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

Q 1 :

Let $α$ and $β$ be the sum and the product of all the non-zero solutions of the equation ${(\bar{z})}^{2} + | z | = 0, z \in C .$ Then $4 (α^{2} + β^{2})$ is equal to [2024]

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(3)

We have, ${(\bar{z})}^{2} + | z | = 0$ ...(i)

Let $z = x + i y \Rightarrow \bar{z} = x - i y$

From (i), ${(x - i y)}^{2} + | x + i y | = 0$

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

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

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

Case I: $x = 0$ and $y \neq 0$

$\Rightarrow - y^{2} + | y | = 0 \Rightarrow | y | = y^{2} \Rightarrow y = \pm 1$

Case II: $x \neq 0$ and $y = 0$

$\Rightarrow x^{2} + | x | = 0$

$\Rightarrow x = 0$ only $(∵ x \in R)$

$\Rightarrow x = 0, y = 0$ (Rejected) $\Rightarrow z = i, - i$ are solutions.

Hence, $α = i + (- i) = 0$ and $β = i (- i) = - i^{2} = 1$

Hence, $4 (α^{2} + β^{2}) = 4 (0 + 1) = 4$

Q 2 :

Consider the following two statements

Statement I : For any two non-zero complex numbers $z_{1}, z_{2},$ $(| z_{1} | + | z_{2} |)$ $| \frac{z_{1}}{| z_{1} |} + \frac{z_{2}}{| z_{2} |} |$ $\leq 2 (| z_{1} | + | z_{2} |),$ and

Statement II : If $x, y, z$ are three distinct complex numbers and a, b, c are three positive real numbers such that $\frac{a}{| y - z |} = \frac{b}{| z - x |} = \frac{c}{| x - y |},$ then $\frac{a^{2}}{y - z} + \frac{b^{2}}{z - x} + \frac{c^{2}}{x - y} = 1 .$

Between the above two statements, [2024]

Statement I is correct but Statement II is incorrect.
Both Statement I and Statement II are incorrect.
Both Statement I and Statement II are correct.
Statement I is incorrect but Statement II is correct.

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(1)

We have, $(| z_{1} | + | z_{2} |)$ $| \frac{z_{1}}{| z_{1} |} + \frac{z_{2}}{| z_{2} |} |$ $\leq (| z_{1} | + | z_{2} |) (\frac{| z_{1} |}{| z_{1} |} + \frac{| z_{2} |}{| z_{2} |})$

$\Rightarrow (| z_{1} | + | z_{2} |)$ $| \frac{z_{1}}{| z_{1} |} + \frac{z_{2}}{| z_{2} |} |$ $\leq 2 (| z_{1} | + | z_{2} |)$

Hence, statement-I is correct.

Now, let $\frac{a}{| y - z |} = \frac{b}{| z - x |} = \frac{c}{| x - y |} = λ$

$\Rightarrow a = λ | y - z |, b = λ | z - x |, c = λ | x - y |$

$\Rightarrow a^{2} = λ^{2} (y - z) (\bar{y} - \bar{z})$ [ $∵$ $| x - y |^{2} = (x - y) (\bar{x - y})$ for any complex number $x$ and $y$ ]

$b^{2} = λ^{2} (z - x) (\bar{z} - \bar{x})$

and $c^{2} = λ^{2} (x - y) (\bar{x} - \bar{y})$

Now, $\frac{a^{2}}{y - z} + \frac{b^{2}}{z - x} + \frac{c^{2}}{x - y} = λ^{2} (\bar{y} - \bar{z} + \bar{z} - \bar{x} + \bar{x} - \bar{y}) = 0$

Hence, statement-II is incorrect.

Q 3 :

Let $z$ be a complex number such that $| z + 2 | = 1$ and $Im$ $(\frac{z + 1}{z + 2})$ $= \frac{1}{5} .$ Then the value of $| R e (\bar{z + 2}) |$ is [2024]

$\frac{2 \sqrt{6}}{5}$
$\frac{\sqrt{6}}{5}$
$\frac{1 + \sqrt{6}}{5}$
$\frac{24}{5}$

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(1)

Let $z = x + i y$

$\Rightarrow | x + i y + 2 | = 1 \Rightarrow {(x + 2)}^{2} + y^{2} = 1$ ...(i)

Also, $Im$ $(\frac{(x + 1) + i y}{(x + 2) + i y})$ $= \frac{1}{5}$

$\Rightarrow Im$ $(\frac{[(x + 1) + i y] [(x + 2) - i y]}{{(x + 2)}^{2} + y^{2}})$ $= \frac{1}{5}$

$\Rightarrow \frac{(x + 2) y - (x + 1) y}{{(x + 2)}^{2} + y^{2}} = \frac{1}{5}$

$\Rightarrow \frac{y}{1} = \frac{1}{5}$ [Using (i)]

$\Rightarrow {(x + 2)}^{2} + \frac{1}{25} = 1 \Rightarrow {(x + 2)}^{2} = 1 - \frac{1}{25}$

$\Rightarrow (x + 2) = \pm \frac{2 \sqrt{6}}{5}$

$∴$ $| Re (\bar{z + 2}) |$ $=$ $| Re[(x + 2) - i y)] |$ $=$ $| (x + 2) |$ $= \frac{2 \sqrt{6}}{5} .$

Q 4 :

The sum of all possible values of $θ \in [- π, 2 π]$ , for which $\frac{1 + i \cos θ}{1 - 2 i \cos θ}$ is purely imaginary, is equal to: [2024]

$2 π$
$4 π$
$3 π$
$5 π$

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(3)

We have, $\frac{1 + i \cos θ}{1 - 2 i \cos θ}$ is purely imaginary.

