 7 2 a 2. 1 | z |2 2 | z |2 = 4 | z |= 2 . 2 z z + +1 = 0 z 2 + z | z | + | z |2 = 0 | z | | z| z = , 2 |z| z = |z |or z = 2 |z| Page | 109 COMPLEX NUMBER MATHEMATICS DPP DPP No. 3 DAILY PRACTICE PAPER 3 or x + iy =| z | 1 i 3 2 2 2 x + iy =| z | 1 + i 2 x = 1 | z |, y =| z | 2 31. (c) 3 2 or x= y + 3 x = 0 or y 3 x = 0 |z| |z| 3 ,y = 2 2 y 2 3x 2 = 0 . cos + cos + cos = 0 and Let sin + sin + sin = 0 a = cos + i sin ; b = cos + i sin c = cos + i sin . and Therefore a + b + c = (cos + cos + cos ) +i (sin + sin + sin ) = 0 + i0 = 0 If a + b + c = 0, then a 3 + b 3 + c 3 = 3abc or (cos + i sin a)3 + (cos + i sin )3 + (cos + i sin )3 = 3(cos + i sin )(cos + i sin )(cos + i sin ) (cos 3 + i sin 3 ) + (cos 3 + i sin 3 ) + (cos 3 + i sin 3 ) = 3[cos( + + ) + i sin( + + )] or 32. (c) cos 3 + cos 3 + cos 3 = 3 cos( + + ). z r = cos z 1 = cos r r + i sin 2 n2 n n2 z n = cos + i sin n n 2 n2 + i sin ; z 2 = cos n n2 2 2 + i sin 2 n2 n ; .... lim (z1 z 2 z 3 ......... z n ) n = lim cos 2 (1 + 2 + 3 + ... + n) n n + i sin 2 (1 + 2 + 3 + ... + n) n n(n + 1) n(n + 1) = lim cos + i sin 2 n 2n 2n 2 = cos 33. (b) 2 + i sin 2 =e i 2 . (x 1)3 = 8 x 1 = ( 8)1 / 3 x 1 = 2, 2 , 2 2 x = 1,1 2 ,1 2 2 Trick : By inspection, we see that (b) satisfies the equation i.e, ( 1 1)3 + 8 = 0, (1 2 1)3 + 8 = 0 (1 2 2 1)3 + 8 = 0 . Page | 110 MATHEMATICS DPP DPP No. 3 DAILY PRACTICE PAPER 34. (c) Since 1, , 2 , 3 ,..... n 1 are the n, nth COMPLEX NUMBER roots of unity, therefore, we have the identity = (x 1)(x )(x 2 ).....( x n 1 ) = x n 1 or (x )( x 2 ).....( x n 1 ) = xn 1 x 1 = x n 1 + x n 2 + ..... + x + 1 Putting x =1 on both sides, we get (1 )(1 2 ).....(1 n 1 ) = n 35. (b) r th term of the given series = r[(r + 1) ][(r + 1) 2 ] = r[(r + 1)2 ( + 2 )(r + 1) + 3 ] = r[(r + 1)2 ( 1)(r + 1) + 1] = r[(r 2 + 3r + 3] = r 3 + 3r 2 + 3r Thus sum of the given series (n 1) = (r 3 + 3r 2 + 3r ) r =1 = 1 1 1 (n 1)2 n2 + 3. (n 1)(n)(2n 1) + 3. (n 1)n 4 6 2 = 1 (n 1)n(n2 + 3n + 4) 4 36. (c) Given equation is 1 3 4 + 5 + i 2 2 334 1 3 + 3 + i 2 2 2 2 = 4 + 5 cos + i sin 3 3 334 365 2 2 + 3 cos + i sin 3 3 365 668 668 = 4 + 5 cos + i sin 3 3 730 730 3 cos + i sin 3 3 2 2 = 4 + 5 cos 222 + + i sin 222 + 3 3 + 3 cos 243 + + i sin 243 + 3 3 2 2 = 4 + 5 cos + i sin + 3 cos i sin 3 3 3 3 1 1 3 3 = 4 + 5 + i + 3 i 2 2 2 2 Page | 111 MATHEMATICS DPP DPP No. 3 DAILY PRACTICE PAPER 3 =i 3 2 = 4 4 + 2i 37. (d) COMPLEX NUMBER . a = cos(2 / 7) + i sin(2 / 7) a 7 = [cos(2 / 7) + i sin(2 / 7)]7 .....