Complex Numbers jee mains questions | JEE Main Maths Complex Numbers Previous Year Question

Q 1 :

Let $A = {θ \in (0, 2 π) : \frac{1 + 2 i \sin θ}{1 - i \sin θ} is purely imaginary} .$ Then the sum of the elements in A is [2023]

$2 π$
$4 π$
$π$
$3 π$

1

2

3

4

(2)

Given, $\frac{1 + 2 i \sin θ}{1 - i \sin θ}$ is purely imaginary, then its real part must be zero.

$\frac{1 + 2 i \sin θ}{1 - i \sin θ} \times \frac{1 + i \sin θ}{1 + i \sin θ} = \frac{(1 + i \sin θ + 2 i \sin θ - 2 \sin^{2} θ)}{1 + \sin^{2} θ}$

$= \frac{1 - 2 \sin^{2} θ}{1 + \sin^{2} θ} + \frac{3 \sin θ}{1 + \sin^{2} θ} i$

Since real part is 0 $\Rightarrow \frac{1 - 2 \sin^{2} θ}{1 + \sin^{2} θ} = 0 \Rightarrow 2 \sin^{2} θ = 1$

$\Rightarrow \sin θ = \pm \frac{1}{\sqrt{2}} \Rightarrow θ = \frac{π}{4}, \frac{3 π}{4}, \frac{5 π}{4}, \frac{7 π}{4}$

Since, $θ \in (0, 2 π)$

Then sum of the elements in A is,

$\frac{π}{4} + \frac{3 π}{4} + \frac{5 π}{4} + \frac{7 π}{4} = \frac{16 π}{4} = 4 π$

Q 2 :

If the set $R = {(a, b) : a + 5 b = 42, a, b \in N}$ has $m$ elements and $\sum_{n = 1}^{m} (1 - i^{n!}) = x + i y,$ where $i = \sqrt{- 1},$ then the value of $m + x + y$ is [2024]

8
5
4
12

1

2

3

4

(4)

Given $R = {(a, b) : a + 5 b = 42, a, b \in N}$

So, $a = 42 - 5 b$

$∴ (a, b) = {(37, 1), (32, 2), (27, 3), (22, 4), (17, 5), (12, 6), (7, 7), (2, 8)}$

$∴ | R | = 8 \Rightarrow m = 8$

Now, $\sum_{n = 1}^{m} (1 - i^{n!}) = x + i y$

Clearly for $n = 4, i^{n!} = i^{4!} = ({{(i)}^{4})}^{6} = {(1)}^{6} = 1$

$∴$ $i^{n!} = 1$ for $n \geq 4$

$∴$ $\sum_{n = 1}^{8} (1 - i^{n!}) = (1 - i) + (1 - i^{2!}) + (1 - i^{3!})$

$= 1 - i + 1 + 1 + 1 + 1 = 5 - i \Rightarrow x = 5 and y = - 1$ .

So, $m + x + y = 8 + 5 - 1 = 12$

Q 3 :

Let $z$ be a complex number such that the real part of $\frac{z - 2 i}{z + 2 i}$ is zero. Then, the maximum value of $| z - (6 + 8 i) |$ is equal to [2024]

$12$
$\infty$
$8$
$10$

1

2

3

4

(1)

Let $z = x + i y, x, y \in R$

Consider $\frac{z - 2 i}{z + 2 i} = \frac{x + i y - 2 i}{x + i y + 2 i} = \frac{x + (y - 2) i}{x + (y + 2) i} \times \frac{x - (y + 2) i}{x - (y + 2) i}$

$= \frac{x^{2} + x (y + 2) i + x (y - 2) i + (y^{2} - 4)}{x^{2} + {(y + 2)}^{2}}$

$\Rightarrow \frac{x^{2} + y^{2} - 4}{x^{2} + {(y + 2)}^{2}} = 0 \Rightarrow x^{2} + y^{2} = 4$

$\Rightarrow | z |^{2} = 4 \Rightarrow | z | = 2$

Now, $| z - (6 + 8 i) | \leq | z | + | - 6 - 8 i | \leq 2 + 10 = 12$

$∴ | z - (6 + 8_{i}) |_{max} = 12$

Q 4 :

If $z$ is a complex number, then the number of common roots of the equations $z^{1985} + z^{100} + 1 = 0$ and $z^{3} + 2 z^{2} + 2 z + 1 = 0,$ is equal to [2024]

2
3
1
0

1

2

3

4

(1)

Given equation, $z^{1985} + z^{100} + 1 = 0$ ...(i)

and $z^{3} + 2 z^{2} + 2 z + 1 = 0$ ...(ii)

From (ii), we have $(z + 1) (z^{2} + z + 1) = 0$

$\Rightarrow z = - 1, z = ω, ω^{2}$

Clearly, $z = - 1$ does not satisfy equation (i).

$∴$ If $z = ω,$ then ${(ω)}^{1985} + {(ω)}^{100} + 1 = ω^{2} + ω + 1 = 0$

Also, $z = ω^{2}$ , then ${(ω^{2})}^{1985} + {(ω^{2})}^{100} + 1 = ω^{2} + ω + 1 = 0$

$∴$ Number of roots = 2

Q 5 :

Among the statements

(S1) : The set ${z \in C - {i} : | z | = 1 and \frac{z - i}{z + i} is purely real}$ contains exactly two elements, and

(S2) : The set ${z \in C - {- 1} : | z | = 1 and \frac{z - 1}{z + 1} is purely imagin ary}$ contains infinitely many elements. [2025]

only (S2) is correct
both are incorrect
only (S1) is correct
both are correct

1

2

3

4

(1)

(S1) : $\frac{z - i}{z + i} = \frac{\bar{z} + i}{\bar{z} - i}$ [ $∵$ Purely real]

$\Rightarrow z \bar{z} - i \bar{z} - i z - 1 = z \bar{z} + z i + i \bar{z} - 1$

$\Rightarrow z + \bar{z} = 0$

$\Rightarrow 2 x = 0$ [ $∵$ Purely real]

$\Rightarrow x = 0$

Also, |z| = 1

$∴ z = i$ ( $∵ z \neq - i i s g i v e n$ )

(S1) is incorrect.

