If $\mathrm{z}=\mathrm{x}+\mathrm{iy}, \mathrm{xy} \neq 0$, satisfies the equation $\mathrm{z}^{2}+\mathrm{i} \overline{\mathrm{z}}=0$, then $\left|\mathrm{z}^{2}\right|$ is equal to

Solution

$\mathrm{z}^{2}=-\mathrm{i} \overline{\mathrm{z}} \Rightarrow\left|\mathrm{z}^{2}\right|=|\mathrm{i} \overline{\mathrm{z}}|=|\mathrm{i} \| \overline{\mathrm{z}}|=|\mathrm{z}| \Rightarrow|\mathrm{z}|^{2}-|\mathrm{z}|=0 \Rightarrow|\mathrm{z}|(|\mathrm{z}|-1)=0$
$|z|=0$ (not acceptable as $\mathrm{xy} \neq 0$ ) $\therefore|\mathrm{z}|=1 \Rightarrow|\mathrm{z}|^{2}=1$