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

Q 1 :

If $z_{1}, z_{2}$ are two distinct complex number such that $| \frac{z_{1} - 2 z_{2}}{\frac{1}{2} - z_{1} {\bar{z}}_{2}} |$ $= 2$ , then [2024]

$z_{1}$ lies on a circle of radius $\frac{1}{2}$ and $z_{2}$ lies on a circle of radius 1.
either $z_{1}$ lies on a circle of radius 1 or $z_{2}$ lies on a circle of radius $\frac{1}{2} .$
both $z_{1}$ and $z_{2}$ lie on the same circle.
either $z_{1}$ lies on a circle of radius $\frac{1}{2}$ or $z_{2}$ lies on a circle of radius 1.

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(2)

We have, $| \frac{z_{1} - 2 z_{2}}{\frac{1}{2} - z_{1} {\bar{z}}_{2}} |$ $= 2$

$\Rightarrow | z_{1} - 2 z_{2} | = | 1 - 2 z_{1} {\bar{z}}_{2} |$

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

$\Rightarrow (z_{1} - 2 z_{2}) ({\bar{z}}_{1} - 2 {\bar{z}}_{2}) = (1 - 2 z_{1} {\bar{z}}_{2}) (1 - 2 {\bar{z}}_{1} z_{2})$

$\Rightarrow | z_{1} |^{2} + 4 | z_{2} |^{2} - 2 {\bar{z}}_{1} z_{2} - 2 {\bar{z}}_{2} z_{1}$

$= 1 + 4 | z_{1} |^{2} | z_{2} |^{2} - 2 z_{1} {\bar{z}}_{2} - 2 {\bar{z}}_{1} z_{2}$

$\Rightarrow | z_{1} |^{2} + 4 | z_{2} |^{2} - 4 | z_{1} |^{2} | z_{2} |^{2} - 1 = 0$

$\Rightarrow (| z_{1} |^{2} - 1) (1 - 4 | z_{2} |^{2}) = 0$

$\Rightarrow | z_{1} | = 1, | z_{2} | = \frac{1}{2}$

Q 2 :

Let $r$ and $θ$ respectively be the modulus and amplitude of the complex number $z = 2 - i (2 \tan \frac{5 π}{8}),$ then $(r, θ)$ is equal to [2024]

$(2 \sec \frac{3 π}{8}, \frac{5 π}{8})$
$(2 \sec \frac{3 π}{8}, \frac{3 π}{8})$
$(2 \sec \frac{11 π}{8}, \frac{11 π}{8})$
$(2 \sec \frac{5 π}{8}, \frac{3 π}{8})$

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(2)

We have, $z = 2 - i (2 \tan \frac{5 π}{8})$

$= 2 + 2 i \tan (π - \frac{5 π}{8}) \Rightarrow z = 2 + 2 i \tan \frac{3 π}{8}$

$\Rightarrow z = 2 (\cos \frac{3 π}{8} + i \sin \frac{3 π}{8}) \sec \frac{3 π}{8}$

So, $r = 2 \sec \frac{3 π}{8}, θ = \frac{3 π}{8}$

Q 3 :

Let the complex numbers $α$ and $\frac{1}{\bar{α}}$ lie on the circles $| z - z_{0} |^{2} = 4$ and $| z - z_{0} |^{2} = 16$ respectively, where $z_{0} = 1 + i .$ Then, the value of $100 | α |^{2}$ is ________. [2024]

(20)

Given $| z - z_{0} |^{2} = 4$ ...(i)

$| z - z_{0} |^{2} = 16$ ...(ii)

$α$ lies on (i), $∴$ $| α - z_{0} |^{2} = 4$

$\Rightarrow (α - z_{0}) (\bar{α} - {\bar{z}}_{0}) = 4 \Rightarrow α \bar{α} - α {\bar{z}}_{0} - z_{0} \bar{α} + | z_{0} |^{2} = 4$

$\Rightarrow | α |^{2} - α {\bar{z}}_{0} - z_{0} \bar{α} + 2 = 4 (∵ | z_{0} | = \sqrt{2})$

$\Rightarrow | α |^{2} - a {\bar{z}}_{0} - z_{0} \bar{α} = 2$ ...(iii)

$\frac{1}{\bar{α}}$ lies on (ii)

$∴$ $| \frac{1}{\bar{α}} - z_{0} |^{2} = 16 \Rightarrow (\frac{1}{\bar{α}} - z_{0}) (\frac{1}{α} - {\bar{z}}_{0}) = 16$

$\Rightarrow (1 - \bar{α} z_{0}) (1 - α {\bar{z}}_{0}) = 16 α \bar{α}$

$\Rightarrow 1 - \bar{α} z_{0} - α {\bar{z}}_{0} + α \bar{α}$ $z_{0} {\bar{z}}_{0} = 16 | α |^{2}$

$\Rightarrow 1 - \bar{α} z_{0} - α {\bar{z}}_{0} = 14 | α |^{2}$ ...(iv)

From (iii) and (iv), we get

$\Rightarrow 15 | α |^{2} = 3 \Rightarrow | α |^{2} = \frac{3}{15} = \frac{1}{5}$

$∴$ $100 | α |^{2} = 100 \times \frac{1}{5} = 20$

Q 4 :

If $α$ denotes the number of solutions of $| 1 - i |^{x}$ $= 2^{x}$ and $β = (\frac{| z |}{\arg (z)}),$ where $z = \frac{π}{4} {(1 + i)}^{4}$ $[\frac{1 - \sqrt{π} i}{\sqrt{π} + i} + \frac{\sqrt{π} - i}{1 + \sqrt{π} i}],$ $i = \sqrt{- 1},$ then the distance of the point $(α, β)$ from the line $4 x - 3 y = 7$ is __________ [2024]

(3)

We have, $| 1 - i |^{x}$ $= 2^{x}$

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

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

$\Rightarrow α = 1$

and $β = (\frac{| z |}{a r g (z)}),$ where $z = \frac{π}{4} {(1 + i)}^{4}$ $[\frac{1 - \sqrt{π} i}{\sqrt{π} + i} + \frac{\sqrt{π} - i}{1 + \sqrt{π} i}]$

