Solve the system of equations

Solution:

Given: $\operatorname{Re}\left(z^{2}\right)=0$ and $|z|=2$

Let $z=x+i y$

$\therefore|z|=\sqrt{x^{2}+y^{2}}$

$\Rightarrow 2=\sqrt{x^{2}+y^{2}}$ [Given]

Squaring both the sides, we get

$x^{2}+y^{2}=4 \ldots$ (i)

Since, $z=x+i y$

$\Rightarrow z^{2}=(x+i y)^{2}$

$\Rightarrow z^{2}=x^{2}+i^{2} y^{2}+2 i x y$

$\Rightarrow z^{2}=x^{2}+(-1) y^{2}+2 i x y$

$\Rightarrow z^{2}=x^{2}-y^{2}+2 i x y$

It is given that $\operatorname{Re}\left(z^{2}\right)=0$

$\Rightarrow \mathrm{x}^{2}-\mathrm{y}^{2}=0 \ldots$ (ii)

Adding eq. (i) and (ii), we get

$x^{2}+y^{2}+x^{2}-y^{2}=4+0$

$\Rightarrow 2 x^{2}=4$

$\Rightarrow x^{2}=2$

$\Rightarrow x=\pm \sqrt{2}$

Putting the value of $x^{2}=2$ in eq. (i), we get

$2+y^{2}=4$

$\Rightarrow y^{2}=2$

$\Rightarrow y=\pm \sqrt{2}$

Hence, $z=\sqrt{2} \pm i \sqrt{2},-\sqrt{2} \pm i \sqrt{2}$