(i) = cos 2 + i sin 2 = 1 S = + = (a + a 2 + a 4 ) + (a 3 + a 5 + a 6 ) S = a + a 2 + a 3 + a 4 + a5 + a6 = S= a(1 a 6 ) 1 a a a7 a 1 = = 1 1 a 1 a .....(ii) P = = (a + a 2 + a 4 )(a 3 + a 5 + a 6 ) = a 4 + a 6 + a7 + a 5 + a7 + a 8 + a7 + a 9 + a10 = a4 + a6 + 1 + a5 + 1 + a + 1 + a2 + a3 (From eqn (i)] = 3 + (a + a 2 + a 3 + a 4 + a 5 + a 6 ) = 3 + S [From (ii)] = 3 1 = 2 Required equation is, x2 + x + 2 = 0 . 38. (d) 11 / n = cos 2r n Let z1 = cos and + i sin 2r n 2r1 2r + i sin 1 n n z 2 = cos Then = x 2 Sx + P = 0 2r2 2r + i sin 2 n n z Z1 OZ 2 = amp 1 z2 2(r1 r2 ) = n 2 . = amp (z1 ) amp (z 2 ) (Given) n = 4(r1 r2 ) =4 integer, so n is of the form 4 k. 39. (a) 2 ( + 1)( 2 + 1) + 3(2 + 1)(2 2 + 1) + ...... + (n + 1)(n + 1) (n 2 + 1) = n (r + 1)(r + 1) (r 2 + 1) r =1 = n (r + 1)(r 2 3 + r + r 2 + 1) r =1 = n r =1 (r + 1) (r 2 r + 1) = n (r 3 r 2 + r + r 2 r + 1) r =1 Page | 112 MATHEMATICS DPP DPP No. 3 DAILY PRACTICE PAPER = n n (r ) + (1) = 3 r =1 r =1 COMPLEX NUMBER 2 n(n + 1) + n. 2 40. (d) We have, (1 + 2 )m = (1 + 4 )m ( 3 = 1) (1 + 2 )m = (1 + )m ( )m = ( 2 )m 2 m = 1 ( 2 )m = 1 = ( ) 2m = ( 3 ) m = 3 2 Hence least positive integral value of m is 3. Page | 113 PHYSICS DPP COMPLEX NUMBER TEST No. 3 DAILY PRACTICE PAPER PRACTICE TEST PAPER 1. If x 1+ i = 1, 1 i then (a) x = 4n, where n is any positive integer (b) x = 2n, where n is any positive integer (c) x = 4n + 1, where n is any positive integer (d) x = 2n + 1, where n is any positive integer 2. 3. 4. 5. If 1 z = x iy and z 3 = p + iq , (a) 2 (b) 1 (c) 2 (d) 1 If x y 2 + /( p + q 2 ) p q is equal to and are different complex numbers with | |= 1 , then (a) 0 (b) 1/2 (c) 1 (d) 2 The complex numbers z1 , z 2 and z3 satisfying z1 z 3 1 i 3 = z2 z3 2 (a) Of area = 0 (b) Right angled isosceles (c) Equilateral (d) Obtuse angled isosceles Let z1 origin, 6. then and z2 be two roots of the equation z 1 and z 2 z 2 + az + b (a) a2 = b (b) a 2 = 2b (c) a 2 = 3b (d) a 2 = 4b A complex number z is such that z 2 arg = 3 z + 2 (a) An ellipse (b) A parabola (c) A circle (d) A straight line The value of 6 sin k =1 8. 2 k 2 k i cos 7 7 (a) 1 (b) 0 (c) i (d) If x n = cos n 3 is equal to are the vertices of a triangle which is = 0, z being complex. Further, assume that form an equilateral triangle. Then . The points representing this complex number will lie on 7. 