$z = 2 π i \Rightarrow \arg (z) = \frac{π}{2}$

and $| z | = 2 π$

$β = \frac{| z |}{\arg (z)} = \frac{2 π}{π / 2} = 4 \Rightarrow β = 4$

Now, the distance of point $(α, β)$ from the line $4 x - 3 y = 7$ is given by

$d = \frac{| 4 \times 1 - 3 \times 4 - 7 |}{\sqrt{16 + 9}} = \frac{| 4 - 12 - 7 |}{5} = \frac{15}{5} = 3$

Q 5 :

The area (in sq. units) of the region

$S = {z \in C : | z - 1 | \leq 2; (z + \bar{z}) + i (z - \bar{z}) \leq 2, Im (z) \geq 0}$ is [2024]

$\frac{7 π}{4}$
$\frac{3 π}{2}$
$\frac{7 π}{3}$
$\frac{17 π}{8}$

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(2)

$| z - 1 |$ $\leq 2 \Rightarrow {(x - 1)}^{2} + y^{2} \leq 4$

$z + \bar{z} + i (z - \bar{z}) \leq 2$

$\Rightarrow x + i y + x - i y + i (x + i y - x + i y) \leq 2$

$\Rightarrow x - y \leq 1$

$Im (z) \geq 0$

$\Rightarrow y \geq 0$

$∴$ $Required area = (\frac{\frac{3 π}{4}}{2 π})$ $(π) {(2)}^{2} = \frac{3}{2} π sq. units$

Q 6 :

Let $S_{1} = {z \in C : | z | \leq 5}$ , $S_{2} = {z \in C : Im (\frac{z + 1 - \sqrt{3} i}{1 - \sqrt{3} i}) \geq 0}$ and $S_{3} = {z \in C : Re (z) \geq 0} .$ Then the area of the region $S_{1} \cap S_{2} \cap S_{3}$ is : [2024]

$\frac{125 π}{12}$
$\frac{125 π}{24}$
$\frac{125 π}{6}$
$\frac{125 π}{4}$

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(1)

$Let z = x + i y be any complex number$

$S_{1} = {z \in C : | z | \leq 5}$

$S_{1} : x^{2} + y^{2} \leq 25$ ...(i)

$S_{2} : Im [\frac{x + iy}{1 - \sqrt{3} i} + 1] \geq 0$

$i.e.,$ $S_{2} : Im [\frac{(x + iy) (1 + \sqrt{3} i)}{4} + 1] \geq 0$

$\Rightarrow \sqrt{3} x + y \geq 0$ ...(ii)

$S_{3} : {z \in C : Re (z) \geq 0}$

$\Rightarrow x \geq 0$ ...(iii)

$Now,$ $\sqrt{3} x + y = 0$

$\Rightarrow y = - \sqrt{3} x$

$\Rightarrow \tan θ = 120 °$

$∴$ $Required area = \frac{25 π}{2} - \frac{1}{12} {(5)}^{2} π$

$= \frac{25 π}{2} - \frac{25 π}{12}$

$= \frac{125 π}{12}$

Q 7 :

The sum of the square of the modulus of the elements in the set ${z = a + i b : a, b \in Z, z \in C, | z - 1 | \leq 1, | z - 5 | \leq | z - 5 i |}$ is ____________. [2024]

(9)

$We have, | z - 1 | \leq 1$

$\Rightarrow {(a - 1)}^{2} + b^{2} \leq 1$

$and | z - 5 | \leq | z - 5 i |$

$\Rightarrow {(a - 5)}^{2} + b^{2} \leq {(a)}^{2} + {(b - 5)}^{2}$

$\Rightarrow a \geq b$

$z = 0 + 0 i, 1 + 0 i, 2 + 0 i, 1 + i, 1 - i$

$\Rightarrow | z | = 0, 1, 2, \sqrt{2}, \sqrt{2}$

$∴$ $Sum of | z |^{2} = 0 + 1 + 4 + 2 + 2 = 9$

Q 8 :

Let z be a complex number such that |z| = 1. If $\frac{2 + k^{2} z}{k + \bar{z}} = k z, k \in R$ , then the maximum distance of $k + i k^{2}$ from the circle |z – (1 + 2i)| = 1 is : [2025]

$\sqrt{3} + 1$
3
$\sqrt{5} + 1$
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(3)

We have, $\frac{2 + k^{2} z}{k + \bar{z}} = k z$

$\Rightarrow 2 + k^{2} z = k^{2} z + k z \bar{z}$

$\Rightarrow | z |^{2} k = 2$ [ $∵ z \bar{z} = | z |^{2}$ ]

$\Rightarrow k = 2$ [ $∵ | z | = 1$ ]

The centre of the given circle is (1, 2) and its radius is 1.

Now, $k + i k^{2} = 2 + 4 i$ .

Maximum distance = $O P + r = \sqrt{1 + 4} + 1 = \sqrt{5} + 1$ .

Q 9 :

If $z_{1}, z_{2}, z_{3} \in C$ are the vertices of an equilateral triangle, whose centroid is $z_{0}$ , then $\sum_{k = 1}^{3} {(z_{k} - z_{0})}^{2}$ is equal to [2025]

0
i
– i
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(1)

Centroid of triangle is $z_{0}$ , then $z_{1} + z_{2} + z_{3} = 3 z_{0}$

$\Rightarrow {(z_{1} + z_{2} + z_{3})}^{2} = 9 z_{0}^{2}$

$\Rightarrow z_{1}^{2} + z_{2}^{2} + z_{3}^{2} + 2 (z_{1}^{2} + z_{2}^{2} + z_{3}^{2}) = 9 z_{0}^{2}$ [ $∵$ For equilateral $△, z_{1} z_{2} + z_{2} z_{3} + z_{3} z_{1} = z_{1}^{2} + z_{2}^{2} + z_{3}^{2}$ ]

$\Rightarrow z_{1}^{2} + z_{2}^{2} + z_{3}^{2} = 3 z_{0}^{2}$

Now, $\sum_{k = 1}^{3} {(z_{k} - z_{0})}^{2} = {(z_{1} - z_{0})}^{2} + {(z_{2} - z_{0})}^{2} + {(z_{3} - z_{0})}^{2}$

$= z_{1}^{2} + z_{2}^{2} + z_{3}^{2} + 3 z_{0}^{2} - 2 (z_{1} + z_{2} + z_{3}) z_{0}$

$= 6 z_{0}^{2} - 6 z_{0}^{2} = 0$ .

