Question

Solve the system of equations $\mathrm{Re}\left(z^2\right)=0,\left|z\right|=2$.

Answer 1

Let $z=x+iy$.
Then ,
$$\begin{gathered} z^2={\left(x+iy\right)}^2 \\ =x^2+i^2{y}^2+2ixy \\ =x^2-y^2+2ixy[\because i^2=-1] \end{gathered}$$

and $\left|z\right|=\sqrt{x^2+y^2}$

According to the question,
$$\begin{gathered} \mathrm{Re}\left(z^2\right)=0\mathrm{and}\left|z\right|=2 \\ \Rightarrow x^2-y^2=0\mathrm{and}\sqrt{x^2+y^2}=2 \\ \Rightarrow x^2-y^2=0\mathrm{and}{x}^2+y^2=4 \\ \mathrm{On}\mathrm{Adding}\mathrm{both}\mathrm{the}\mathrm{equations},\mathrm{we}\mathrm{get} \\ 2x^2=4 \\ \Rightarrow x^2=2 \\ \Rightarrow x=\pm \sqrt{2} \\ \Rightarrow y^2=2 \\ \Rightarrow y=\pm \sqrt{2} \end{gathered}$$

Thus, $x=\pm \sqrt{2}\mathrm{and}y=\pm \sqrt{2}$.