1 + i sin n 3 , is i then x1 .x 2 .x 3 .... x (a) 1 (b) 1 (c) i (d) i is equal to 9. 10. Common roots of the equations (a) , 2 (b) (c) 2, 3 (d) None of these 12. 13. 14. (c) 1,2 + 3 ,2 + 3 2 (d) 1, 2 3 ,2 3 2 Let 1 ix = a ib 1 + ix a 2 + b2 = 1 , (a) 2a (1 + a)2 + b 2 (b) 2b (1 + a)2 + b 2 (c) 2a (1 + b)2 + a 2 (d) 2b (1 + b)2 + a 2 If | z + 4 | 3, 16. 17. and where a and (a) 6, 6 (b) 6, 0 (c) 7, 2 (d) 0, 1 k = 1, 2,.... n (b) Greater than one (c) Zero (d) Less than one z1 are real, then and z2 be two complex numbers with then principal | z + 1| arg (z1 z2 ) (x 2)3 + 27 = 0 are x= are and 1 + 2 + ... + n = 1, then the value of (a) Equal to one Let b then the greatest and the least value of If | ak | 1, k 0 for z1985 + z100 + 1 = 0 are then the roots of the equation 1, , 2 + , 15. 1, , 2 (b) 1, 1, 1 and , 3 If the cube roots of unity are (a) 11. z 3 + 2z 2 + 2z + 1 = 0 | 1 a1 + 2 a 2 + .... n a n | and as their principal arguments such that is given by (a) + + (b) + (c) + 2 (d) + Let z be a complex number satisfying | z 5i | 1 such that amp z is minimum. Then 2 6 24i + 5 5 (b) (c) 2 6 24i 5 5 (d) None of these z = x + iy is taken such that the amplitude of fraction (a) x 2 + y 2 + 2y = 1 (b) x 2 + y 2 2y = 0 (c) x 2 + y 2 + 2y = 1 (d) x 2 + y 2 2y = 1 If z = i log (2 3 ), then z is equal to 24 2 6i + 5 5 (a) If complex number is z 1 z +1 is always 4 , then cos z = (a) i (b) 2 i (c) 1 (d) 2 Page | 1 18. If the complex numbers | z1 |=| z2 |=| z3 |, 19. then z1, z2 , z3 represent the vertices of an equilateral triangle such that z1 + z2 + z3 = (a) 0 (b) 1 (c) (d) None of these 1 The value of infinite product (cos + i sin ) (cos 2 + i sin ) (cos 2 + i sin 2 ).... 2 2 2 is 20. (a) cos 2 i sin 2 (b) cos 2 + i sin 2 (c) sin 2 i cos 2 (d) sin 2 + i cos 2 If 1 is any nth root of unity, then (a) 2n 1 (b) 2n 1 (c) n 1 (d) n 1 S = 1 + 3 + 5 2 .. upto n terms, is equal to Page | 2 ANSWER SHEET 1. (a) x x (1 + i)2 1+ i =1 =1 2 1 i 1 i x 1 + i 2 + 2i x 1 + 1 = 1 i = 1 x = 4n, n I + 2. (a) . z = x iy, z 1 / 3 = p + iq z = ( p + iq)3 = p 3 iq 3 + 3 p 2 qi 3 pq 2 z = ( p 3 3 pq 2 ) + i(3 p 2 q q 3 ) Equating real and imaginary part we get x = p 3 3 pq2 , y = (3 p 2q q3 ) x = p( p 2 3q 2 ) , y = q(q 2 3 p 2 ) x = p 2 3q 2 p y = q 2 3p 2 q ..(i) (ii) Adding (i) and (ii), we get x y + = p 2 + q 2 3(q 2 + p 2 ) = 2 p 2 2q 2 p q x y + = 2( p 2 + q 2 ) . p q 3. (c) 1 = 4. (c) = Hence x y + p q p2 + q2 = 2 . = ( ) 1 1 = =1 | | ( ) | | z1 z 3 1 3 = i = z2 z3 2 2 { | z |=| z |} 1 3 + = 1. 