Q 10 :

If the locus of $z \in C$ , such that $Re (\frac{z - 1}{2 z + i}) + R e (\frac{\bar{z} - 1}{2 \bar{z} - i}) = 2$ , is a circle of radius r and center (a, b), then $\frac{15 a b}{r^{2}}$ is equal to: [2025]

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(4)

We have, $Re (\frac{z - 1}{2 z + i}) + R e (\frac{\bar{z} - 1}{2 \bar{z} - i}) = 2$

$\Rightarrow Re (\frac{z - 1}{2 z + i}) + R e (\frac{\bar{z - 1}}{2 z + i}) = 2$ $[∵ (\frac{\bar{z - 1}}{2 z + i}) = \frac{\bar{z} - 1}{2 \bar{z} - i}]$

$\Rightarrow 2 Re (\frac{z - 1}{2 z + i}) = 2 \Rightarrow Re (\frac{z - 1}{2 z + i}) = 1$

Let z = x + iy

$Re (\frac{(x - 1) + i y}{2 x + i (2 y + 1)}) = 1$

$\Rightarrow R e [\frac{((x - 1) + i y) (2 x - i (2 y + 1))}{(2 x + i (2 y + 1)) (2 x - i (2 y + 1))}] = 1$

$\Rightarrow \frac{2 x (x - 1) + y (2 y + 1)}{4 x^{2} + {(2 y + 1)}^{2}} = 1$

$\Rightarrow 2 x^{2} - 2 x + 2 y^{2} + y = 4 x^{2} + 4 y^{2} + 1 + 4 y$

$\Rightarrow 2 x^{2} + 2 y^{2} + 3 y + 2 x + 1 = 0$

$\Rightarrow x^{2} + y^{2} + x + \frac{3}{2} y + \frac{1}{2} = 0$

$∴ Centre = (\frac{- 1}{2}, \frac{- 3}{4}) and r = \sqrt{\frac{1}{4} + \frac{9}{16} - \frac{1}{2}} = \frac{\sqrt{5}}{4}$

$\Rightarrow a = \frac{- 1}{2}, b = \frac{- 3}{4}, r^{2} = \frac{5}{16}$

$∴ \frac{15 a b}{r^{2}} = 15 \times (\frac{- 1}{2}) \times (\frac{- 3}{4}) \times \frac{16}{5} = 18$ .

Q 11 :

Let $z_{1}, z_{2}$ and $z_{3}$ be three complex number on the circle |z| = 1 with $\arg (z_{1}) = \frac{- π}{4}$ , $\arg (z_{2}) = 0$ and $\arg (z_{3}) = \frac{π}{4}$ . If $| z_{1} z_{2} + z_{2} z_{3} + z_{3} z_{1} |^{2} = α + β \sqrt{2}, α, β \in Z$ , then the value of $α^{2} + β^{2}$ is : [2025]

24
41
29
31

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(3)

Given, |z| = 1

Now, $z_{1} = e^{i \frac{π}{4}} = \frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}} = \frac{1 - i}{\sqrt{2}}; z_{2} = e^{i 0} = 1$ ;

$z_{3} = e^{i \frac{π}{4}} = \frac{1 + i}{\sqrt{2}}$

Now, $| z_{1} {\bar{z}}_{2} + z_{2} {\bar{z}}_{3} + z_{3} {\bar{z}}_{1} |^{2}$ $=$ $| \frac{1 - i}{\sqrt{2}} \cdot 1 + 1 \cdot \frac{1 - i}{\sqrt{2}} + {(\frac{1 + i}{2})}^{2} |^{2}$

$=$ $| \sqrt{2} + (1 - \sqrt{2}) i |^{2}$ $= 5 - 2 \sqrt{2} = α + β \sqrt{2}$

$\Rightarrow α = 5 and β = - 2$ $∴ α^{2} + β^{2} = 29$

Q 12 :

Let the curve $z (1 + i) + \bar{z} (1 - i) = 4, z \in C$ , divide the region $| z - 3 | \leq 1$ into two parts of area $α$ and $β$ . Then $| α - β |$ equals : [2025]

$1 + \frac{π}{3}$
$1 + \frac{π}{6}$
$1 + \frac{π}{4}$
$1 + \frac{π}{2}$

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(4)

Let z = x + iy, then from given equation, we have

(x + iy)(1 + i) + (x – iy)(1 – i) = 4

$\Rightarrow$ x + ix + iy – y + x – ix – iy – y = 4

$\Rightarrow$ 2x – 2y = 4 $\Rightarrow$ x – y = 2

Now, $| z - 3 | \leq 1 \Rightarrow {(x - 3)}^{2} + y^{2} \leq 1$

$β$ = Area of shaded region = $\frac{π {(1)}^{2}}{4} - \frac{1}{2} \times 1 \times 1$

= $(\frac{π}{4} - \frac{1}{2})$ sq. units

$α$ = Area of unshaded region inside the circle

$= \frac{3}{4} π {(1)}^{2} + \frac{1}{2} \times 1 \times 1 = (\frac{3 π}{4} + \frac{1}{2})$ sq. units

Now, $| α - β |$ = difference of area

= $(\frac{3 π}{4} + \frac{1}{2}) - (\frac{π}{4} - \frac{1}{2}) = \frac{π}{2} + 1$ .