$∴ \frac{1 + i \cos θ}{1 - 2 i \cos θ} + \frac{1 - i \cos θ}{1 + 2 i \cos θ} = 0$ $(∵ z + \bar{z} = 0)$

$\Rightarrow \frac{1 + i \cos θ + 2 i \cos θ - 2 \cos^{2} θ + 1 - i \cos θ - 2 i \cos θ - 2 \cos^{2} θ}{1 + 4 \cos^{2} θ} = 0$

$\Rightarrow 2 - 4 \cos^{2} θ = 0 \Rightarrow \cos^{2} θ = \frac{1}{2}$

$θ = n π \pm \frac{π}{4}$

$θ$ can be $\frac{π}{4}, \frac{- π}{4}, \frac{3 π}{4}, \frac{- 3 π}{4}, \frac{5 π}{4}, \frac{7 π}{4}$ $(∵ θ \in [- π, 2 π])$

$∴$ Required $= \frac{π}{4} - \frac{π}{4} + \frac{3 π}{4} - \frac{3 π}{4} + \frac{5 π}{4} + \frac{7 π}{4} = 3 π$

Q 5 :

Let $S = {z \in C : | z - 1 | = 1$ and $(\sqrt{2} - 1) (z + \bar{z}) - i (z - \bar{z}) = 2 \sqrt{2}} .$ Let $z_{1}, z_{2} \in S$ be such that $| z_{1} |$ $=$ $\max_{z \in S}$ $| z |$ and $| z_{2} |$ $=$ $\min_{z \in S}$ $| z | .$ Then $| \sqrt{2} z_{1} - z_{2} |^{2}$ equals: [2024]

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(2)

We have, $S = {z \in C : | z - 1 | = 1$ and $(\sqrt{2} - 1) (z + \bar{z}) - i (z - \bar{z}) = 2 \sqrt{2}}$

Let $z = x + i y$ and $\bar{z} = x - i y$

${(x - 1)}^{2} + y^{2} = 1$ ...(i)

$(\sqrt{2} - 1) x + y = \sqrt{2}$ ...(ii)

Equation (i) and (ii) intersect at (1, 1) and $(1 + \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})$

$∴$ $S = {(1, 1), (1 + \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})}$

$| z_{1} |$ $= \max_{z \in S}$ $| z |$ at $(1 + \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})$

$∴$ $z_{1} = 1 + \frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}}$

$z_{2} = 1 + i$

$∴$ $\sqrt{2} z + 1 + i - 1 - i = \sqrt{2}$

$∴$ $| \sqrt{2} z_{1} - z_{2} |^{2} = 2$

Q 6 :

If $S = {z \in C :$ $| z - i |$ $=$ $| z + i |$ $=$ $| z - 1 |$ $}$ then $n (S)$ is: [2024]

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(1)

Given, $| z - i |$ $=$ $| z + i |$ $=$ $| z - 1 |$ ...(i)

Putting $z = x + i y$ in (i), we get $| x + i y - i |$ $=$ $| x + i y + i |$ $=$ $| x + i y - 1 |$

Now, $| x + i (y - 1) |$ $=$ $| x - 1 + i y |$

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

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

$\Rightarrow x - y = 0$ ...(ii)

Again, $| x + i (y + 1) |$ $=$ $| x - 1 + i y |$

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

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

$\Rightarrow x + y = 0$ ...(iii)

From (ii) and (iii), we get $x = 0$ and $y = 0$

Thus, $z = 0 + i 0 ∴ n (S) = 1$

Q 7 :

If $z = \frac{1}{2} - 2 i$ is such that $| z + 1 |$ $= α z + β (1 + i),$ $i = \sqrt{- 1}$ and $α, β \in R,$ then $α + β$ is equal to [2024]

- 1
- 4
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(3)

Given, $z = \frac{1}{2} - 2 i$ ...(i)

and $| z + 1 | = α z + β (1 + i)$ ...(ii)

From (i) and (ii), we get:

$| \frac{1}{2} - 2 i + 1 |$ $= α (\frac{1}{2} - 2 i) + β (1 + i)$

$\Rightarrow$ $| \frac{3}{2} - 2 i |$ $= α (\frac{1}{2} - 2 i) + β (1 + i)$

$\Rightarrow \sqrt{\frac{9}{4} + 4} = \frac{α}{2} - 2 α i + β + β i$ $[∵$ $| z |$ $= \sqrt{x^{2} + y^{2}}]$

