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]

$2 π$
$4 π$
$π$
$3 π$

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(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

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(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]

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

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(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

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(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

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(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

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(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

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(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]

0%

(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 $α^{2} = (\frac{- 13}{11} - 1) (\frac{- 13}{11} - 2)$

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

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

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]

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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]

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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]