Q 13 :

Let $| \frac{\bar{z} - i}{2 \bar{z} + i} |$ $= \frac{1}{3}, z \in C$ , be the equation of a circle with center at C. If the area of the triangle, whose vertices are at the point (0, 0), C and $(α, 0)$ is 11 square units, then $α^{2}$ equals : [2025]

100
$\frac{81}{25}$
$\frac{121}{25}$
50

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(1)

Given, $| \frac{\bar{z} - i}{2 \bar{z} + i} |$ $= \frac{1}{3} and z \in C$

$\Rightarrow$ $| \frac{\bar{z} - i}{\bar{z} + \frac{i}{2}} |$ $= \frac{2}{3} \Rightarrow 3$ $| x - i y - i |$ $= 2$ $| x - i y + \frac{i}{2} |$

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

$\Rightarrow 3 \sqrt{x^{2} + {(y + 1)}^{2}} = 2 \sqrt{{(x)}^{2} + {(y - \frac{1}{2})}^{2}}$

Squaring on both sides, we get

$9 (x^{2} + y^{2} + 1 + 2 y) = 4 (x^{2} + y^{2} + \frac{1}{4} - y)$

$\Rightarrow 9 x^{2} + 9 y^{2} + 9 + 18 y = 4 x^{2} + 4 y^{2} + 1 - 4 y$

$\Rightarrow 5 x^{2} + 5 y^{2} + 22 y + 8 = 0$

$\Rightarrow x^{2} + y^{2} + \frac{22}{5} y + \frac{8}{5} = 0$

Centre $(C) = (0, - \frac{11}{5})$

Now, $\frac{1}{2}$ $| 0 + 0 + α (0 + \frac{11}{5}) |$ $= 11$ [Given]

$\Rightarrow \frac{11 α}{5} = 22 \Rightarrow α = 10$ $∴ α^{2} = 100$ .

Q 14 :

The number of complex numbers z, satisfying $| z |$ $= 1 and$ $| \frac{z}{\bar{z}} + \frac{\bar{z}}{z} |$ $= 1$ , is: [2025]

8
10
6
4

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(1)

Let $z = e^{i θ}$ , then $\bar{z} = e^{- i θ} \Rightarrow \frac{z}{\bar{z}} = e^{i 2 θ}$

Also, |z| = 1

$∵$ $| \frac{z}{\bar{z}} + \frac{\bar{z}}{z} |$ $= 1 \Rightarrow$ $| e^{i 2 θ} + e^{- i 2 θ} |$ $= 1 \Rightarrow$ $| \cos 2 θ |$ $= \frac{1}{2}$

$∴$ Number of solutions = 8.

Q 15 :

Let O be the origin, the point A be $z_{1} = \sqrt{3} + 2 \sqrt{2} i$ , the point $B (z_{2})$ be such that $\sqrt{3} | z_{2} | = | z_{1} |$ and $\arg (z_{2}) = \arg (z_{1}) + \frac{π}{6}$ . Then [2025]

area of triangle ABO is $\frac{11}{\sqrt{3}}$
ABO is an obtuse angled isosceles triangle
ABO is a scalene triangle
area of triangle ABO is $\frac{11}{4}$

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(2)

We have, $z_{1} = \sqrt{3} + 2 \sqrt{2} i$ ; $\sqrt{3} | z_{2} | = | z_{1} |$ and $\arg (z_{2}) = \arg (z_{1}) + \frac{π}{6}$

$\Rightarrow \arg (z_{2}) - \arg (z_{1}) = \frac{π}{6}$

$∴ z_{2} = \frac{| z_{2} |}{| z_{1} |} \cdot z_{1} e^{i (π / 6)}$

$= \frac{1}{\sqrt{3}} [\frac{(\sqrt{3} + 2 \sqrt{2} i) (\sqrt{3} + i)}{2}]$

$\Rightarrow z_{2} = \frac{1}{2 \sqrt{3}} [3 - 2 \sqrt{2} + i (2 \sqrt{6} + \sqrt{3})]$

Now, $z_{1} - z_{2} = \sqrt{3} + 2 \sqrt{2} i - \frac{(3 - 2 \sqrt{2} + i) (2 \sqrt{6} + \sqrt{3})}{2 \sqrt{3}}$

$= \frac{6 + 4 \sqrt{6} i - 3 + 2 \sqrt{2} - 2 \sqrt{6} i - i \sqrt{3}}{2 \sqrt{3}}$

$= \frac{3 + 2 \sqrt{2} + i (2 \sqrt{6} - \sqrt{3})}{2 \sqrt{3}}$ $∴$ $| z_{1} - z_{2} |$ $=$ $| z_{2} |$

$\Rightarrow$ $△$ ABO is isosceles with angles $\frac{π}{6}, \frac{π}{6}$ and $\frac{2 π}{3}$ .

$∴$ Area of $△$ ABO = $\frac{1}{2} \sqrt{11} \times \frac{\sqrt{11}}{\sqrt{3}} \sin \frac{π}{6} = \frac{11}{4 \sqrt{3}}$ .

Q 16 :

Let $| z_{1} - 8 - 2 i |$ $\leq 1$ and $| z_{2} - 2 + 6 i |$ $\leq 2, z_{1}, z_{2} \in C$ . Then the minimum value of $| z_{1} - z_{2} |$ is : [2025]

13
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7
10

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(3)

$A B = \sqrt{{(8 - 2)}^{2} + {(2 + 6)}^{2}} = \sqrt{100} = 10$

$∴$ $| z_{1} - z_{2} |_{\min}$ $= 10 - 2 - 1 = 7$

Q 17 :

Let $w_{1}$ be the point obtained by the rotation of $z_{1} = 5 + 4 i$ about the origin through a right angle in the anticlockwise direction, and $w_{2}$ be the point obtained by the rotation of $z_{2} = 3 + 5 i$ about the origin through a right angle in the clockwise direction. Then the principal argument of $w_{1} - w_{2}$ is equal to [2023]

$- π + \tan^{- 1} \frac{33}{5}$
$π - \tan^{- 1} \frac{33}{5}$
$- π + \tan^{- 1} \frac{8}{9}$
$π - \tan^{- 1} \frac{8}{9}$

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(4)

$w_{1} = z_{1} i = (5 + 4 i) i = - 4 + 5 i$

$w_{2} = z_{2} (- i) = (3 + 5 i) (- i) = 5 - 3 i$

$w_{1} - w_{2} = - 9 + 8 i$

$Principal argument = π - \tan^{- 1} (\frac{8}{9})$

Q 18 :