$\Rightarrow \sqrt{\frac{25}{4}} = \frac{α}{2} + β + i (- 2 α + β) \Rightarrow \frac{5}{2} = \frac{α}{2} + β + i (- 2 α + β)$

$\Rightarrow \frac{5}{2} = \frac{α}{2} + β$ ...(iii) and $- 2 α + β = 0 \Rightarrow β = 2 α$ ...(iv)

Solving (iii) and (iv), we get $α = 1$ and $β = 2$

$∴$ $α + β = 1 + 2 = 3$

Q 8 :

If $z = x + i y, x y \neq 0,$ satisfies the equation $z^{2} + i \bar{z} = 0,$ then $| z^{2} |$ is equal to [2024]

$9$
$\frac{1}{4}$
$1$
$4$

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(3)

We have, $z^{2} + i \bar{z} = 0$

$\Rightarrow z^{2} = - i \bar{z} \Rightarrow$ $| z^{2} |$ $=$ $| - i \bar{z} |$

$\Rightarrow | z^{2} | = | - i | | \bar{z} | \Rightarrow | z^{2} | = 1 | z |$

$\Rightarrow | z |^{2} - | z | = 0$ $[∴ | \bar{z} | = | z |]$

$\Rightarrow$ $| z |$ $(| z | - 1) = 0$

$\Rightarrow | z | = 0 (not acceptable) or | z | = 1 \Rightarrow | z |^{2} = 1$

Q 9 :

Let $z_{1}$ and $z_{2}$ be two complex numbers such that $z_{1} + z_{2} = 5$ and $z_{1}^{3} + z_{2}^{3} = 20 + 15 i .$ Then $| z_{1}^{4} + z_{2}^{4} |$ equals [2024]

$30 \sqrt{3}$
$15 \sqrt{15}$
$75$
$25 \sqrt{3}$

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(3)

Given, $z_{1} + z_{2} = 5 \Rightarrow {(z_{1} + z_{2})}^{3} = 5^{3}$

$\Rightarrow 20 + 15 i + 3 z_{1} z_{2} (5) = 125$

$\Rightarrow z_{1} z_{2} = 7 - i$

Now $z_{1}^{4} + z_{2}^{4} = {[{(z_{1} + z_{2})}^{2} - 2 z_{1} z_{2}]}^{2} - 2 z_{1}^{2} z_{2}^{2}$

$= {[25 - 2 (7 - i)]}^{2} - 2 {(7 - i)}^{2}$

$= 625 + 4 {(7 - i)}^{2} - 100 (7 - i) - 2 {(7 - i)}^{2}$

$= 625 + (7 - i) [2 (7 - i) - 100]$

$= 625 + (7 - i) [14 - 2 i - 100]$

$= 625 + (7 - i) [- 86 - 2 i]$

$= 625 - 602 - 14 i + 86 i - 2 = 21 + 72 i$

$∴$ $| z_{1}^{4} + z_{2}^{4} | = \sqrt{441 + 5184} = \sqrt{5625} = 75$

Q 10 :

Let $P = {z \in C : | z + 2 - 3 i | \leq 1}$ and

$Q = {z \in C : z (1 + i) + \bar{z} (1 - i) \leq - 8} .$

Let in $P \cap Q, | z - 3 + 2 i |$ be maximum and minimum at $z_{1}$ and $z_{2}$ respectively. If $| z_{1} |^{2} + 2 | z_{2} |^{2} = α + β \sqrt{2}$ , where $α, β$ are integers, then $α + β$ equals________ . [2024]

(36)

Q 11 :

Let $A = {z \in C : | z - 2 - i | = 3}$ , $B = {z \in C : R e (z - i z) = 2}$ and $S = A \cap B$ . Then $\sum_{z \in S} | z |^{2}$ is equal to __________. [2025]

(22)

Let z = x + iy.

We have, A : |z – 2 – i| = 3

$\Rightarrow$ $| (x - 2) + i (y - 1) |$ $= 3 \Rightarrow {(x - 2)}^{2} + {(y - 1)}^{2} = 9$ ... (i)

Also, B : Re(z – iz) = 2

$R e ((x + y) + i (y - x)) = 2 \Rightarrow x + y = 2$ ... (ii)

From (i) and (ii), we get $x = \frac{3 \pm \sqrt{17}}{2}, y = \frac{1 \mp \sqrt{17}}{2}$

$∴ \sum_{z \in S} | z |^{2} = \frac{1}{4} [2 \times 26 + 2 \times 18] = 22$ .

