# Solve the system of equations Re

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