4 4 so, | z1 z 3 |=| z 2 z 3 | Also, or 3 / 2 z z3 = tan 1 ( 3 ) = = tan 1 amp 1 z z 1 / 2 3 3 2 z z3 = amp 2 z1 z 3 3 or z 2 z 3 z1 = 60 o The triangle has two sides equal and the angle between the equal sides equilateral. (c) Let z1, z 2 be the roots of z 2 + az + b = 0 . 5. Then + z 22 So, it is z1 + z 2 = a, z1z 2 = b Since z12 = 60o. z1 , z 2 + z 32 and z3 = 0 form an equilateral triangle, therefore as we know = z1 z 2 + z 2 z 3 + z 3 z1 Page | 3 z12 + z 22 = z1 z 2 6. (c) ( z 3 = 0) (z1 + z 2 )2 = 3z1 z 2 ( a)2 = 3b a 2 = 3b . (x 2) + iy z 2 tan 1 arg = = z + 2 3 (x + 2) + iy 3 (x 2)2 + y 2 = tan( / 3)[ (x + 2)2 + y 2 ] Squaring both sides, (x 2) 2 + y 2 = 3[ x + 2] 2 + y 2 ] x 2 + y 2 + 4 4 x = 3 x 2 + 3y 2 + 12x + 12 2 x 2 + 2y 2 + 16 x + 8 = 0 x 2 + y 2 + 8 x + 4 = 0 which is a equation of circle. 7. (d) Let z = cos 2 2 + i sin , 7 7 then by De Moivre's theorem z k = cos 2 k 2 k + i sin 7 7 Now the given sum 6 S= sin k =1 6 2 k 2 k i cos 7 7 7 = ( i) cos 2 k + i sin 2 k k =1 7 6 6 2 k 2 k + i sin zk cos = ( i) 7 7 k =1 k =1 = ( i) Which is a G.P. of which the first term is z = cos 2 2 + i sin 1 7 7 z, number of terms is 6 and the common ratio is So summing up the G.P., we have S = ( i) z(1 z 6 ) z z7 z 1 = ( i) = ( i) =i 1 z 1 z 1 z 7 2 2 [ z 7 = cos + i sin = cos 2 + i sin 2 = 1 7 7 8. (c) x1.x 2 .x 3 .... x = cos + i sin cos 2 + i sin 2 cos 3 + i sin 3 .... 3 3 3 3 3 3 = cos + 2 + 3 + ..... + i sin + 2 + 3 + .... 3 3 3 3 3 3 / 3 / 3 = cos + i sin = cos + i sin = i. 1 1 2 2 1 1 3 3 9. (a) The first equation can be written as (z + 1)(z 2 + z + 1) = 0 . Its roots are Now, let f (z) = z1985 + z100 + 1 We have f ( 1) = ( 1)1985 + ( 1)100 + 1 0 Therefore 1 is not a root of the equation f (z) = 0 Again f ( ) = 1985 + 100 + 1 = ( 3 )661 2 + ( 3 )33 + 1 = 2 + + 1 = 0 Therefore is a root of the equation f (z) = 0 . Similarly, we can show that f ( 2 ) = 0 1, and 2 Page | 4 10. Hence and 2 are the common roots. Trick : Obviously and 2 satisfy both the equations but equations. (d) Here 11 / 3 = 1, , 2 For the equation 11. (b) (x 2)3 = 27 = 3 3 x 2 = 3(1)1 / 3 = 3(1, , 2 ) = 3, 3 ,3 2 x = 1,2 3 ,2 3 2 . (1 ix)(1 ix) 1 ix = a ib = a ib (1 + ix)(1 ix) 1 + ix 1 x2 1 x 2 2ix 2x = a ib = a and =b 2 2 1+ x 1 + x2 1+ x 2x Now we can write x as x = 1+ x 2 2x 2 1+ x2 = 13. = 1+ x2 1 x2 +1 1+ x2 b 2b 2b 2b = = = 1 + a 1 + 1 + 2a 1 + (a 2 + b 2 ) + 2a (1 + a)2 + b 2 Trick : 12. 