Let C be the circle in the complex plane with centre $z_{0} = \frac{1}{2} (1 + 3 i)$ and radius $r$ = 1. Let $z_{1} = 1 + i$ and the complex number $z_{2}$ be outside the circle C such that $| z_{1} - z_{0} |$ $| z_{2} - z_{0} |$ $= 1 .$ If $z_{0}, z_{1}$ , and $z_{2}$ are collinear, then the smaller value of $| z_{2} |^{2}$ is equal to [2023]

$\frac{5}{2}$
$\frac{7}{2}$
$\frac{3}{2}$
$\frac{13}{2}$

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(1)

$z_{0} = (\frac{1}{2} + \frac{3 i}{2})$

$z_{1} = 1 + i$

$z_{1} - z_{0} = (1 + i) - (\frac{1}{2} + \frac{3 i}{2}) = \frac{1}{2} - \frac{i}{2} = \frac{1 - i}{2}$

$| z_{1} - z_{0} |$ $= \sqrt{\frac{1}{4} + \frac{1}{4}} = \frac{1}{\sqrt{2}}$

$As | z_{1} - z_{0} |$ $| z_{2} - z_{0} |$ $= 1$

$\Rightarrow \frac{1}{\sqrt{2}}$ $| z_{2} - z_{0} |$ $= 1 \Rightarrow$ $| z_{2} - z_{0} |$ $= \sqrt{2}$

Slope of line forming $z_{0}, z_{2} =$ $\frac{\frac{3}{2} - 1}{\frac{1}{2} - 1} = \frac{\frac{1}{2}}{\frac{- 1}{2}} = - 1$

$\Rightarrow \tan θ = - 1 = \tan 135 ° \Rightarrow θ = 135 °$

$| z_{2} - z_{0} |$ $= \sqrt{2}$

$z_{2} (\frac{1}{2} + \sqrt{2} \cos 135 °, \frac{3}{2} + \sqrt{2} \sin 135 °)$

or

$z_{2} (\frac{1}{2} - \sqrt{2} \cos 135 °, \frac{3}{2} - \sqrt{2} \sin 135 °)$

$z_{2} (\frac{- 1}{2}, \frac{5}{2})$ or $z_{2} (\frac{3}{2}, \frac{1}{2})$

$| z_{2} |^{2} = \frac{1}{4} + \frac{25}{4} = \frac{26}{4}$ or $| z_{2} |^{2} = \frac{9}{4} + \frac{1}{4} = \frac{5}{2};$ $\Rightarrow | z_{2} |_{\min}^{2} = \frac{5}{2}$

Q 19 :

If the center and radius of the circle $| \frac{z - 2}{z - 3} |$ $= 2$ are respectively $(α, β)$ and $γ$ , then $3 (α + β + γ)$ is equal to [2023]

12
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10
9

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(1)

$Centre of circle$ $| \frac{z - a}{z - b} |$ $= k (\neq 1) is$

$(\frac{k^{2} b - a}{k^{2} - 1}, 0)$ and radius = $\frac{k | a - b |}{k^{2} - 1}$

Now, centre of circle $| \frac{z - 2}{z - 3} |$ $= 2$ is $(\frac{4 \times 3 - 2}{4 - 1}, 0)$ i.e., $(\frac{10}{3}, 0)$

and radius = $\frac{2 \times 1}{3} = \frac{2}{3}$ $∴$ $α = \frac{10}{3}, β = 0, γ = \frac{2}{3}$

$3 (α + β + γ) = 3 (\frac{10}{3} + 0 + \frac{2}{3}) = 12$

Q 20 :

For all $z \in C$ on the curve $C_{1} :$ $| z |$ $= 4$ , let the locus of the point $z + \frac{1}{z}$ be the curve $C_{2}$ . Then: [2023]

the curve $C_{1}$ lies inside $C_{2}$
the curves $C_{1}$ and $C_{2}$ intersect at 4 points
the curve $C_{2}$ lies inside $C_{1}$
the curves $C_{1}$ and $C_{2}$ intersect at 2 points

1

2

3

4

(2)

We have, curve $C_{1} :$ $| z |$ $= 4 ∀ z \in C$ , which is a circle with centre (0, 0) and radius 4, therefore,

$x^{2} + y^{2} = 16$ $\dots (i)$

$Now, z = 4 e^{i θ}$

$z + \frac{1}{z} = 4 e^{i θ} + \frac{1}{4 e^{i θ}} = 4 e^{i θ} + \frac{1}{4} e^{- i θ}$

$= 4 (\cos θ + i \sin θ) + \frac{1}{4} (\cos θ - i \sin θ) = \frac{17}{4} \cos θ + i \frac{15}{4} \sin θ$

$To eliminate θ, let α = \frac{17}{4} \cos θ and β = \frac{15}{4} \sin θ$

$\frac{α^{2}}{{(\frac{17}{4})}^{2}} + \frac{β^{2}}{{(\frac{15}{4})}^{2}} = \frac{{(\frac{17}{4})}^{2} \cos^{2} θ}{{(\frac{17}{4})}^{2}} + \frac{{(\frac{15}{4})}^{2} \sin^{2} θ}{{(\frac{15}{4})}^{2}} = 1$

$∴$ $The curve C_{2} is \frac{x^{2}}{{(\frac{17}{4})}^{2}} + \frac{y^{2}}{{(\frac{15}{4})}^{2}} = 1$ $\dots (i i)$

which is an ellipse with centre (0, 0).

From (i) and (ii),

Hence, the curves $C_{1}$ and $C_{2}$ intersect at 4 points.