(S2) : $\frac{z - 1}{z + 1} + \frac{\bar{z} - 1}{\bar{z} + 1} = 0$ [ $∵$ Purely imaginary]

$\Rightarrow z \bar{z} - \bar{z} + z - 1 + z \bar{z} - z + \bar{z} - 1 = 0 \Rightarrow z \bar{z} = 1 \Rightarrow$ $| z | = 1$ ,

which represents a circle with radius 1 and centre (0, 0).

(S2) is correct.

Q 6 :

Let $a \neq b$ be two non-zero real numbers. Then the number of elements in the set $X = {z \in ℂ : Re (a z^{2} + b z) = a and Re (b z^{2} + a z) = b}$ is equal to [2023]

2
0
3
1

1

2

3

4

(2)

We have, $Re (a z^{2} + b z) = a$

$Re (a (x^{2} - y^{2} + 2 i x y) + b (x + i y)) = a$

$a (x^{2} - y^{2}) + b x = a ...(i)$

Also, $Re (b z^{2} + a z) = b$

$b (x^{2} - y^{2}) + a x = b ...(ii)$

From (i) and (ii), we get $(a - b) (x^{2} - y^{2}) - (a - b) x = a - b$

$\Rightarrow x^{2} - y^{2} - x = 1 ...(iii)$

Also, adding (i) and (ii), we get $(a + b) (x^{2} - y^{2}) + (a + b) x = a + b$

Case 1: If $a + b \neq 0$ , then $x^{2} - y^{2} + x = 1$ $...(iv)$

From (iii) and (iv), we get $x = 0, y^{2} = - 1$ , which is not possible.

Case 2: If $a + b = 0$ , then infinite number of solutions exist.

So, set X has infinite number of elements.

So, number of elements in the set X is 0.

Q 7 :

For two non-zero complex numbers $z_{1}$ and $z_{2}$ , if $Re (z_{1} z_{2}) = 0$ and $Re (z_{1} + z_{2}) = 0$ , then which of the following are possible?

(A) $Im (z_{1}) > 0$ and $Im (z_{2}) > 0$
(B) $Im (z_{1}) < 0$ and $Im (z_{2}) > 0$
(C) $Im (z_{1}) > 0$ and $Im (z_{2}) < 0$
(D) $Im (z_{1}) < 0$ and $Im (z_{2}) < 0$

Choose the correct answer from the options given below: [2023]

A and B
B and C
B and D
A and C

1

2

3

4

(2)

Let $z_{1} = x_{1} + i y_{1}$ ; $z_{2} = x_{2} + i y_{2}$

$Re (z_{1} z_{2}) \Rightarrow x_{1} x_{2} - y_{1} y_{2} = 0 ...(i)$

$Re (z_{1} + z_{2}) \Rightarrow x_{1} + x_{2} = 0 ...(ii)$

$\Rightarrow x_{1}^{2} + y_{1} y_{2} = 0 \Rightarrow y_{1} y_{2} = - x_{1}^{2}$

So, $Im (z_{1})$ and $Im (z_{2})$ are opposite in sign

So, the correct statements are B and C.

Q 8 :

$Let S = {z \in ℂ - {i, 2 i} : \frac{z^{2} + 8 i z - 15}{z^{2} - 3 i z - 2} \in ℝ} .$

$If α - \frac{13}{11} i \in S, α \in ℝ - {0}, then 242 α^{2} is equal to$ __________. [2023]

(1680)

We have, $\frac{z^{2} + 8 i z - 15}{z^{2} - 3 i z - 2} = \frac{z^{2} - 3 i z - 2 + 11 i z - 13}{z^{2} - 3 i z - 2}$

$= 1 + (\frac{11 i z - 13}{z^{2} - 3 i z - 2}) = 1 + \frac{11 i (z + \frac{13 i}{11})}{z^{2} - 3 i z - 2} \in R$

Put $z = α - \frac{13 i}{11}$

Thus, $z^{2} - 3 i z - 2$ is imaginary.

Put $z = x + i y$

$\Rightarrow {(x + i y)}^{2} - 3 i (x + i y) - 2 is imaginary.$

$\Rightarrow x^{2} - y^{2} + 2 x y i - 3 i x + 3 y - 2 is imaginary.$

$\Rightarrow Re (x^{2} - y^{2} + 3 y - 2 + (2 x y - 3 x) i) = 0$

$\Rightarrow x^{2} - y^{2} + 3 y - 2 = 0 \Rightarrow x^{2} = y^{2} - 3 y + 2$

$\Rightarrow x^{2} = (y - 1) (y - 2)$

Put $x = α$ and $y = \frac{- 13}{11}$ , we get $α^{2} = (\frac{- 13}{11} - 1) (\frac{- 13}{11} - 2)$

$= (\frac{- 13 - 11}{11}) (\frac{- 13 - 22}{11}) = (\frac{- 24}{11}) (\frac{- 35}{11}) = \frac{24 \times 35}{121}$

$∴$ $242 α^{2} = \frac{24 \times 35 \times 242}{121} = 1680$

Q 9 :

Let $S = {z : 3 \leq | 2 z - 3 (1 + i) | \leq 7}$ be a set of complex numbers. Then $\min_{z \in S}$ $| (z + \frac{1}{2} (5 + 3 i)) |$ is equal to: [2026]

$\frac{5}{2}$
$\frac{1}{2}$
$\frac{3}{2}$
$2$

1

2

3

4

(3)

