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