Q 21 :

$For α, β, z \in ℂ and λ > 1,$ $if \sqrt{λ - 1}$ is the radius of the circle $| z - α |^{2} + | z - β |^{2} = 2 λ,$ then $| α - β |$ is equal to _________. [2023]

(2)

For a circle, we have, $| z - z_{1} |^{2} + | z - z_{2} |^{2} = | z_{1} - z_{2} |^{2}$

$r = \frac{| z_{1} - z_{2} |}{2} = \frac{| α - β |}{2} = \sqrt{λ - 1}$

$Also, 2 λ = | α - β |^{2}$

$\Rightarrow 2 λ = 4 (λ - 1) \Rightarrow 2 λ = 4 λ - 4 \Rightarrow 2 λ = 4 \Rightarrow λ = 2$

$∴$ $| α - β |$ $= 2$

Q 22 :

$Let w = z \bar{z} + k_{1} z + k_{2} i z + λ (1 + i), k_{1}, k_{2} \in ℝ .$ $Let Re (w) = 0 be the circle C of radius 1$ in the first quadrant touching the line $y = 1$ and the y-axis. If the curve $Im (w) = 0$ intersects C at A and B, then $30 {(A B)}^{2}$ is equal to ______. [2023]

(24)

We have, $w = z \bar{z} + k_{1} z + k_{2} i z + λ (1 + i)$

Put $z = x + i y$

$w = x^{2} + y^{2} + k_{1} (x + i y) + k_{2} i (x + i y) + λ (1 + i)$

$= x^{2} + y^{2} + k_{1} x + k_{1} i y + k_{2} i x - k_{2} y + λ + λ i$

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

Given $Re (w) = 0$

$\Rightarrow x^{2} + y^{2} + k_{1} x - k_{2} y + λ = 0$ $...(i)$

$Centre = (\frac{- k_{1}}{2}, \frac{k_{2}}{2})$

$Radius = \sqrt{\frac{k_{1}^{2}}{4} + \frac{k_{2}^{2}}{4} - λ}$

Now, $\frac{- k_{1}}{2} = 1 \Rightarrow k_{1} = - 2$

and $\frac{k_{2}}{2} = 2 \Rightarrow k_{2} = 4$

Also, $\sqrt{\frac{k_{1}^{2}}{4} + \frac{k_{2}^{2}}{4} - λ} = 1$

$\Rightarrow \frac{k_{1}^{2}}{4} + \frac{k_{2}^{2}}{4} - λ = 1$

$\Rightarrow 1 + 4 - λ = 1 \Rightarrow λ = 4$

Put $k_{1} = - 2, k_{2} = 4, λ = 4$ in (i)

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

Now, $Im (w) = 0,$ $k_{2} x + k_{1} y + λ = 0$

$4 x - 2 y + 4 = 0$ $...(iii)$

From (ii) and (iii), $A \equiv (0, 2)$ and $B \equiv (0.4, 2.8)$

Now, $A B = \sqrt{{(0.4 - 0)}^{2} + {(2.8 - 2)}^{2}}$

$\Rightarrow A B^{2} = {(0.4)}^{2} + {(0.8)}^{2} = 0.8$

$∴$ $30 {(A B)}^{2} = 30 (0.8) = 24$

Q 23 :

$Let z = 1 + i and z_{1} = \frac{1 + i \bar{z}}{\bar{z} (1 - z) + \frac{1}{z}} . Then \frac{12}{π} \arg (z_{1}) is equal to$ __________ . [2023]

(9)

We have $z = 1 + i$ and $z_{1} = \frac{1 + i \bar{z}}{\bar{z} (1 - z) + \frac{1}{z}}$

$= \frac{1 + i (1 - i)}{(1 - i) [1 - (1 + i)] + \frac{1}{1 + i}} = - 1 + i$

Now, $\arg (z_{1}) = π - \tan^{- 1} (1) = π - \frac{π}{4} = \frac{3 π}{4}$

$\frac{12}{π} \arg (z_{1}) = \frac{12}{π} \times \frac{3 π}{4} = 9$

Topic Question Set

Q 1 :

If $z_{1}, z_{2}$ are two distinct complex number such that $| \frac{z_{1} - 2 z_{2}}{\frac{1}{2} - z_{1} {\bar{z}}_{2}} |$ $= 2$ , then [2024]

Q 2 :

Let $r$ and $θ$ respectively be the modulus and amplitude of the complex number $z = 2 - i (2 \tan \frac{5 π}{8}),$ then $(r, θ)$ is equal to [2024]

Q 3 :

Let the complex numbers $α$ and $\frac{1}{\bar{α}}$ lie on the circles $| z - z_{0} |^{2} = 4$ and $| z - z_{0} |^{2} = 16$ respectively, where $z_{0} = 1 + i .$ Then, the value of $100 | α |^{2}$ is ________. [2024]

Q 5 :

The area (in sq. units) of the region

$S = {z \in C : | z - 1 | \leq 2; (z + \bar{z}) + i (z - \bar{z}) \leq 2, Im (z) \geq 0}$ is [2024]

Q 6 :

Let $S_{1} = {z \in C : | z | \leq 5}$ , $S_{2} = {z \in C : Im (\frac{z + 1 - \sqrt{3} i}{1 - \sqrt{3} i}) \geq 0}$ and $S_{3} = {z \in C : Re (z) \geq 0} .$ Then the area of the region $S_{1} \cap S_{2} \cap S_{3}$ is : [2024]

Q 7 :

The sum of the square of the modulus of the elements in the set ${z = a + i b : a, b \in Z, z \in C, | z - 1 | \leq 1, | z - 5 | \leq | z - 5 i |}$ is ____________. [2024]

Q 8 :

Let z be a complex number such that |z| = 1. If $\frac{2 + k^{2} z}{k + \bar{z}} = k z, k \in R$ , then the maximum distance of $k + i k^{2}$ from the circle |z – (1 + 2i)| = 1 is : [2025]

Q 9 :

If $z_{1}, z_{2}, z_{3} \in C$ are the vertices of an equilateral triangle, whose centroid is $z_{0}$ , then $\sum_{k = 1}^{3} {(z_{k} - z_{0})}^{2}$ is equal to [2025]

Q 10 :

If the locus of $z \in C$ , such that $Re (\frac{z - 1}{2 z + i}) + R e (\frac{\bar{z} - 1}{2 \bar{z} - i}) = 2$ , is a circle of radius r and center (a, b), then $\frac{15 a b}{r^{2}}$ is equal to: [2025]

Q 12 :