$\frac{3}{2} \leq$ $| z - \frac{3}{2} (1 + i) |$ $\leq \frac{7}{2}$

$\min_{z \in S}$ $| z - (\frac{- 5}{2} - \frac{3}{2} i) |$ $= P B$

$P B = P C - \frac{7}{2} \Rightarrow 5 - \frac{7}{2} = \frac{3}{2}$

$Option (3)$

Q 10 :

If $x^{2} + x + 1 = 0$ , then the value of ${(x + \frac{1}{x})}^{4} + {(x^{2} + \frac{1}{x^{2}})}^{4} + {(x^{3} + \frac{1}{x^{3}})}^{4} + . . . + {(x^{25} + \frac{1}{x^{25}})}^{4}$ is [2026]

128
162
145
175

1

2

3

4

(3)

$x^{2} + x + 1 = 0$

$\Rightarrow x = ω or ω^{2}$

$∴$ $α = ω, β = ω^{2}$

$= {(ω + ω^{2})}^{4} + {(ω^{2} + ω^{4})}^{4} + {(ω^{3} + ω^{6})}^{4} + \dots + {(ω^{25} + ω^{50})}^{4}$

$= [{(ω + ω^{2})}^{4} + {(ω^{2} + ω^{4})}^{4} + {(ω^{4} + ω^{8})}^{4} + \dots + {(ω^{25} + ω^{50})}^{4}]$ $+ [{(ω^{3} + ω^{6})}^{4} + {(ω^{6} + ω^{12})}^{4} + {(ω^{9} + ω^{18})}^{4} + \dots + {(ω^{24} + ω^{48})}^{4}]$

$= \underset{17 times}{\underset{⏟}{[1 + 1 + 1 + \dots + 1]}} + \underset{8 times}{\underset{⏟}{[{(1 + 1)}^{4} + {(1 + 1)}^{4} + \dots + {(1 + 1)}^{4}]}}$

$= 17 + 128$

$= 145$

Q 11 :

Let $S = {z \in ℂ : | \frac{z - 6 i}{z - 2 i} | = 1 and | \frac{z - 8 + 2 i}{z + 2 i} | = \frac{3}{5}} .$

Then $\sum_{z \in S} | z |^{2}$ is equal to [2026]

423
385
413
398

1

2

3

4

(2)

$Solving$ $| \frac{z - 6 i}{z - 2 i} |$ $= 1 \Rightarrow y = 4 \dots (1)$

$(where z = x + i y)$

$Now solving$ $| \frac{z - 8 + 2 i}{z + 2 i} |$ $= \frac{3}{5}$

$\Rightarrow x^{2} + y^{2} - 25 x + 4 y + 104 = 0 \dots (2)$

$Solving (1) & (2) \Rightarrow z = 17 + 4 i$ $&$ $8 + 4 i$

$\Rightarrow \sum | z |^{2} = {(17)}^{2} + {(4)}^{2} + {(8)}^{2} + {(4)}^{2} = 385$

Q 12 :

Let $α = \frac{- 1 + i \sqrt{3}}{2} and β = \frac{- 1 - i \sqrt{3}}{2}, i = \sqrt{- 1} .$

If ${(7 - 7 α + 9 β)}^{20} + {(9 + 7 α - 7 β)}^{20} + {(- 7 + 9 α + 7 β)}^{20} + {(14 + 7 α + 7 β)}^{20} = m^{10},$ then $m$ is __________ . [2026]

(49)

$(9 + 7 ω - 7 ω^{2}) + ω^{20} {(9 + 7 ω - 7 ω^{2})}^{20} + ω^{40} {(9 + 7 ω - 7 ω^{2})}^{20} + (14 + 7 {(ω + ω^{2}))}^{20}$

${(9 + 7 ω - 7 ω^{2})}^{20} (1 + ω + ω^{2}) + {(14 - 7)}^{20}$

$= 7^{20}$

$= {(49)}^{10}$

$Hence, M = 49$

Q 13 :

$Let A = {z \in ℂ : | z - 2 | \leq 4} and$

$B = {z \in ℂ : | z - 2 | + | z + 2 | = 5} .$

Then the max ${| z_{1} - z_{2} | : z_{1} \in A and z_{2} \in B}$ is: [2026]

9
8
$\frac{15}{2}$
$\frac{17}{2}$

1

2

3

4

(4)

$| z - 2 |$ $\leq 4 \Rightarrow {(x - 2)}^{2} + y^{2} \leq 16$

$| z - 2 |$ $+$ $| z + 2 |$ $= 5 \Rightarrow \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$

$\Rightarrow \frac{4 x^{2}}{25} + \frac{4 y^{2}}{9} = 1$

$Maximum value of$ $| z_{1} - z_{2} |$ $= 6 + \frac{5}{2} = \frac{17}{2}$

Q 14 :

Let $z = (1 + i) (1 + 2 i) (1 + 3 i)$ ... $(1 + n i),$ where $i = \sqrt{- 1}$ . If $| z |^{2} = 44200$ , then $n$ is equal to _________. [2026]

(5)

$| z |^{2}$ $= 2^{3} \cdot 5^{2} \cdot 13 \cdot 17$

$\prod_{r = 1}^{n} (1 + r^{2}) = 2^{3} \cdot 5^{2} \cdot 13 \cdot 17 = (2) \cdot (5) \cdot (2 \cdot 5) \cdot (17) \cdot (2 \cdot 13) = 2 \cdot 5 \cdot 10 \cdot 17 \cdot 26$

$so n = 5$

Q 15 :

If $z = \frac{\sqrt{3}}{2} + \frac{i}{2}, i = \sqrt{- 1}$ then ${(z^{201} - i)}^{8}$ is equal to [2026]

0
1
256
-1

1

2

3

4

(3)

$z = \cos \frac{π}{6} + i \sin \frac{π}{6}$

$z^{201} = \cos (201 \frac{π}{6}) + i \sin (201 \frac{π}{6}) = - i$

${(z^{201} - i)}^{8} = {(- 2 i)}^{8} = 256$

Q 16 :

Let z be the complex number satisfying |z − 5| ≤ 3 and having maximum positive principal argument.

