Question

Solve the system of equations Re ($z^2$)=0, |z| = 2.

Answer 1

Re ($z^2$)=0, |z| = 2

$\text{Let}z=x+iy{z}^2=0\Rightarrow (x+iy{)}^2=0\Rightarrow x^2+2ixy-y^2=0\Rightarrow x^2-y^2=0...(i).\text{which is the real part of}(x+iy{)}^2.|z|=x\Rightarrow \sqrt{x^2+y^2}=2\Rightarrow x^2+y^2=4...\left(ii\right)\text{Adding (i) and (ii), we get}2x^2=4\Rightarrow x^2=2\Rightarrow x=\pm \sqrt{2},y=\pm \sqrt{2}x+iy=\sqrt{2}+i\sqrt{2}=-\sqrt{2}-i\sqrt{2}=\sqrt{2}-i\sqrt{2}=-\sqrt{2}+i\sqrt{2}$