### Complex number

• 1. GRAViitY COMPLEX NUMBER MATHEMATICS-PI SECTION – I Straight Objective Type This section contains 8 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct. 21. If z and  are two non-zero complex numbers such that |z| = 1 and Arg z – Arg  = z  (A) 1 (C) i 22. 23. 24.  , then 2 (B) 1 (D)  i Let z and  be complex numbers such that z  i  0 and arg z = , then arg z =   (A) (B) 4 2 3 5 (C) (D) 4 4 i z 1 e If the imaginary part of the expression i   be zero, then locus of z is z 1 e (A) straight line (B) parabola (C) unit circle (D) ellipse If  is a complex number such that || = r  1 then z    between the foci is (A) 2 (C) 3 1 describes a conic. The distance  (B) 2( 2  1) (D) 4 z lie on 1  z2 (A) a line not passing through the origin (B) | z | 2 (C) the x-axis (D) the y-axis 25. If |z| = 1 and z  ±1, then all the values of 26. The number of solutions of the system of equations given by |z| = 3 and | z  1  i | 2 is equal to (A) 4 (B) 2 (C) 1 (D) no solution 27. Let z = cos  + isin. Then the value of 15 1 Im(z2m1 ) m 1 sin 2 1 (C) 2sin 2 (A) 28. 1 3sin 2 1 (D) 4sin 2 (B) In geometrical progression first term and common ratio are both value of the nth term of the progression is (A) 2n (C) 1 BY at  = 2º is RAJESH SIR (B) 4n (D) 3n 1 ( 3  i). Then the absolute 2
• 2. GRAViitY COMPLEX NUMBER SECTION – II Multiple Correct Answer Type This section contains 4 multiple correct answer(s) type questions. Each question has 4 choices (A), (B), (C) and (D), out of which ONE OR MORE is/are correct. 29. If z  20i  21  21  20i , then the principal value of arg z can be (A)  4 (C)  3 4 3 (D)  4 (B)  4 30. If z1 = a + ib and z2 = c + id are complex numbers such that |z1| = |z2| = 1 and Re(z1z2) = 0 then the pair of complex numbers 1 = a + ic and 2 = b + id satisfies. (A) |1| = 1 (B) |2| = 1 (C) Re (1 2) = 0 (D) |1| = 2 31. If z1 = 5 + 12i and |z2| = 4 then (A) maximum (|z1 + iz2|) = 17 (C) minimum 32. z1 4 z2  z2  (B) minimum (|z1 + (1 + i)z2|) = 13  9 2 13 4 (D) maximum z1 4 z2  z2  13 3 If z is a complex number satisfying |z – i Re(z)| = |z – Im (z)| then z lies on (A) y = x (B) y = x (C) y = x + 1 (D) y = x + 1 SECTION – III Linked Comprehension Type This section contains 2 paragraphs. Based upon each paragraph, 3 multiple choice questions have to be answered. Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct. Paragraph for Question Nos. 33 to 35 Let A, B, C be three sets of complex numbers as defined below A = {z : Im z  1} B = {z : |z – 2 – i| = 3} C = {z : Re((1 – i)z) = 2 } 33. The number of elements in the set ABC is (A) 0 (C) 2 (B) 1 (D)  34. Let z be any point in ABC. Then, |z + 1 – i|2 + |z – 5 – i|2 lies between (A) 25 and 29 (B) 30 and 34 (C) 35 and 39 (D) 40 and 44 35. Let z be any point in ABC and let w be any point satisfying |  2 – i| < 3. Then, |z|  |w| + 3 lies between (A) – 6 and 3 (B) – 4 and 6 (C) – 6 and 6 (D) – 3 and 9 BY RAJESH SIR
