Question

If $z_1$ and $z_2$ are two complex numbers such that $\left|z_1\right|=\left|z_2\right|$ and $arg{z}_1+arg{z}_2=\pi$ then $z_1$ and $z_2$ are

• A. 1
• B. $z_1=z_2$
• C. $z_1=-{\overline{z}}_2$
• D. none of these

Answer 1

The correct option is C $z_1=-{\overline{z}}_2$

Let $z_1=\left|z_1\right|(\cos {\theta }_1+i\sin {\theta }_1)$ and $z_2=\left|z_2\right|(\cos {\theta }_2+i\sin {\theta }_2)$

Given:$\left|z_1\right|=\left|z_2\right|$

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

$\Rightarrow {\theta }_1+{\theta }_2=\pi$

$\Rightarrow {\theta }_1=\pi -{\theta }_2$

Now,$z_1=\left|z_2\right|[\cos (\pi -{\theta }_2)+i\sin (\pi -{\theta }_2)]$

$\Rightarrow z_1=\left|z_2\right|[-\cos {\theta }_2+i\sin {\theta }_2]$

$\Rightarrow z_1=-\left|z_2\right|[\cos {\theta }_2-i\sin {\theta }_2]$

$\Rightarrow z_1=-\left[\left|z_2\right|(\cos {\theta }_2-i\sin {\theta }_2)\right]$

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