1 does not satisfy the (x 2)3 + 27 = 0 3 i.e., 1 ix 1 x 2 2ix = = a ib 1 + ix 1 + x2 Let a = 0 x = 1 and b = 1 . Also option (b) gives 1 . (b) | z + 4 | 3 3 z + 4 +3 6 z + 1 0 0 | z + 1| 6 , (d) We have 0 (z + 1) 6 Hence greatest and least values of | z + 1|are 6 and 0 respectively. | 1a1 + 2a2 + ..... + nan | | 1a1 | + | 2a2 | +..... + | nan | =| 1 || a1 | +..... + | n || an | = 1 | a1 | +..... + n | an | [ each k . 0 ] 1 + ..... + n [ | ak | 1and so k | ak | k for all k = 1,2,.... n ] Hence | 1a1 + 2a2 + ..... + nan | 1 . Thus | 1a1 + ..... + n an | 1 . 14. (c) We know that principal arguments of a complex number lie between therefore principal arg (z1z2 ) = arg z1 + arg z2 = + , is given by + 2 . 15. (a) We have Y C and , but + , OC = 5, CA = 1 (0,5) A(Z) O X Page | 5 = AOX = min .amp z, AOC = 90 o 1 1 cos = 5 5 sin(90o ) = z = OA cos + iOA sin 1 1 z = 5 2 1 + i 5 2 1 1 2 5 5 2 6 (1 + i 2 6 ) . 5 z 1 (x + iy) 1 (x 1) + iy = = z + 1 (x + iy) + 1 (x + 1) + iy = 16. (d) = = {(x 1) + iy} {(x + 1) iy} {(x + 1) + iy} {(x + 1) iy} {(x 2 1) + y 2 } + i{y(x + 1) y(x 1)} (x + 1)2 + y 2 ( x 2 1) + y 2 2y = + i 2 2 ( x + 1) 2 + y 2 ( x + 1 ) + y 2y (x 2 1) + y 2 z 1 1 amp = tan 2 2 (x + 1) + y (x + 1) 2 + y 2 z +1 17. 2y 2y = tan 1 2 tan = 2 2 4 x + y2 1 4 x + y 1 1= x 2 + y 2 2y = 1 . 2y x 2 + y 2 1 = 2y x2 + y2 1 (d) Given, complex function e iz = e i or 2 log( 2 3 ) = e log( 2 e iz = (2 + 3 ). 3) z = i log(2 3 ). = e log( 2 Similarly, 3) The given equation may be written as 1 e iz = (2 3 ). We know that cos z = 18. e iz + e iz (2 + 3 ) + (2 3 ) = = 2. 2 2 (a) Let the complex number if O z1 , z 2 , z 3 be the origin, we have denote the vertices A, B, C of an equilateral triangle ABC . Then, OA = z1 , OB = z 2 , OC = z 3 Therefore | z1 |=| z 2 |=| z3 | OA = OB = OC i.e., O is the circumcentre of ABC . Hence z1 + z 2 + z 3 = 0 Note : Students should remember this question as a fact. 19. (b) (cos + i sin )(cos 2 + i sin )(cos 2 + i sin 2 ) .... 2 2 2 = cos + + 2 + ..... + i sin + + 2 + ..... 2 2 2 2 + i sin = cos 2 + i sin 2 = cos 1 ( 1 / 2 ) 1 ( 1 / 2 ) 20. (b) We have S = 1 + 3 + 5 2 + ..... + (2n 1) n 1 S = + 3 + 5 + ..... + (2n 1) Subtracting (ii) from (i), we get 2 3 n . ..(i) ..(ii) Page | 6 (1 )S = 1 + 2 + 2 2 + ..... + 2 n 1 (2n 1) n = 2(1 + + 2 ..... + n 1 ) 1 (2n 1) n = 2(1 n ) 2n 2n = 2n ( n = 1) S = 1 (1 ) . Page | 7

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