• 3. GRAViitY COMPLEX NUMBER Paragraph for Question Nos. 36 to 38 Suppose z and w be two complex numbers such that |z|  1, |w|  1 and |z + iw| = |z – i w | = 2. Use the result | z |2  zz and |z + w|  |z| + |w|, answer the following 36. Which of the following is true about |z| and || 1 2 3 (C) | z | | w | 4 (A) | z || w | 37. 38. (B) | z | 1 3 , | w | 2 4 (D) |z| = |w| = 1 Which of the following is true for z and  (A) Re(z) = Re(w) (C) Re(z) = Im(w) (B) Im(z) = Im(w) (D) Im(z) = Re(w) Number of complex numbers satisfying the above conditions is (A) 1 (B) 2 (C) 4 (D) indeterminate SECTION – IV 39. 40. Matrix Match Type Match the statements/expressions in Column I with the open intervals in Column II Column I Column II 10 (A) (P)   2  sin 0  (r  )(r   )   900  r 1  (B) If roots of t2 + t + 1 = 0 be ,  then 4 + 4 + – (Q) 4 1 –1  = 4 (C)  1  cos   isin   If    cos n  isin n, then n (R) i  sin   i(1  cos )  = (D) (S)   If z r  cos r  isin r , r = 1,2,3,…., then value 1 3 3 of z1z2z3 ……  = Number of solutions of Column I (A) (B) (C) (D) BY 2 z |z|  0 2 2 z z 0 z 2  8z  0 | z  2 | 1 and | z  1| 2 RAJESH SIR Column II (P) 1 (Q) 3 (R) (S) 4 Infinite
• 4. GRAViitY COMPLEX NUMBER MATHEMATICS-PII SECTION – I Straight Objective Type This section contains 4 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct. 20. The roots of 1 + z + z3 + z4 = 0 are represented by the vertices of (A) a square (B) an equilateral triangle (C) a rhombus (D) a rectangle 21. If |z – 1| + |z + 3|  8, then the range of values of |z – 4| is, (A) (0, 8) (B) [1, 9] (C) [0, 8] (D) [5, 9] 22. If z1, z2 and z3 be the vertices of ABC, taken in anti-clock wise direction and z0 be the  z 0  z1  sin 2A  z 0  z 3  sin 2C is equal to    z 0  z 2  sin 2B  z 0  z 2  sin 2B  circumcentre, then  (A) 0 (C) – 1 23. (B) 1 (D) 2 If a, b, c, a1,b1,c1 are non zero complex numbers satisfying a b c    1  i and a1 b1 c1 a1 b1 c1 a2 b2 c 2    0 , then 2  2  2 is equal to a b c a1 b1 c1 (A) 2i (C) 2 (B) 2 + 2i (D) 2 – 2i SECTION – II Multiple Correct Answer Type This section contains 5 multiple correct answer(s) type questions. Each question has 4 choices (A), (B), (C) and (D), out of which ONE OR MORE is/are correct. 24. A, B, C are the points representing the complex numbers z1, z2, z3 respectively on the complex plane and the circumcentre of the triangle ABC lies at the origin. If the altitude AD of the triangle ABC meets the circumcircle again at P, then P represents the complex number zz zz z z (A)  z1z 2 z 3 (B)  1 2 (C)  1 3 (D)  2 3 z3 z2 z1 25. If points A and B are represented by the non-zero complex numbers z1 and z2 on the Argand plane such that |z1 + z2| = |z1 – z2| and 0 is the origin, then z z (A) orthocentre of OAB lies at 0 (B) circumcentre of AOB is 1 2 2 z   (C) arg  1    (D) OAB is isosceles 2  z2  26. If f(x) and g(x) are two polynomials such that the polynomial h(x) = xf(x3) + x2g(x6) is divisible by x2 + x + 1, then (A) f(1) = g(1) (B) f(1) = g(1) (C) h(1) = 0 (D) none of these BY RAJESH SIR
• 5. GRAViitY 27. 28. COMPLEX NUMBER If  (  1) is the fifth root of unity then (A) |1     2  3   4 | 0  (C) |1     2 |  2cos 5 (B) |1     2  3 |  1  (D) |1   |  2cos 10 If the lines az  az  b  0 and cz  cz  d  0 are mutually perpendicular, where a and c are non-zero complex numbers and b and d are real numbers, then (A) aa  cc  0 (B) ac is purely imaginary  a c a (C) arg     (D)  2 a c c SECTION – III Matrix Match Type 29. Match the statements/expressions in Column I with the open intervals in Column II Column I (A) (P) z  3 z2 Let z1, z2 be complex numbers such that 1  1 and |z2|  1, then