Question

If $z_1$ and $z_2$ be two distinct complex numbers satisfying $|z_1^2-z_2^2|=|{\overline{z_1}}^2+{\overline{z_2}}^2-2\overline{z_1}\overline{z_2}|$ and if $(\arg z_1-\arg z_2)=\frac{a\pi }{b}$ , then the least possible value of $\frac{|a+b|}{2}$ is $(a,b\in I)$

Answer 1

$|z_1^2-z_2^2|=|(\overline{z_1}-\overline{z_2}{)}^2|$

$\Rightarrow |z_1+z_2|=|{\overline{z}}_1-{\overline{z}}_2|$
$\Rightarrow |z_1+z_2|=|z_1-z_2|$

$z_1\perp z_2$

$\Rightarrow \arg \left(\frac{z_1}{z_2}\right)=2n\pi \pm \frac{\pi }{2}\Rightarrow -\frac{\pi }{2},\frac{\pi }{2}$

The minimum value of $|a+b|=1$
where $a=-1$ and $b=2$