Question:
Solve the system of equations $\operatorname{Re}\left(z^{2}\right)=0,|z|=2$.
Solution:
Let $z=x+i y$.
Then.
$z^{2}=(x+i y)^{2}$
$=x^{2}+i^{2} y^{2}+2 i x y$
$=x^{2}-y^{2}+2 i x y$ $\left[\because i^{2}=-1\right]$
and $|z|=\sqrt{x^{2}+y^{2}}$
According to the question,
$\operatorname{Re}\left(z^{2}\right)=0$ and $|z|=2$
$\Rightarrow x^{2}-y^{2}=0$ and $\sqrt{x^{2}+y^{2}}=2$
$\Rightarrow x^{2}-y^{2}=0$ and $x^{2}+y^{2}=4$
On Adding both the equations, we get
$2 x^{2}=4$
$\Rightarrow x^{2}=2$
$\Rightarrow x=\pm \sqrt{2}$
$\Rightarrow y^{2}=2$
$\Rightarrow y=\pm \sqrt{2}$
Thus, $x=\pm \sqrt{2}$ and $y=\pm \sqrt{2}$.