Question

If $z_1$ and $z_2$ are two distinct complex numbers satisfying the relation $\left|z_1^2-z_2^2\right|=\left|{\bar{z}}_1^2+{\bar{z}}_2^2-2{\bar{z}}_1{\bar{z}}_2\right|$ and $\left(\arg z_1-\arg z_2\right)=\frac{a\pi }{b}$, then the least possible value of $|a-b|$ is equal to (where, $a\&b$ are integers)

Answer 1

Answer: 1

$\left|z_1^2-z_2^2\right|=\left|{\bar{z}}_1^2+{\bar{z}}_2^2-2{\bar{z}}_1{\bar{z}}_2\right|$
$\left|z_1^2-z_2^2\right|=\left|z_1^2+z_2^2-2z_1{z}_2\right|$
$\left|z_1+z_2\right|\left|z_1-z_2\right|=\left|z_1-z_2\right|\left|z_1-z_2\right|$
$\Rightarrow \left|z_1+z_2\right|=\left|z_1-z_2\right|$
$\overline{z_1}\perp \overline{z_2}$
$\Rightarrow arg\left(\frac{z_1}{z_2}\right)=2n\pi \pm \frac{\pi }{2}$
i.e. $-\frac{\pi }{2},\frac{\pi }{2},\frac{3\pi }{2},\dots \dots ..$
The minimum value of $\left|a-b\right|=1$ (when $a=1\&b=2$ )