Column II 3  z1 z2 6 |z1| is equal to (B) (C) (D) 30. Number of non-zero complex number satisfying z  iz 2 Let a, b  (0, 1) and z1 = a + i, z2 = 1 + bi and z3 = 0 be the vertices of an equilateral triangle then value of a  b  2 3 is equal to Consider a circle having OP as diameter where O being origin and P be z1. Take two points Q(z2) and R(z3) on the circle and also on the same 4 (R) 3 (S) 3 2 side of OP. If POQ =/2k, QOR = /k and 8 z2  (5  3 3) z1 z3 then k is equal to Let the complex numbers z1, z2 and z3 represent the vertices A, B and C of triangle ABC respectively, which is inscribed in the circle of radius unity and centre at origin. The internal bisector of the angle A meets the circumcircle again at the point D, which is represent by the complex number z4, and altitude from A to BC meets the circumcircle at E, given by z5. Now match the entries from the following columns Column I Column II (A) (P) z z  arg  2 2 3  is equal to   z4  (B) (C) (D)  z4  arg   is equal to  z 2  z3  zz  arg  1 3  is equal to  z 2 z5  (Q)  z2  arg  4   z1z 5  (S) (R) RAJESH SIR  2  4 0 (T) BY (Q)  /2 5
• 6. GRAViitY COMPLEX NUMBER SECTION – IV Integer Answer Type This section contains 8 questions. The answer to each of the question is a single digit integer, ranging from 0 to 9. The appropriate bubbles below the respective question numbers in the ORS have to be darkened. For example, if the correct answers to question numbers X, Y, Z and W (say) are 6, 0, 9 and 2, respectively, then the correct darkening of bubbles will look like the following: a b c    0 where a, b, c are three distinct complex numbers, then the value of bc ca a b a2 b2 c2   is equal to (b  c) 2 (c  a) 2 (a  b) 2 31. If 32. If |z1| = 1, |z2| = 2, |z3| = 3 and |9 z1z2 + 4z1z3 + z2z3| = 12 then |z1 + z2 + z3| is equal to 33.  3z  6  3i   If the complex numbers z for which arg  and |z – 3 + i| = 3, are   2z  8  6i  4 4   2  4   2    k    i 1   and  k    i 1   then k must be equal to 5  5 5  5   20 34. If   e 2  i/7 and f  x   A 0   A k x k then the value of f(x) + f(x) + f(2x) + …. + f(6x) k 1 is k(A0 + A7x7 + A14x14), then k must be equal to 35. If magnitude of a complex number 4 – 3i is tripled and rotated by an angle  anticlockwise about origin then resulting complex number would – 12 + i then  must be equal to 36. The maximum value of |z| when z satisfies the condition z  37. Let z1, z2 be the roots of the equation z2 + az + b = 0 where a and b may be complex. Let A and B represent z1 and z2 in the Argand’s plane. If AOB    0 and OA = OB.  Then 2 = b cos2   . where  value is …… 2 38. z1, z2 are roots of the equation z2 + az + b = 0. If AOB (0 is origin), A and B represent z1 and a2 z2 is equilateral, then is equal to …….. b BY RAJESH SIR 2  2 is   1 z
• 7. GRAViitY COMPLEX NUMBER PAPER – I KEY MATHS 21. (D) 22. (C) 23. (C) 24. (D) 25. (D) 26. (D) 27. (D) 28. (C) 29. (ABCD) 30. (ABC) 31. (AD) 32. (AB) 33. (B) 34. (C) 35. (D) 36. (D) 37. (D) 38. (B) 39. (A - S), (B - P), (C - Q), (D - R) 40. (A- Q), (B - S), (C - R), (D – P) PAPER – II KEY MATHS 20. (B) 21. (B) 22. (C) 23. (A) 24. (BCD) 25. (ABC) 26. (ABC) 27. (ABC) 28. (BC) 29. (A-R), (B-R), (C-Q), (D-P) 30. (A –S), (B –QT), (C –S), (D –P) 31. 2 32. 2 33. 4 34. 7 35. 9 36. 3 37. 4 38. 3 BY RAJESH SIR