Question

For complex numbers $z$ and $w$,prove that ${\left|z\right|}^{2}w-{\left|w\right|}^{2}z=z-w$ if and only if $z=w$ or $z\overline{w}=1$.

Answer 1

Note the following points ${\left|z\right|}^2=z\overline{z}$, when $z$ is real then $z=\overline{z}$. From the given relation
$z(1+{\left|w\right|}^2)=w(1+{\left|z\right|}^2)$
$\therefore \frac{z}{w}=\frac{1+{\left|z\right|}^2}{1+{\left|w\right|}^2}=$Real
Hence $\frac{z}{w}=\left(\frac{\overline{z}}{w}\right)=\frac{\overline{z}}{\overline{w}}$
or $z\overline{w}=\overline{z}w$ ......$\left(1\right)$
Now $\left(z\overline{z}\right)w-\left(w\overline{w}\right)z=z-w$ as given
or $z(\overline{z}w-1)-w(\overline{w}z-1)=0$
or $(z\overline{w}-1)(z-w)=0$ by$\left(1\right)$
$\therefore z=w$ or $z\overline{w}=1$