Then $34 {|| \frac{5 z - 12}{5 i z + 16} ||}^{2}$ is equal to: [2026]

16
12
20
26

1

2

3

4

(3)

$| z - 5 |$ $\leq 3$

For $\arg (z)$ to be maximum, $z$ lies at $P$ .

$z \equiv (4 \cos θ, 4 \sin θ)$

$\equiv (4 \cdot \frac{4}{5}, 4 \cdot \frac{3}{5}) = (\frac{16}{5}, \frac{12}{5}) = \frac{16}{5} + \frac{12 i}{5}$

Now, $34 {|| \frac{5 z - 12}{5 i z + 16} ||}^{2} = 34 {|| \frac{(16 + 12 i) - 12}{(16 i - 12) + 16} ||}^{2}$

$= 34 {|| \frac{4 + 12 i}{16 i + 4} ||}^{2}$

$= 34 (\frac{16 + 144}{256 + 16})$

$= 34 (\frac{160}{272}) = 20$

Q 17 :

Let $S = {z \in ℂ : 4 z^{2} + \bar{z} = 0} .$ Then $\sum_{z \in S} | z |^{2}$ is equal to: [2026]

7/64
1/16
5/64
3/16

1

2

3

4

(4)

$4 z^{2} + \bar{z} = 0$

$Let z = x + i y$

$4 {(x + i y)}^{2} + x - i y = 0$

$4 x^{2} - 4 y^{2} + 8 x y i + x - i y = 0$

$4 x^{2} - 4 y^{2} + x = 0 & y (8 x - 1) = 0$

$\Rightarrow y = 0 or x = \frac{1}{8}$

$If y = 0, 4 x^{2} + x = 0$

$x = 0, \frac{- 1}{4}$

$∴$ $z_{1} = 0 + 0 i, | z_{1} |^{2} = 0$

$z_{2} = 0 - \frac{1}{4} i, | z_{2} |^{2} = \frac{1}{16}$

$If x = \frac{1}{8},$

$4 \cdot \frac{1}{64} - 4 y^{2} + \frac{1}{8} = 0$

$\Rightarrow 4 y^{2} = \frac{3}{16} \Rightarrow y = \pm \frac{\sqrt{3}}{8}$

$∴$ $z_{3} = \frac{1}{8} + \frac{\sqrt{3}}{8} i, | z_{3} |^{2} = \frac{1}{64} + \frac{3}{64} = \frac{1}{16}$

$z_{4} = \frac{1}{8} - \frac{\sqrt{3}}{8} i, | z_{4} |^{2} = \frac{1}{64} + \frac{3}{64} = \frac{1}{16}$

$∴$ $\sum_{i = 1}^{n} | z_{i} |^{2} = 0 + \frac{1}{16} + \frac{1}{16} + \frac{1}{16} = \frac{3}{16}$

Q 18 :

Let z be a complex number such that $| z - 6 | = 5 and | z + 2 - 6 i | = 5$ . Then the value of $z^{3} + 3 z^{2} - 15 z + 141$ is equal to [2026]

42
50
37
61

1

2

3

4

(2)

$Center of first circle C_{1} (6, 0), r_{1} = 5$

$Center of second circle C_{2} (- 2, 6), r_{2} = 5$

$∵$ $C_{1} C_{2} = r_{1} + r_{2}$

$∴$ $common point Z is mid point of C_{1} & C_{2}$

$∴$ $z = 2 + 3 i$

$∴$ $z^{2} = 4 z - 13$

$∴$ $z^{3} = 3 z - 52$

$∴$ $z^{3} + 3 z^{2} - 15 z + 141 = 50$

Q 19 :

Let $A = {z \in ℂ : 1 \leq | z - (1 + i) | \leq 2}$ and $B = {z \in A : | z - (1 - i) | = 1}$ . Then B:

is an empty set
contains exactly two elements
contains exactly three elements
is an infinite set

1

2

3

4

(4)

$A = {z \in C : 1 \leq | z - (1 + i) | \leq 2}$

$B = {z \in A : | z - (1 - i) | = 1}$ $A \cap B has infinite set.$

Q 20 :

$If \arg (\frac{z - 2}{z - 2 i}) = \frac{π}{4} then which of the following is correct?$

$| z |_{\min} = 0$
$| z |_{\min} = 2 (\sqrt{2} - 1)$
$| z |_{\max} = 2 (\sqrt{2} + 1)$
$| z |_{\max} = 4$

1

2

3

4

(3)

Q 21 :

Let $α \in R, z_{1}, z_{2}, z_{3}$ be three distinct complex numbers such that $| z_{1} |$ $=$ $| z_{2} |$ $=$ $| z_{3} |$ $= 3$ & $| (k z_{1} + z_{2}) - (k z_{2} + z_{3}) |_{\min}$ $= α$ $| z_{3} - z_{2} |$ $| z_{3} - z_{1} |,$ $\forall k \in R - {0}$ then $36 α =____$

(6)

$| k z_{1} + (1 - k) z_{2} - z_{3} |$ $= height from vertex C .$

$∴$ $Area = \frac{1}{2} (A B) (height) = \frac{1}{2}$ $| z_{1} - z_{3} |$ $| z_{2} - z_{3} |$ $\sin C$

$α = \frac{\sin C}{A B} = \frac{1}{2 R} = \frac{1}{6}$

Q 22 :

Let $z_{1}$ and $z_{2}$ be two complex numbers such that $| z_{1} |$ $= 1$ and $| z_{2} |$ $= 10$ . If $θ = \arg (\frac{z_{1} - z_{2}}{z_{2}})$ then maximum value of $\tan^{2} θ$ is