Q 12 :

$If for z = α + i β, | z + 2 | = z + 4 (1 + i), then α + β and α β are the roots of the equation$ [2023]

$x^{2} + 7 x + 12 = 0$
$x^{2} + x - 12 = 0$
$x^{2} + 3 x - 4 = 0$
$x^{2} + 2 x - 3 = 0$

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(1)

We have, $| z + 2 |$ $= z + 4 (1 + i)$

$\Rightarrow$ $| α + i β + 2 |$ $= α + i β + 4 + 4 i$

$\Rightarrow$ $\sqrt{{(α + 2)}^{2} + β^{2}} = (α + 4) + i (β + 4)$

Comparing real and imaginary parts, we get

$β + 4 = 0 \Rightarrow β = - 4$

Also, ${(α + 2)}^{2} + 4^{2} = {(α + 4)}^{2}$

$\Rightarrow α^{2} + 4 + 4 α + 16 = α^{2} + 16 + 8 α \Rightarrow 4 α = 4 \Rightarrow α = 1$

Now, $α + β = - 3, α β = - 4$

Equation having roots - 3 and - 4 is $(x + 3) (x + 4) = 0$

i.e. $x^{2} + 7 x + 12 = 0$

Q 13 :

Let the complex number $z = x + i y$ be such that $\frac{2 z - 3 i}{2 z + i}$ is purely imaginary. If $x + y^{2} = 0$ , then $y^{4} + y^{2} - y$ is equal to [2023]

$\frac{3}{2}$
$\frac{2}{3}$
$\frac{3}{4}$
$\frac{4}{3}$

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(3)

$Given, \frac{2 z - 3 i}{2 z + i} is purely imaginary.$

$∴$ $\frac{2 z - 3 i}{2 z + i} + \frac{2 \bar{z} + 3 i}{2 \bar{z} - i} = 0$

$\Rightarrow (2 z - 3 i) (2 \bar{z} - i) + (2 \bar{z} + 3 i) (2 z + i) = 0$

$\Rightarrow 4 z \bar{z} - 2 i z - 6 i \bar{z} - 3 + 4 z \bar{z} + 2 i \bar{z} + 6 i z - 3 = 0$

$\Rightarrow 8 z \bar{z} + 4 i z - 4 i \bar{z} - 6 = 0$ $[∵ z = x + i y]$

$\Rightarrow 4 (x^{2} + y^{2}) + 2 i (x + i y) - 2 i (x - i y) - 3 = 0$

$\Rightarrow 4 x^{2} + 4 y^{2} - 4 y - 3 = 0$

$\Rightarrow 4 y^{4} + 4 y^{2} - 4 y - 3 = 0$ $[∵ x + y^{2} = 0 (Given)]$

$\Rightarrow y^{4} + y^{2} - y = \frac{3}{4}$

Q 14 :

Let $S = {z = x + i y : \frac{2 z - 3 i}{4 z + 2 i} is a real number} .$ Then which of the following is NOT correct? [2023]

$(x, y) = (0, - \frac{1}{2})$
$y + x^{2} + y^{2} \neq - \frac{1}{4}$
$y \in (- \infty, - \frac{1}{2}) \cup (- \frac{1}{2}, \infty)$
$x = 0$

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(1)

$As, \frac{2 z - 3 i}{4 z + 2 i} is a real number.$

$∴$ $\frac{2 z - 3 i}{4 z + 2 i} = \frac{\bar{2 z - 3 i}}{4 z + 2 i} \Rightarrow \frac{2 z - 3 i}{4 z + 2 i} = \frac{2 \bar{z} + 3 i}{4 \bar{z} - 2 i}$

$\Rightarrow (2 z - 3 i) (4 \bar{z} - 2 i) = (2 \bar{z} + 3 i) (4 z + 2 i)$

$\Rightarrow 8 z \bar{z} - 4 i z - 12 i \bar{z} - 6 = 8 z \bar{z} + 4 i \bar{z} + 12 i z - 6$

$\Rightarrow - 4 i (z + \bar{z}) - 12 i (z + \bar{z}) = 0$

$\Rightarrow 16 i (z + \bar{z}) = 0 \Rightarrow z + \bar{z} = 0 \Rightarrow 2 x = 0 \Rightarrow x = 0$

$As, y can not be \frac{- 1}{2} .$

$∴$ $Option (1) is not possible.$

Q 15 :

For $a \in ℂ$ , let $A = {z \in ℂ : R e (a + \bar{z}) > I m (\bar{a} + z)} and B = {z \in ℂ : R e (a + \bar{z}) < I m (\bar{a} + z)} .$ Then among the two statements:

(S1) : If $R e (a), I m (a) > 0$ , then the set A contains all the real numbers.

(S2) : If $R e (a), I m (a) < 0$ , then the set B contains all the real numbers. [2023]

only (S2) is true
both are true
only (S1) is true
both are false

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(4)

$Let a = x_{1} + i y_{1} and z = x + i y$

$Now Re (a + \bar{z}) > Im (\bar{a} + z)$

$\Rightarrow Re (x_{1} + i y_{1} + x - i y) > Im (x_{1} - i y_{1} + x + i y)$

$\Rightarrow Re (x_{1} + x + i (y_{1} - y)) > Im (x_{1} + x + i (y - y_{1}))$

$∴ x_{1} + x > - y_{1} + y$

$Let x_{1} = 2, y_{1} = 10, x = - 12, y = 0$

Given inequality is not valid for these values.

