Question

$z_1$ and $z_2$ are two complex numbers such that $\left|z_1\right|=\left|z_2\right|$ and $arg\left(z_1\right)+arg\left(z_2\right)=\pi$, then $z_1$ is equal to:

• A. $2{\overline{z}}_2$
• B. ${\overline{z}}_2$
• C. $-{\overline{z}}_2$
• D. None of these

Answer 1

The correct option is A $2{\overline{z}}_2$

Let $|z_1|=|z_2|=r$

let $arg\left(z_1\right)=0$

Given,

$arg\left(z_1\right)+arg\left(z_2\right)=\pi$

$\Rightarrow a+arg\left(z_2\right)=\pi$

$\Rightarrow arg\left(z_2\right)=\pi -\theta$

$z_1=r(\cos \theta +i\sin \theta )$

$z_2=r\left(\cos \right(\pi -\theta )+i\sin (\pi -\theta \left)\right)$

$z_2=r(-\cos \theta +i\sin \theta )$

$z_2=-r(\cos \theta -i\sin \theta )$

$z_2=-\overline{z_1}$

$\therefore \overline{z_2}=-\left(\overline{z_1}\right)$

$\Rightarrow \overline{z_2}=-z_1$

$\therefore z_1=-\overline{z_2}$