$\frac{1}{10}$
$\frac{1}{100}$
$\frac{1}{99}$
$\frac{10}{99}$

1

2

3

4

(3)

$Put \frac{z_{1}}{z_{2}} = z$

$∴$ $| \frac{z_{1}}{z_{2}} |$ $=$ $| z |$ $\Rightarrow$ $| z |$ $= \frac{1}{10}$

$Also, θ = \arg (\frac{z_{1}}{z_{2}} - 1) = \arg (z - 1)$

$∴$ $\tan^{2} θ_{\max} = \frac{1 / 100}{99 / 100} = \frac{1}{99}$

Q 23 :

Let x and y be real numbers such that $50 (\frac{2 x}{1 + 3 i} - \frac{y}{1 - 2 i}) = 31 + 17 i, i = \sqrt{- 1}$ . Then the value of 10(x – 3y) is: [2026]

20
31
35
75

1

2

3

4

(4)

Given : $50 (\frac{2 x}{1 + 3 i} - \frac{y}{1 - 2 i}) = 31 + 17 i$

$\Rightarrow 50 (\frac{2 x (1 - 3 i)}{1 + 9} - \frac{y (1 + 2 i)}{5}) = 31 + 17 i$

$\Rightarrow 10 (x (1 - 3 i) - y (1 + 2 i) = 31 + 17 i$

$\Rightarrow 10 (x - y) + 10 i (- 3 x - 2 y) = 31 + 17 i$

On comparing real and imaginary parts, we get

$x - y = \frac{31}{10}$ ... (i)

$- 3 x - 2 y = \frac{17}{10}$ ... (ii)

Solving (i) & (ii), we get

$x = \frac{9}{10} and y = \frac{- 11}{5}$

Now, $10 (x - 3 y) = 10 (\frac{9}{10} - 3 (- \frac{11}{5}))$

= 9 + 66 = 75

Q 24 :

The value of $\sum_{r = 1}^{5} {(x^{r} + \frac{1}{x^{r}})}^{2},$ where $x$ satisfies the equation $x^{2} + x + 1 = 0$ , is

5
6
7
8

1

2

3

4

(4)

$x^{2} + x + 1 = 0 \Rightarrow x + \frac{1}{x} = - 1$

$∴$ $\sum_{r = 1}^{5} {(x^{r} + \frac{1}{x^{r}})}^{2} = 8$

Q 25 :

Let $ω$ be a complex number such that $2 ω + 1 = z$ , where $z = \sqrt{- 3}$ . If $| \begin{matrix} 1 & 1 & 1 \\ 1 & - ω^{2} - 1 & ω^{2} \\ 1 & ω^{2} & ω^{7} \end{matrix} |$ $= 3 k$ , then $3 k$ is equal to

$- 3 z$
$3 z$
$- 3$
$+ 3$

1

2

3

4

(1)

$Given, 2 ω + 1 = z$ $2 ω + 1 = \sqrt{- 3}$ $[∴ z = \sqrt{- 3}] \Rightarrow ω = \frac{- 1 + \sqrt{3} i}{2}$

$Since, ω is cube root of unity.$ $∴$ $ω^{2} = \frac{- 1 - \sqrt{3} i}{2} and ω^{3 n} = 1$

$Now,$ $| \begin{matrix} 1 & 1 & 1 \\ 1 & - ω^{2} - 1 & ω^{2} \\ 1 & ω^{2} & ω^{7} \end{matrix} |$ $= 3 k$

$\Rightarrow$ $| \begin{matrix} 1 & 1 & 1 \\ 1 & ω & ω^{2} \\ 1 & ω^{2} & ω \end{matrix} |$ $= 3 k$

$[∴ 1 + ω + ω^{2} = 0 and ω^{7} = {(ω^{3})}^{2} \cdot ω = ω]$

$On applying R_{1} \to R_{1} + R_{2} + R_{3}, we get$ $| \begin{matrix} 3 & 1 + ω + ω^{2} & 1 + ω + ω^{2} \\ 1 & ω & ω^{2} \\ 1 & ω^{2} & ω \end{matrix} |$ $= 3 k$

$=$ $| \begin{matrix} 3 & 0 & 0 \\ 1 & ω & ω^{2} \\ 1 & ω^{2} & ω \end{matrix} |$ $= 3 k$

$\Rightarrow 3 (ω^{2} - ω^{4}) = 3 k$

$\Rightarrow (ω^{2} - ω) = k$

$∴$ $k = (\frac{- 1 - \sqrt{3} i}{2}) - (\frac{- 1 + \sqrt{3} i}{2}) = - \sqrt{3} i = - z$

Q 26 :

Let $A = {θ \in (0, π) : \frac{1 + 2 i \sin θ}{1 - i \sin θ} is purely imaginary}$ . Then sum of the elements of A is _____

$π$
$2 π$
$4 π$
$3 π$

1

2

3

4

(1)

$Real part = 0 \Rightarrow 1 - 2 \sin^{2} θ = 0 \Rightarrow \sin θ = \pm \frac{1}{\sqrt{2}},$ $θ = \frac{π}{4}, \frac{3 π}{4}$

Topic Question Set

Q 1 :

Let $A = {θ \in (0, 2 π) : \frac{1 + 2 i \sin θ}{1 - i \sin θ} is purely imaginary} .$ Then the sum of the elements in A is [2023]

Q 2 :

If the set $R = {(a, b) : a + 5 b = 42, a, b \in N}$ has $m$ elements and $\sum_{n = 1}^{m} (1 - i^{n!}) = x + i y,$ where $i = \sqrt{- 1},$ then the value of $m + x + y$ is [2024]

Q 3 :

Let $z$ be a complex number such that the real part of $\frac{z - 2 i}{z + 2 i}$ is zero. Then, the maximum value of $| z - (6 + 8 i) |$ is equal to [2024]