Let the curve $z (1 + i) + \bar{z} (1 - i) = 4, z \in C$ , divide the region $| z - 3 | \leq 1$ into two parts of area $α$ and $β$ . Then $| α - β |$ equals : [2025]

Q 13 :

Let $| \frac{\bar{z} - i}{2 \bar{z} + i} |$ $= \frac{1}{3}, z \in C$ , be the equation of a circle with center at C. If the area of the triangle, whose vertices are at the point (0, 0), C and $(α, 0)$ is 11 square units, then $α^{2}$ equals : [2025]

Q 14 :

The number of complex numbers z, satisfying $| z |$ $= 1 and$ $| \frac{z}{\bar{z}} + \frac{\bar{z}}{z} |$ $= 1$ , is: [2025]

Q 15 :

Let O be the origin, the point A be $z_{1} = \sqrt{3} + 2 \sqrt{2} i$ , the point $B (z_{2})$ be such that $\sqrt{3} | z_{2} | = | z_{1} |$ and $\arg (z_{2}) = \arg (z_{1}) + \frac{π}{6}$ . Then [2025]

Q 16 :

Let $| z_{1} - 8 - 2 i |$ $\leq 1$ and $| z_{2} - 2 + 6 i |$ $\leq 2, z_{1}, z_{2} \in C$ . Then the minimum value of $| z_{1} - z_{2} |$ is : [2025]

Q 19 :

If the center and radius of the circle $| \frac{z - 2}{z - 3} |$ $= 2$ are respectively $(α, β)$ and $γ$ , then $3 (α + β + γ)$ is equal to [2023]

Q 20 :

For all $z \in C$ on the curve $C_{1} :$ $| z |$ $= 4$ , let the locus of the point $z + \frac{1}{z}$ be the curve $C_{2}$ . Then: [2023]

Q 21 :

$For α, β, z \in ℂ and λ > 1,$ $if \sqrt{λ - 1}$ is the radius of the circle $| z - α |^{2} + | z - β |^{2} = 2 λ,$ then $| α - β |$ is equal to _________. [2023]

Q 22 :

$Let w = z \bar{z} + k_{1} z + k_{2} i z + λ (1 + i), k_{1}, k_{2} \in ℝ .$ $Let Re (w) = 0 be the circle C of radius 1$ in the first quadrant touching the line $y = 1$ and the y-axis. If the curve $Im (w) = 0$ intersects C at A and B, then $30 {(A B)}^{2}$ is equal to ______. [2023]

Q 23 :

$Let z = 1 + i and z_{1} = \frac{1 + i \bar{z}}{\bar{z} (1 - z) + \frac{1}{z}} . Then \frac{12}{π} \arg (z_{1}) is equal to$ __________ . [2023]

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Topic Question Set

Q 1 : If z1,z2 are two distinct complex number such that |z1-2z212-z1z¯2|=2, then [2024]

Q 2 : Let r and θ respectively be the modulus and amplitude of the complex number z=2-i(2tan5π8), then (r,θ) is equal to [2024]

Q 3 : Let the complex numbers α and 1α¯ lie on the circles |z-z0|2=4 and |z-z0|2=16 respectively, where z0=1+i. Then, the value of 100|α|2 is ________. [2024]

Q 4 : If α denotes the number of solutions of |1-i|x=2x and β=(|z|arg(z)), where z=π4(1+i)4[1-πiπ+i+π-i1+πi],i=-1, then the distance of the point (α,β) from the line 4x-3y=7 is __________ [2024]

Q 5 : The area (in sq. units) of the region S={z∈C:|z-1|≤2; (z+z¯)+i(z-z¯)≤2, Im(z)≥0} is [2024]

Q 6 : Let S1={z∈C:|z|≤5}, S2={z∈C:Im(z+1-3i1-3i)≥0} and S3={z∈C:Re(z)≥0}. Then the area of the region S1∩S2∩S3 is : [2024]

Q 7 : The sum of the square of the modulus of the elements in the set {z=a+ib:a,b∈Z,z∈C,|z-1|≤1,|z-5|≤|z-5i|} is ____________. [2024]

Q 8 : Let z be a complex number such that |z| = 1. If 2+k2zk+z–=kz, k∈R, then the maximum distance of k+ik2 from the circle |z – (1 + 2i)| = 1 is : [2025]

Q 9 : If z1,z2,z3∈C are the vertices of an equilateral triangle, whose centroid is z0, then ∑k=13(zk–z0)2 is equal to [2025]

Q 10 : If the locus of z∈C, such that Re(z–12z+i)+Re(z––12z–-i)=2, is a circle of radius r and center (a, b), then 15abr2 is equal to: [2025]

Q 11 : Let z1, z2 and z3 be three complex number on the circle |z| = 1 with arg(z1)=–π4, arg(z2)=0 and arg(z3)=π4. If |z1z2+z2z3+z3z1|2=α+β2, α,β∈Z, then the value of α2+β2 is : [2025]

Q 12 : Let the curve z(1+i)+z¯(1–i)=4, z∈C, divide the region |z–3|≤1 into two parts of area α and β. Then |α–β| equals : [2025]

Q 13 : Let |z¯–i2z¯+i|=13, z∈C, be the equation of a circle with center at C. If the area of the triangle, whose vertices are at the point (0, 0), C and (α,0) is 11 square units, then α2 equals : [2025]

Q 14 : The number of complex numbers z, satisfying |z|=1 and |zz¯+z¯z|=1, is: [2025]

Q 15 : Let O be the origin, the point A be z1=3+22i, the point B(z2) be such that 3|z2|=|z1| and arg(z2)=arg(z1)+π6. Then [2025]

Q 16 : Let |z1–8–2i|≤1 and |z2–2+6i|≤2, z1,z2∈C. Then the minimum value of |z1–z2| is : [2025]

Q 18 : Let C be the circle in the complex plane with centre z0=12(1+3i) and radius r = 1. Let z1=1+i and the complex number z2 be outside the circle C such that |z1-z0||z2-z0|=1. If z0,z1, and z2 are collinear, then the smaller value of |z2|2 is equal to [2023]

