Let $\mathrm{z}_{1}$ and $\mathrm{z}_{2}$ be two complex numbers such that $\overline{\mathrm{z}}_{1}=\mathrm{i} \overline{\mathrm{z}}_{2}$ and $\arg \left(\frac{\mathrm{z}_{1}}{\overline{\mathrm{z}}_{2}}\right)=\pi$. Then

(C) $\arg \mathrm{z}_{1}=\frac{\pi}{4}$

(A) $\arg \mathrm{z}_{2}=\frac{\pi}{4}$

(B) $\arg \mathrm{z}_{2}=-\frac{3 \pi}{4}$

(D) $\arg \mathrm{z}_{1}=-\frac{3 \pi}{4}$

JEE Main · 2022 · 25 Jun 2022 · Shift 2 · Complex Numbers

Let

z_{1}

and

z_{2}

be two complex numbers such that

\overline{z}_{1} = i \overline{z}_{2}

and

ar g (\frac{z _{1}}{z _{2}}) = π

. Then

Answer: (C)

Given

\overline{z}_{1} = i \overline{z}_{2}

, taking conjugate:

z_{1} = - i z_{2}

.

Now,

ar g (\frac{z _{1}}{z _{2}}) = π

becomes

ar g (\frac{- i z _{2}}{z _{2}}) = π

.

Let

ar g (z_{2}) = θ

. Then:

ar g (- i) + ar g (z_{2}) - ar g (\overline{z}_{2}) = π

- \frac{π}{2} + θ - (- θ) = π

- \frac{π}{2} + 2 θ = π

2 θ = \frac{3 π}{2}

θ = \frac{3 π}{4}

ar g (z_{2}) = \frac{3 π}{4}

.

Since

z_{1} = - i z_{2}

, we have

ar g (z_{1}) = ar g (- i) + ar g (z_{2}) = - \frac{π}{2} + \frac{3 π}{4} = \frac{π}{4}

JEE Main · 2022 · 25 Jun 2022 · Shift 2 · Single correct option