Q 4 :

If $z$ is a complex number, then the number of common roots of the equations $z^{1985} + z^{100} + 1 = 0$ and $z^{3} + 2 z^{2} + 2 z + 1 = 0,$ is equal to [2024]

Q 5 :

Among the statements

(S1) : The set ${z \in C - {i} : | z | = 1 and \frac{z - i}{z + i} is purely real}$ contains exactly two elements, and

(S2) : The set ${z \in C - {- 1} : | z | = 1 and \frac{z - 1}{z + 1} is purely imagin ary}$ contains infinitely many elements. [2025]

Q 6 :

Let $a \neq b$ be two non-zero real numbers. Then the number of elements in the set $X = {z \in ℂ : Re (a z^{2} + b z) = a and Re (b z^{2} + a z) = b}$ is equal to [2023]

Q 8 :

$Let S = {z \in ℂ - {i, 2 i} : \frac{z^{2} + 8 i z - 15}{z^{2} - 3 i z - 2} \in ℝ} .$

$If α - \frac{13}{11} i \in S, α \in ℝ - {0}, then 242 α^{2} is equal to$ __________. [2023]

Q 9 :

Let $S = {z : 3 \leq | 2 z - 3 (1 + i) | \leq 7}$ be a set of complex numbers. Then $\min_{z \in S}$ $| (z + \frac{1}{2} (5 + 3 i)) |$ is equal to: [2026]

Q 10 :

If $x^{2} + x + 1 = 0$ , then the value of ${(x + \frac{1}{x})}^{4} + {(x^{2} + \frac{1}{x^{2}})}^{4} + {(x^{3} + \frac{1}{x^{3}})}^{4} + . . . + {(x^{25} + \frac{1}{x^{25}})}^{4}$ is [2026]

Q 11 :

Let $S = {z \in ℂ : | \frac{z - 6 i}{z - 2 i} | = 1 and | \frac{z - 8 + 2 i}{z + 2 i} | = \frac{3}{5}} .$

Then $\sum_{z \in S} | z |^{2}$ is equal to [2026]

Q 12 :

Let $α = \frac{- 1 + i \sqrt{3}}{2} and β = \frac{- 1 - i \sqrt{3}}{2}, i = \sqrt{- 1} .$

If ${(7 - 7 α + 9 β)}^{20} + {(9 + 7 α - 7 β)}^{20} + {(- 7 + 9 α + 7 β)}^{20} + {(14 + 7 α + 7 β)}^{20} = m^{10},$ then $m$ is __________ . [2026]

Q 13 :

$Let A = {z \in ℂ : | z - 2 | \leq 4} and$

$B = {z \in ℂ : | z - 2 | + | z + 2 | = 5} .$

Then the max ${| z_{1} - z_{2} | : z_{1} \in A and z_{2} \in B}$ is: [2026]

Q 14 :

Let $z = (1 + i) (1 + 2 i) (1 + 3 i)$ ... $(1 + n i),$ where $i = \sqrt{- 1}$ . If $| z |^{2} = 44200$ , then $n$ is equal to _________. [2026]

Q 15 :

If $z = \frac{\sqrt{3}}{2} + \frac{i}{2}, i = \sqrt{- 1}$ then ${(z^{201} - i)}^{8}$ is equal to [2026]

Q 16 :

Let z be the complex number satisfying |z − 5| ≤ 3 and having maximum positive principal argument.

Then $34 {|| \frac{5 z - 12}{5 i z + 16} ||}^{2}$ is equal to: [2026]

Q 17 :

Let $S = {z \in ℂ : 4 z^{2} + \bar{z} = 0} .$ Then $\sum_{z \in S} | z |^{2}$ is equal to: [2026]

Q 18 :

Let z be a complex number such that $| z - 6 | = 5 and | z + 2 - 6 i | = 5$ . Then the value of $z^{3} + 3 z^{2} - 15 z + 141$ is equal to [2026]

Q 19 :

Let $A = {z \in ℂ : 1 \leq | z - (1 + i) | \leq 2}$ and $B = {z \in A : | z - (1 - i) | = 1}$ . Then B:

Q 20 :

$If \arg (\frac{z - 2}{z - 2 i}) = \frac{π}{4} then which of the following is correct?$

Q 21 :

Let $α \in R, z_{1}, z_{2}, z_{3}$ be three distinct complex numbers such that $| z_{1} |$ $=$ $| z_{2} |$ $=$ $| z_{3} |$ $= 3$ & $| (k z_{1} + z_{2}) - (k z_{2} + z_{3}) |_{\min}$ $= α$ $| z_{3} - z_{2} |$ $| z_{3} - z_{1} |,$ $\forall k \in R - {0}$ then $36 α =____$

Q 22 :

Let $z_{1}$ and $z_{2}$ be two complex numbers such that $| z_{1} |$ $= 1$ and $| z_{2} |$ $= 10$ . If $θ = \arg (\frac{z_{1} - z_{2}}{z_{2}})$ then maximum value of $\tan^{2} θ$ is

Q 23 :

Let x and y be real numbers such that $50 (\frac{2 x}{1 + 3 i} - \frac{y}{1 - 2 i}) = 31 + 17 i, i = \sqrt{- 1}$ . Then the value of 10(x – 3y) is: [2026]

Q 24 :

The value of $\sum_{r = 1}^{5} {(x^{r} + \frac{1}{x^{r}})}^{2},$ where $x$ satisfies the equation $x^{2} + x + 1 = 0$ , is

Q 25 :

Let $ω$ be a complex number such that $2 ω + 1 = z$ , where $z = \sqrt{- 3}$ . If $| \begin{matrix} 1 & 1 & 1 \\ 1 & - ω^{2} - 1 & ω^{2} \\ 1 & ω^{2} & ω^{7} \end{matrix} |$ $= 3 k$ , then $3 k$ is equal to

Q 26 :

Let $A = {θ \in (0, π) : \frac{1 + 2 i \sin θ}{1 - i \sin θ} is purely imaginary}$ . Then sum of the elements of A is _____

Want Us to Email you About Special Offers & Updates?