Q 19 : If the center and radius of the circle |z-2z-3|=2 are respectively (α,β) and γ, then 3(α+β+γ) is equal to [2023]

Q 20 : For all z∈C on the curve C1:|z|=4, let the locus of the point z+1z be the curve C2. Then: [2023]

Q 21 : For α,β,z∈ℂ and λ>1, if λ-1 is the radius of the circle |z-α|2+|z-β|2=2λ, then |α-β| is equal to _________. [2023]

Q 22 : Let w=zz¯+k1z+k2iz+λ(1+i), k1,k2∈ℝ. Let Re(w)=0 be the circle C of radius 1 in the first quadrant touching the line y=1 and the y-axis. If the curve Im(w)=0 intersects C at A and B, then 30(AB)2 is equal to ______. [2023]

Q 23 : Let z=1+i and z1=1+iz¯z¯(1-z)+1z. Then 12πarg(z1) is equal to __________ . [2023]

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Q 1 :

If $z_{1}, z_{2}$ are two distinct complex number such that $| \frac{z_{1} - 2 z_{2}}{\frac{1}{2} - z_{1} {\bar{z}}_{2}} |$ $= 2$ , then [2024]

Q 2 :

Let $r$ and $θ$ respectively be the modulus and amplitude of the complex number $z = 2 - i (2 \tan \frac{5 π}{8}),$ then $(r, θ)$ is equal to [2024]

Q 3 :

Let the complex numbers $α$ and $\frac{1}{\bar{α}}$ lie on the circles $| z - z_{0} |^{2} = 4$ and $| z - z_{0} |^{2} = 16$ respectively, where $z_{0} = 1 + i .$ Then, the value of $100 | α |^{2}$ is ________. [2024]

Q 5 :

The area (in sq. units) of the region

$S = {z \in C : | z - 1 | \leq 2; (z + \bar{z}) + i (z - \bar{z}) \leq 2, Im (z) \geq 0}$ is [2024]

Q 6 :

Let $S_{1} = {z \in C : | z | \leq 5}$ , $S_{2} = {z \in C : Im (\frac{z + 1 - \sqrt{3} i}{1 - \sqrt{3} i}) \geq 0}$ and $S_{3} = {z \in C : Re (z) \geq 0} .$ Then the area of the region $S_{1} \cap S_{2} \cap S_{3}$ is : [2024]

Q 7 :

The sum of the square of the modulus of the elements in the set ${z = a + i b : a, b \in Z, z \in C, | z - 1 | \leq 1, | z - 5 | \leq | z - 5 i |}$ is ____________. [2024]

Q 8 :

Let z be a complex number such that |z| = 1. If $\frac{2 + k^{2} z}{k + \bar{z}} = k z, k \in R$ , then the maximum distance of $k + i k^{2}$ from the circle |z – (1 + 2i)| = 1 is : [2025]

Q 9 :

If $z_{1}, z_{2}, z_{3} \in C$ are the vertices of an equilateral triangle, whose centroid is $z_{0}$ , then $\sum_{k = 1}^{3} {(z_{k} - z_{0})}^{2}$ is equal to [2025]

Q 10 :

If the locus of $z \in C$ , such that $Re (\frac{z - 1}{2 z + i}) + R e (\frac{\bar{z} - 1}{2 \bar{z} - i}) = 2$ , is a circle of radius r and center (a, b), then $\frac{15 a b}{r^{2}}$ is equal to: [2025]

Q 12 :

Let the curve $z (1 + i) + \bar{z} (1 - i) = 4, z \in C$ , divide the region $| z - 3 | \leq 1$ into two parts of area $α$ and $β$ . Then $| α - β |$ equals : [2025]

Q 13 :

Let $| \frac{\bar{z} - i}{2 \bar{z} + i} |$ $= \frac{1}{3}, z \in C$ , be the equation of a circle with center at C. If the area of the triangle, whose vertices are at the point (0, 0), C and $(α, 0)$ is 11 square units, then $α^{2}$ equals : [2025]

Q 14 :

The number of complex numbers z, satisfying $| z |$ $= 1 and$ $| \frac{z}{\bar{z}} + \frac{\bar{z}}{z} |$ $= 1$ , is: [2025]

Q 15 :

Let O be the origin, the point A be $z_{1} = \sqrt{3} + 2 \sqrt{2} i$ , the point $B (z_{2})$ be such that $\sqrt{3} | z_{2} | = | z_{1} |$ and $\arg (z_{2}) = \arg (z_{1}) + \frac{π}{6}$ . Then [2025]

Q 16 :

Let $| z_{1} - 8 - 2 i |$ $\leq 1$ and $| z_{2} - 2 + 6 i |$ $\leq 2, z_{1}, z_{2} \in C$ . Then the minimum value of $| z_{1} - z_{2} |$ is : [2025]

Q 19 :

If the center and radius of the circle $| \frac{z - 2}{z - 3} |$ $= 2$ are respectively $(α, β)$ and $γ$ , then $3 (α + β + γ)$ is equal to [2023]

Q 20 :

For all $z \in C$ on the curve $C_{1} :$ $| z |$ $= 4$ , let the locus of the point $z + \frac{1}{z}$ be the curve $C_{2}$ . Then: [2023]

Q 21 :

$For α, β, z \in ℂ and λ > 1,$ $if \sqrt{λ - 1}$ is the radius of the circle $| z - α |^{2} + | z - β |^{2} = 2 λ,$ then $| α - β |$ is equal to _________. [2023]

Q 22 :

$Let w = z \bar{z} + k_{1} z + k_{2} i z + λ (1 + i), k_{1}, k_{2} \in ℝ .$ $Let Re (w) = 0 be the circle C of radius 1$ in the first quadrant touching the line $y = 1$ and the y-axis. If the curve $Im (w) = 0$ intersects C at A and B, then $30 {(A B)}^{2}$ is equal to ______. [2023]

Q 23 :

$Let z = 1 + i and z_{1} = \frac{1 + i \bar{z}}{\bar{z} (1 - z) + \frac{1}{z}} . Then \frac{12}{π} \arg (z_{1}) is equal to$ __________ . [2023]