$S 1 is false.$

$Now Re (a + \bar{z}) < Im (\bar{a} + z)$

$\Rightarrow Re (x_{1} + x + i (y_{1} - y)) < Im (x_{1} + x + i (y - y_{1}))$

$\Rightarrow x_{1} + x < - y_{1} + y$

$Let x_{1} = - 2, y_{1} = - 10, x = 12, y = 0$

Given inequality is not valid for these values.

$S 2 is false.$

Q 16 :

Let $S = {z \in ℂ : \bar{z} = i (z^{2} + R e (\bar{z}))} .$ Then $\sum_{z \in S} | z |^{2}$ is equal to [2023]

4
$\frac{5}{2}$
$\frac{7}{2}$
3

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(1)

$\bar{z} = i (z^{2} + Re (\bar{z}))$

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

$\Rightarrow x = - 2 x y and y = y^{2} - x^{2} - x$

$x = 0 \Rightarrow y = y^{2} \Rightarrow y = 0 or 1$

$y = \frac{- 1}{2} \Rightarrow \frac{- 1}{2} = \frac{1}{4} - x^{2} \Rightarrow x = \frac{1}{2} or \frac{- 3}{2}$

$Possible places of z = 0 + i 0, 0 + i, \frac{1}{2} - \frac{1}{2} i, \frac{- 3}{2} - \frac{i}{2}$

$\Rightarrow \sum | z |^{2} = 0 + 1 + (\frac{1}{4} + \frac{1}{4}) + (\frac{9}{4} + \frac{1}{4}) = 1 + 3 = 4$

Q 17 :

If the set ${R e (\frac{z - \bar{z} + z \bar{z}}{2 - 3 z + 5 \bar{z}}) : z \in C, R e (z) = 3}$ is equal to the interval $(α, β]$ , then 24 $(β - α)$ is equal to [2023]

27
36
42
30

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(4)

Let $z_{1} = (\frac{z - \bar{z} + z \bar{z}}{2 - 3 z + 5 \bar{z}})$

Let $z = 3 + i y$ $[∵ Re (z) = 3]$

$\Rightarrow \bar{z} = 3 - i y$

$z_{1} = \frac{2 i y + (9 + y^{2})}{2 - 3 (3 + i y) + 5 (3 - i y)}$

$\frac{9 + y^{2} + i (2 y)}{8 - 8 i y} = \frac{(9 + y^{2}) + i (2 y)}{8 (1 - i y)} \times \frac{(1 + i y)}{(1 + i y)}$

$= \frac{9 + 9 i y + y^{2} - 2 y^{2} + i y^{3} + 2 i y}{8 (1 + y^{2})}$

$So, Re (z_{1}) = \frac{(9 + y^{2}) - 2 y^{2}}{8 (1 + y^{2})} = \frac{9 - y^{2}}{8 (1 + y^{2})}$

$= \frac{1}{8} [\frac{10 - (1 + y^{2})}{1 + y^{2}}] = \frac{1}{8} [\frac{10}{1 + y^{2}} - 1]$

Thus, $1 + y^{2} \in [1, \infty) \Rightarrow \frac{1}{1 + y^{2}} \in (0, 1]$

$\Rightarrow \frac{10}{1 + y^{2}} \in (0, 10] \Rightarrow \frac{10}{1 + y^{2}} - 1 \in (- 1, 9]$

$∴ Re (z_{1}) \in (\frac{- 1}{8}, \frac{9}{8}]$ ; $α = \frac{- 1}{8}, β = \frac{9}{8}$

$∴$ $24 (β - α) = 24 (\frac{9}{8} + \frac{1}{8}) = 30$

Q 18 :

Let $z_{1} = 2 + 3 i$ and $z_{2} = 3 + 4 i$ . The set $S = {Z \in ℂ : | z - z_{1} |^{2} - | z - z_{2} |^{2} = | z_{1} - z_{2} |^{2}}$ represents a [2023]

straight line with the sum of its intercepts on the coordinate axes equals −18
hyperbola with eccentricity 2
hyperbola with the length of the transverse axis 7
straight line with the sum of its intercepts on the coordinate axes equals 14

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(4)

Let $z = x + i y$ ; $z_{1} = 2 + 3 i$ ; $z_{2} = 3 + 4 i$

$| (x + i y) - (2 + 3 i) |^{2} - | (x + i y) - (3 + 4 i) |^{2} = | (2 + 3 i) - (3 + 4 i) |^{2}$

${(x - 2)}^{2} + {(y - 3)}^{2} - {(x - 3)}^{2} - {(y - 4)}^{2} = 1^{2} + 1^{2}$

$2 x - 5 + 2 y - 7 = 2 \Rightarrow 2 x + 2 y = 14 \Rightarrow x + y = 7;$

$x intercept = y intercept = 7; sum of intercepts = 14$

Q 19 :

Let $z$ be a complex number such that $| \frac{z - 2 i}{z + i} |$ $= 2, z \neq - i .$ Then $z$ lies on the circle of radius 2 and centre [2023]