Topic Question Set

Q 1 : Let A={θ∈(0,2π):1+2isinθ1-isinθis purely imaginary}. Then the sum of the elements in A is [2023]

Q 2 : If the set R={(a,b):a+5b=42, a,b∈N} has m elements and ∑n=1m(1-in!)=x+iy, where i=-1, then the value of m+x+y is [2024]

Q 3 : Let z be a complex number such that the real part of z-2iz+2i is zero. Then, the maximum value of |z-(6+8i)| is equal to [2024]

Q 4 : If z is a complex number, then the number of common roots of the equations z1985+z100+1=0 and z3+2z2+2z+1=0, is equal to [2024]

Q 5 : Among the statements (S1) : The set {z∈C–{i} : |z|=1 and z–iz+i is purely real} contains exactly two elements, and (S2) : The set {z∈C–{–1} : |z|=1 and z–1z+1 is purely imaginary} contains infinitely many elements. [2025]

Q 6 : Let a≠b be two non-zero real numbers. Then the number of elements in the set X={z∈ℂ:Re(az2+bz)=a and Re(bz2+az)=b} is equal to [2023]

Q 7 : For two non-zero complex numbers z1 and z2, if Re(z1z2)=0 and Re(z1+z2)=0, then which of the following are possible? (A) Im(z1)>0 and Im(z2)>0 (B) Im(z1)<0 and Im(z2)>0 (C) Im(z1)>0 and Im(z2)<0 (D) Im(z1)<0 and Im(z2)<0 Choose the correct answer from the options given below: [2023]

Q 8 : Let S={z∈ℂ-{i,2i}: z2+8iz-15z2-3iz-2∈ℝ}. If α-1311i∈S, α∈ℝ-{0}, then 242α2 is equal to __________. [2023]

Q 9 : Let S={z:3≤|2z-3(1+i)|≤7} be a set of complex numbers. Then minz∈S|(z+12(5+3i))| is equal to: [2026]

Q 10 : If x2+x+1=0 , then the value of (x+1x)4+(x2+1x2)4+(x3+1x3)4+...+(x25+1x25)4 is [2026]

Q 11 : Let S={z∈ℂ:|z-6iz-2i|=1 and |z-8+2iz+2i|=35}. Then ∑z∈S|z|2 is equal to [2026]

Q 12 : Let α=-1+i32 and β=-1-i32, i=-1. If (7-7α+9β)20+(9+7α-7β)20+(-7+9α+7β)20+(14+7α+7β)20=m10, then m is __________ . [2026]

Q 13 : Let A={z∈ℂ:|z-2|≤4} and B={z∈ℂ: |z-2|+|z+2|=5}. Then the max {|z1-z2|:z1∈A and z2∈B} is: [2026]

Q 14 : Let z=(1+i)(1+2i)(1+3i)...(1+ni), where i=-1. If |z|2=44200, then n is equal to _________. [2026]

Q 15 : If z=32+i2, i=-1 then (z201-i)8 is equal to [2026]

Q 16 : Let z be the complex number satisfying |z − 5| ≤ 3 and having maximum positive principal argument. Then 34|5z-125iz+16|2 is equal to: [2026]

Q 17 : Let S={z∈ℂ:4z2+z¯=0}. Then ∑z∈S|z|2 is equal to: [2026]

Q 18 : Let z be a complex number such that |z-6|=5 and |z+2-6i|=5. Then the value of z3+3z2-15z+141 is equal to [2026]

Q 19 : Let A={z∈ℂ:1≤|z-(1+i)|≤2} and B={z∈A:|z-(1-i)|=1}. Then B:

Q 20 : If arg(z-2z-2i)=π4 then which of the following is correct?

Q 21 : Let α∈R, z1,z2,z3 be three distinct complex numbers such that |z1|=|z2|=|z3|=3 & |(kz1+z2)-(kz2+z3)|min=α|z3-z2||z3-z1|,∀k∈R-{0} then 36α=____

Q 22 : Let z1 and z2 be two complex numbers such that |z1|=1 and |z2|=10. If θ=arg(z1-z2z2) then maximum value of tan2θ is

Q 23 : Let x and y be real numbers such that 50(2x1+3i–y1–2i)=31+17i, i=–1. Then the value of 10(x – 3y) is: [2026]

Q 24 : The value of ∑r=15(xr+1xr)2, where x satisfies the equation x2+x+1=0, is

Q 25 : Let ω be a complex number such that 2ω+1=z, where z=-3. If |1111-ω2-1ω21ω2ω7|=3k, then 3k is equal to

Q 26 : Let A={θ∈(0,π):1+2isinθ1-isinθ is purely imaginary}. Then sum of the elements of A is _____

Want Us to Email you About Special Offers & Updates?

Q 1 :

Let $A = {θ \in (0, 2 π) : \frac{1 + 2 i \sin θ}{1 - i \sin θ} is purely imaginary} .$ Then the sum of the elements in A is [2023]

Q 2 :

If the set $R = {(a, b) : a + 5 b = 42, a, b \in N}$ has $m$ elements and $\sum_{n = 1}^{m} (1 - i^{n!}) = x + i y,$ where $i = \sqrt{- 1},$ then the value of $m + x + y$ is [2024]

Q 3 :

Let $z$ be a complex number such that the real part of $\frac{z - 2 i}{z + 2 i}$ is zero. Then, the maximum value of $| z - (6 + 8 i) |$ is equal to [2024]

Q 4 :

If $z$ is a complex number, then the number of common roots of the equations $z^{1985} + z^{100} + 1 = 0$ and $z^{3} + 2 z^{2} + 2 z + 1 = 0,$ is equal to [2024]

