Let z _ 1 and z _ 2 be two complex numbers such that ¯ z 1 = i ¯ z 2 and arg( z _ 1 ¯ z 2 ) = π , then the argument of z _ 1 is

Correct answer: C. arg z _ 1 = (π)/(4) ¯ z 1 = i ¯ z 2 ⇒ z _ 1 = - i z _ 2 ⇒ i z _ 1 = z _ 2 ⇒ e(iπ)/(2) z _ 1 = z _ 2 Given, arg( z _ 1 ¯ z 2 ) = π Let z _ 1 = r e ^ iθ ⇒ ¯ z _ 2 = r e ^ - i((π)/(2) + θ) So arg( e ^ i(2θ + (π)/(2)) ) = π ⇒ 2θ + (π)/(2) = π ⇒ θ = (π)/(4)

JEE Main 202225 JuneEvening Shift

Exam: JEE Main
Year: 2022
Date: 25 June
Shift: Evening Shift
Subject: Mathematics
Chapter: Complex Number

JEE Main 2022 · 25 June · Evening Shift · Mathematics · Complex Number

Let z _ 1 and z _ 2 be two complex numbers such that ¯ z 1 = i ¯ z...

Let

{z}_{1}

and

{z}_{2}

be two complex numbers such that

\overset{¯}{z}1 = i\overset{¯}{z}2

and

arg\left(\dfrac{{z}_{1}}{\overset{¯}{z}2}\right) = \pi

, then the argument of

{z}_{1}

Options

A $\text{arg}{z}_{2} = \dfrac{\pi}{4}$
B $\text{arg}{z}_{2} = - \dfrac{3\pi}{4}$
C $\text{arg}{z}_{1} = \dfrac{\pi}{4}$
D $\text{arg}{z}_{1} = - \dfrac{3\pi}{4}$

Correct answer

C. $\text{arg}{z}_{1} = \dfrac{\pi}{4}$

Step-by-step solution

$\overset{¯}{z}1 = i\overset{¯}{z}2 ⇒ {z}_{1} = - i{z}_{2} ⇒ i{z}_{1} = {z}_{2} ⇒ e\dfrac{i\pi}{2}{z}_{1} = {z}_{2}$

Given, $arg\left(\dfrac{{z}_{1}}{\overset{¯}{z}2}\right) = \pi$

Let ${z}_{1} = r{e}^{i\theta} ⇒ \overset{¯}{{z}_{2}} = r{e}^{- i\left(\dfrac{\pi}{2} + \theta\right)}$

So $arg\left({e}^{i\left(2\theta + \dfrac{\pi}{2}\right)}\right) = \pi ⇒ 2\theta + \dfrac{\pi}{2} = \pi$
$⇒ \theta = \dfrac{\pi}{4}$

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