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

$\left|z_{1}^{2} - z_{2}^{2}\right|=\left|\bar{z}_{1}^{2} + \bar{z}_{2}^{2} - 2 \bar{z}_{1} \bar{z}_{2}\right|$
$\left|z_{1}^{2} - z_{2}^{2}\right|=\left|z_{1}^{2} + z_{2}^{2} - 2 z_{1} z_{2}\right|$
$\left|z_{1} + z_{2}\right|\left|z_{1} - z_{2}\right|=\left|z_{1} - z_{2}\right|\left|z_{1} - z_{2}\right|$
$\Rightarrow \left|z_{1} + z_{2}\right|=\left|z_{1} - z_{2}\right|$
$\bar{z_{1}}\bot\bar{z_{2}}$
$\Rightarrow arg \left(\frac{z_{1}}{z_{2}}\right)=2n\pi \pm\frac{\pi }{2}$
i.e. $-\frac{\pi }{2},\frac{\pi }{2},\frac{3 \pi }{2},\ldots \ldots ..$
The minimum value of $\left|a - b\right|=1$ (when $a=1\&b=2$ )