Q 5 :

Among the statements

(S1) : The set ${z \in C - {i} : | z | = 1 and \frac{z - i}{z + i} is purely real}$ contains exactly two elements, and

(S2) : The set ${z \in C - {- 1} : | z | = 1 and \frac{z - 1}{z + 1} is purely imagin ary}$ contains infinitely many elements. [2025]

Q 6 :

Let $a \neq b$ be two non-zero real numbers. Then the number of elements in the set $X = {z \in ℂ : Re (a z^{2} + b z) = a and Re (b z^{2} + a z) = b}$ is equal to [2023]

Q 8 :

$Let S = {z \in ℂ - {i, 2 i} : \frac{z^{2} + 8 i z - 15}{z^{2} - 3 i z - 2} \in ℝ} .$

$If α - \frac{13}{11} i \in S, α \in ℝ - {0}, then 242 α^{2} is equal to$ __________. [2023]

Q 9 :

Let $S = {z : 3 \leq | 2 z - 3 (1 + i) | \leq 7}$ be a set of complex numbers. Then $\min_{z \in S}$ $| (z + \frac{1}{2} (5 + 3 i)) |$ is equal to: [2026]

Q 10 :

If $x^{2} + x + 1 = 0$ , then the value of ${(x + \frac{1}{x})}^{4} + {(x^{2} + \frac{1}{x^{2}})}^{4} + {(x^{3} + \frac{1}{x^{3}})}^{4} + . . . + {(x^{25} + \frac{1}{x^{25}})}^{4}$ is [2026]

Q 11 :

Let $S = {z \in ℂ : | \frac{z - 6 i}{z - 2 i} | = 1 and | \frac{z - 8 + 2 i}{z + 2 i} | = \frac{3}{5}} .$

Then $\sum_{z \in S} | z |^{2}$ is equal to [2026]

Q 12 :

Let $α = \frac{- 1 + i \sqrt{3}}{2} and β = \frac{- 1 - i \sqrt{3}}{2}, i = \sqrt{- 1} .$

If ${(7 - 7 α + 9 β)}^{20} + {(9 + 7 α - 7 β)}^{20} + {(- 7 + 9 α + 7 β)}^{20} + {(14 + 7 α + 7 β)}^{20} = m^{10},$ then $m$ is __________ . [2026]

Q 13 :

$Let A = {z \in ℂ : | z - 2 | \leq 4} and$

$B = {z \in ℂ : | z - 2 | + | z + 2 | = 5} .$

Then the max ${| z_{1} - z_{2} | : z_{1} \in A and z_{2} \in B}$ is: [2026]

Q 14 :

Let $z = (1 + i) (1 + 2 i) (1 + 3 i)$ ... $(1 + n i),$ where $i = \sqrt{- 1}$ . If $| z |^{2} = 44200$ , then $n$ is equal to _________. [2026]

Q 15 :

If $z = \frac{\sqrt{3}}{2} + \frac{i}{2}, i = \sqrt{- 1}$ then ${(z^{201} - i)}^{8}$ is equal to [2026]

Q 16 :

Let z be the complex number satisfying |z − 5| ≤ 3 and having maximum positive principal argument.

Then $34 {|| \frac{5 z - 12}{5 i z + 16} ||}^{2}$ is equal to: [2026]

Q 17 :

Let $S = {z \in ℂ : 4 z^{2} + \bar{z} = 0} .$ Then $\sum_{z \in S} | z |^{2}$ is equal to: [2026]

Q 18 :

Let z be a complex number such that $| z - 6 | = 5 and | z + 2 - 6 i | = 5$ . Then the value of $z^{3} + 3 z^{2} - 15 z + 141$ is equal to [2026]

Q 19 :

Let $A = {z \in ℂ : 1 \leq | z - (1 + i) | \leq 2}$ and $B = {z \in A : | z - (1 - i) | = 1}$ . Then B:

Q 20 :

$If \arg (\frac{z - 2}{z - 2 i}) = \frac{π}{4} then which of the following is correct?$

Q 21 :

Let $α \in R, z_{1}, z_{2}, z_{3}$ be three distinct complex numbers such that $| z_{1} |$ $=$ $| z_{2} |$ $=$ $| z_{3} |$ $= 3$ & $| (k z_{1} + z_{2}) - (k z_{2} + z_{3}) |_{\min}$ $= α$ $| z_{3} - z_{2} |$ $| z_{3} - z_{1} |,$ $\forall k \in R - {0}$ then $36 α =____$

Q 22 :

Let $z_{1}$ and $z_{2}$ be two complex numbers such that $| z_{1} |$ $= 1$ and $| z_{2} |$ $= 10$ . If $θ = \arg (\frac{z_{1} - z_{2}}{z_{2}})$ then maximum value of $\tan^{2} θ$ is

Q 23 :

Let x and y be real numbers such that $50 (\frac{2 x}{1 + 3 i} - \frac{y}{1 - 2 i}) = 31 + 17 i, i = \sqrt{- 1}$ . Then the value of 10(x – 3y) is: [2026]

Q 24 :

The value of $\sum_{r = 1}^{5} {(x^{r} + \frac{1}{x^{r}})}^{2},$ where $x$ satisfies the equation $x^{2} + x + 1 = 0$ , is

Q 25 :

Let $ω$ be a complex number such that $2 ω + 1 = z$ , where $z = \sqrt{- 3}$ . If $| \begin{matrix} 1 & 1 & 1 \\ 1 & - ω^{2} - 1 & ω^{2} \\ 1 & ω^{2} & ω^{7} \end{matrix} |$ $= 3 k$ , then $3 k$ is equal to

Q 26 :

Let $A = {θ \in (0, π) : \frac{1 + 2 i \sin θ}{1 - i \sin θ} is purely imaginary}$ . Then sum of the elements of A is _____