Question

Solve the system of equation
$Re\left(z^2\right)=0,\left|z\right|=2.$

Answer 1

Let$z=x+iy$
$\therefore z^2=(x+iy{)}^2$
$\Rightarrow z^2=x^2+i^2{y}^2+2xyi$
$\Rightarrow z^2=x^2-y^2+2xyi$
$\because Re\left(z^2\right)=0$
$\Rightarrow x^2-y^2=0\dots \left(1\right)$
$And\overline{z}=2$
$\Rightarrow |x+iy|=2$
$\Rightarrow \sqrt{(x^2+y^2)}=2$
$\Rightarrow x^2+y^2=4\dots \left(2\right)$
Add equations (1) & (2)
$\Rightarrow 2x^2=4$
$\Rightarrow x=\pm \sqrt{2}$