(2, 0)
(0, 2)
(0, 0)
(0, -2)

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(4)

Q 20 :

The complex number $z = \frac{i - 1}{\cos \frac{π}{3} + i \sin \frac{π}{3}}$ is equal to [2023]

$\sqrt{2} i (\cos \frac{5 π}{12} - i \sin \frac{5 π}{12})$
$\cos \frac{π}{12} - i \sin \frac{π}{12}$
$\sqrt{2} (\cos \frac{π}{12} + i \sin \frac{π}{12})$
$\sqrt{2} (\cos \frac{5 π}{12} + i \sin \frac{5 π}{12})$

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(4)

$z = \frac{i - 1}{\cos \frac{π}{3} + i \sin \frac{π}{3}} = \frac{\sqrt{2} e^{i \frac{3 π}{4}}}{e^{i \frac{π}{3}}}$

$= \sqrt{2} e^{i (\frac{3 π}{4} - \frac{π}{3})} = \sqrt{2} e^{i (\frac{5 π}{12})} = \sqrt{2} (\cos \frac{5 π}{12} + i \sin \frac{5 π}{12})$

Q 21 :

Let $α = 8 - 14 i,$ $A = {z \in ℂ : \frac{α z - \bar{α} \bar{z}}{z^{2} - {(\bar{z})}^{2} - 112 i} = 1}$ and $B = {z \in ℂ : | z + 3 i | = 4} .$ Then $\sum_{z \in A \cap B} (Re z - Im z)$ is equal to ________. [2023]

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Given, $α = 8 - 14 i$

Let $z = x + i y$

$α z = (8 - 14 i) (x + i y) = (8 x + 14 y) + i (- 14 x + 8 y)$

Now, $z + \bar{z} = 2 x$ and $z - \bar{z} = 2 i y$

For $A,$ $\frac{2 i (- 14 x + 8 y)}{i (4 x y - 112)} = 1$

$\Rightarrow (x - 4) (y + 7) = 0 \Rightarrow x = 4 or y = - 7$

For $B,$ $x^{2} + {(y + 3)}^{2} = 16$

When $x = 4$ , then $y = - 3$ ; when $y = - 7$ , then $x = 0$

$∴ A \cap B = {4 - 3 i, 0 - 7 i}$

So, $\sum_{z \in A \cap B} (Re z - Im z) = 4 - (- 3) + (0 - (- 7)) = 7 + 7 = 14$

Topic Question Set

Q 1 :

Let $α$ and $β$ be the sum and the product of all the non-zero solutions of the equation ${(\bar{z})}^{2} + | z | = 0, z \in C .$ Then $4 (α^{2} + β^{2})$ is equal to [2024]

Q 3 :

Let $z$ be a complex number such that $| z + 2 | = 1$ and $Im$ $(\frac{z + 1}{z + 2})$ $= \frac{1}{5} .$ Then the value of $| R e (\bar{z + 2}) |$ is [2024]

Q 4 :

The sum of all possible values of $θ \in [- π, 2 π]$ , for which $\frac{1 + i \cos θ}{1 - 2 i \cos θ}$ is purely imaginary, is equal to: [2024]

Q 5 :

Let $S = {z \in C : | z - 1 | = 1$ and $(\sqrt{2} - 1) (z + \bar{z}) - i (z - \bar{z}) = 2 \sqrt{2}} .$ Let $z_{1}, z_{2} \in S$ be such that $| z_{1} |$ $=$ $\max_{z \in S}$ $| z |$ and $| z_{2} |$ $=$ $\min_{z \in S}$ $| z | .$ Then $| \sqrt{2} z_{1} - z_{2} |^{2}$ equals: [2024]

Q 6 :

If $S = {z \in C :$ $| z - i |$ $=$ $| z + i |$ $=$ $| z - 1 |$ $}$ then $n (S)$ is: [2024]

Q 7 :

If $z = \frac{1}{2} - 2 i$ is such that $| z + 1 |$ $= α z + β (1 + i),$ $i = \sqrt{- 1}$ and $α, β \in R,$ then $α + β$ is equal to [2024]

Q 8 :

If $z = x + i y, x y \neq 0,$ satisfies the equation $z^{2} + i \bar{z} = 0,$ then $| z^{2} |$ is equal to [2024]

Q 9 :

Let $z_{1}$ and $z_{2}$ be two complex numbers such that $z_{1} + z_{2} = 5$ and $z_{1}^{3} + z_{2}^{3} = 20 + 15 i .$ Then $| z_{1}^{4} + z_{2}^{4} |$ equals [2024]

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Q 11 :

Let $A = {z \in C : | z - 2 - i | = 3}$ , $B = {z \in C : R e (z - i z) = 2}$ and $S = A \cap B$ . Then $\sum_{z \in S} | z |^{2}$ is equal to __________. [2025]

Q 12 :

$If for z = α + i β, | z + 2 | = z + 4 (1 + i), then α + β and α β are the roots of the equation$ [2023]

Q 13 :

Let the complex number $z = x + i y$ be such that $\frac{2 z - 3 i}{2 z + i}$ is purely imaginary. If $x + y^{2} = 0$ , then $y^{4} + y^{2} - y$ is equal to [2023]

Q 14 :

Let $S = {z = x + i y : \frac{2 z - 3 i}{4 z + 2 i} is a real number} .$ Then which of the following is NOT correct? [2023]

Q 16 :

Let $S = {z \in ℂ : \bar{z} = i (z^{2} + R e (\bar{z}))} .$ Then $\sum_{z \in S} | z |^{2}$ is equal to [2023]

Q 17 :

If the set ${R e (\frac{z - \bar{z} + z \bar{z}}{2 - 3 z + 5 \bar{z}}) : z \in C, R e (z) = 3}$ is equal to the interval $(α, β]$ , then 24 $(β - α)$ is equal to [2023]

Q 18 :

Let $z_{1} = 2 + 3 i$ and $z_{2} = 3 + 4 i$ . The set $S = {Z \in ℂ : | z - z_{1} |^{2} - | z - z_{2} |^{2} = | z_{1} - z_{2} |^{2}}$ represents a [2023]

Q 19 :

Let $z$ be a complex number such that $| \frac{z - 2 i}{z + i} |$ $= 2, z \neq - i .$ Then $z$ lies on the circle of radius 2 and centre [2023]

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Q 20 :

The complex number $z = \frac{i - 1}{\cos \frac{π}{3} + i \sin \frac{π}{3}}$ is equal to [2023]

Q 21 :

Let $α = 8 - 14 i,$ $A = {z \in ℂ : \frac{α z - \bar{α} \bar{z}}{z^{2} - {(\bar{z})}^{2} - 112 i} = 1}$ and $B = {z \in ℂ : | z + 3 i | = 4} .$ Then $\sum_{z \in A \cap B} (Re z - Im z)$ is equal to ________. [2023]

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Topic Question Set

Q 1 : Let α and β be the sum and the product of all the non-zero solutions of the equation (z¯)2+|z|=0, z∈C. Then 4(α2+β2) is equal to [2024]

Q 3 : Let z be a complex number such that |z+2|=1 and Im(z+1z+2)=15. Then the value of |Re(z+2¯)| is [2024]

Q 4 : The sum of all possible values of θ∈[-π,2π], for which 1+icosθ1-2icosθ is purely imaginary, is equal to: [2024]

Q 5 : Let S={z∈C:|z-1|=1 and (2-1)(z+z¯)-i(z-z¯)=22}. Let z1,z2∈S be such that |z1|=maxz∈S|z| and |z2|=minz∈S|z|. Then |2z1-z2|2 equals: [2024]

Q 6 : If S={z∈C:|z-i|=|z+i|=|z-1|} then n(S) is: [2024]

Q 7 : If z=12-2i is such that |z+1|=αz+β(1+i),i=-1 and α,β∈R, then α+β is equal to [2024]

Q 8 : If z=x+iy,xy≠0, satisfies the equation z2+iz¯=0, then |z2| is equal to [2024]

Q 9 : Let z1 and z2 be two complex numbers such that z1+z2=5 and z13+z23=20+15i. Then |z14+z24| equals [2024]

Q 10 : Let P={z∈C:|z+2-3i|≤1} and Q={z∈C:z(1+i)+z¯(1-i)≤-8}. Let in P∩Q,|z-3+2i| be maximum and minimum at z1 and z2 respectively. If |z1|2+2|z2|2=α+β2, where α,β are integers, then α+β equals________ . [2024]

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Q 11 : Let A={z∈C:|z–2–i|=3}, B={z∈C:Re(z–iz)=2} and S=A∩B. Then ∑z∈S|z|2 is equal to __________. [2025]

Q 12 : If for z=α+iβ, |z+2|=z+4(1+i), then α+β and αβ are the roots of the equation [2023]

Q 13 : Let the complex number z=x+iy be such that 2z-3i2z+i is purely imaginary. If x+y2=0, then y4+y2-y is equal to [2023]

Q 14 : Let S={z=x+iy:2z-3i4z+2i is a real number}. Then which of the following is NOT correct? [2023]

Q 15 : For a∈ℂ, let A={z∈ℂ:Re(a+z¯)>Im(a¯+z)} and B={z∈ℂ:Re(a+z¯)<Im(a¯+z)}. Then among the two statements: (S1) : If Re(a),Im(a)>0, then the set A contains all the real numbers. (S2) : If Re(a),Im(a)<0, then the set B contains all the real numbers. [2023]

Q 16 : Let S={z∈ℂ:z¯=i(z2+Re(z¯))}. Then ∑z∈S|z|2 is equal to [2023]

Q 17 : If the set {Re(z-z¯+zz¯2-3z+5z¯) :z∈C,Re(z)=3} is equal to the interval (α,β], then 24 (β-α) is equal to [2023]

Q 18 : Let z1=2+3i and z2=3+4i. The set S={Z∈ℂ:|z-z1|2-|z-z2|2=|z1-z2|2} represents a [2023]

Q 19 : Let z be a complex number such that |z-2iz+i|=2, z≠-i. Then z lies on the circle of radius 2 and centre [2023]

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Q 20 : The complex number z=i-1cosπ3+isinπ3 is equal to [2023]

Q 21 : Let α=8-14i, A={z∈ℂ: αz-α¯z¯z2-(z¯)2-112i=1} and B={z∈ℂ:|z+3i|=4}. Then ∑z∈A∩B(Rez-Imz) is equal to ________. [2023]

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Q 1 :

Let $α$ and $β$ be the sum and the product of all the non-zero solutions of the equation ${(\bar{z})}^{2} + | z | = 0, z \in C .$ Then $4 (α^{2} + β^{2})$ is equal to [2024]

Q 3 :

Let $z$ be a complex number such that $| z + 2 | = 1$ and $Im$ $(\frac{z + 1}{z + 2})$ $= \frac{1}{5} .$ Then the value of $| R e (\bar{z + 2}) |$ is [2024]

Q 4 :

The sum of all possible values of $θ \in [- π, 2 π]$ , for which $\frac{1 + i \cos θ}{1 - 2 i \cos θ}$ is purely imaginary, is equal to: [2024]

Q 5 :

Let $S = {z \in C : | z - 1 | = 1$ and $(\sqrt{2} - 1) (z + \bar{z}) - i (z - \bar{z}) = 2 \sqrt{2}} .$ Let $z_{1}, z_{2} \in S$ be such that $| z_{1} |$ $=$ $\max_{z \in S}$ $| z |$ and $| z_{2} |$ $=$ $\min_{z \in S}$ $| z | .$ Then $| \sqrt{2} z_{1} - z_{2} |^{2}$ equals: [2024]

Q 6 :

If $S = {z \in C :$ $| z - i |$ $=$ $| z + i |$ $=$ $| z - 1 |$ $}$ then $n (S)$ is: [2024]

Q 7 :

If $z = \frac{1}{2} - 2 i$ is such that $| z + 1 |$ $= α z + β (1 + i),$ $i = \sqrt{- 1}$ and $α, β \in R,$ then $α + β$ is equal to [2024]

Q 8 :

If $z = x + i y, x y \neq 0,$ satisfies the equation $z^{2} + i \bar{z} = 0,$ then $| z^{2} |$ is equal to [2024]

Q 9 :

Let $z_{1}$ and $z_{2}$ be two complex numbers such that $z_{1} + z_{2} = 5$ and $z_{1}^{3} + z_{2}^{3} = 20 + 15 i .$ Then $| z_{1}^{4} + z_{2}^{4} |$ equals [2024]

Q 11 :

Let $A = {z \in C : | z - 2 - i | = 3}$ , $B = {z \in C : R e (z - i z) = 2}$ and $S = A \cap B$ . Then $\sum_{z \in S} | z |^{2}$ is equal to __________. [2025]

Q 12 :

$If for z = α + i β, | z + 2 | = z + 4 (1 + i), then α + β and α β are the roots of the equation$ [2023]

Q 13 :

Let the complex number $z = x + i y$ be such that $\frac{2 z - 3 i}{2 z + i}$ is purely imaginary. If $x + y^{2} = 0$ , then $y^{4} + y^{2} - y$ is equal to [2023]

Q 14 :

Let $S = {z = x + i y : \frac{2 z - 3 i}{4 z + 2 i} is a real number} .$ Then which of the following is NOT correct? [2023]

Q 16 :

Let $S = {z \in ℂ : \bar{z} = i (z^{2} + R e (\bar{z}))} .$ Then $\sum_{z \in S} | z |^{2}$ is equal to [2023]

Q 17 :

If the set ${R e (\frac{z - \bar{z} + z \bar{z}}{2 - 3 z + 5 \bar{z}}) : z \in C, R e (z) = 3}$ is equal to the interval $(α, β]$ , then 24 $(β - α)$ is equal to [2023]

Q 18 :

Let $z_{1} = 2 + 3 i$ and $z_{2} = 3 + 4 i$ . The set $S = {Z \in ℂ : | z - z_{1} |^{2} - | z - z_{2} |^{2} = | z_{1} - z_{2} |^{2}}$ represents a [2023]

Q 19 :

Let $z$ be a complex number such that $| \frac{z - 2 i}{z + i} |$ $= 2, z \neq - i .$ Then $z$ lies on the circle of radius 2 and centre [2023]

Q 20 :

The complex number $z = \frac{i - 1}{\cos \frac{π}{3} + i \sin \frac{π}{3}}$ is equal to [2023]

Q 21 :

Let $α = 8 - 14 i,$ $A = {z \in ℂ : \frac{α z - \bar{α} \bar{z}}{z^{2} - {(\bar{z})}^{2} - 112 i} = 1}$ and $B = {z \in ℂ : | z + 3 i | = 4} .$ Then $\sum_{z \in A \cap B} (Re z - Im z)$ is equal to ________. [2023]