Question
Solve the system of equations Re(z2) = 0, |z| = 2.

Answer

Given that, Re(z2) = 0, |z| = 2
Let $\text{z}=\text{x}+\text{iy}$ Then $|\text{z}|=\sqrt{\text{x}^2+\text{y}^2}$
Given that $\sqrt{\text{x}^2+\text{y}^2}=2$
$\Rightarrow\text{x}^2+\text{y}^2=4$
Also, $\text{z}^2=\text{x}^2+2\text{ixy}-\text{y}^2=(\text{x}^2-\text{y}^2)+2\text{ixy}$
Now, $\text{Re}(\text{z}^2)=0$
$\Rightarrow\text{x}^2-\text{y}^2=0$
Solving (i) and (ii) we get
$\Rightarrow\text{x}^2=\text{y}^2=2$
$\Rightarrow\text{x}=\pm\sqrt{2}$ and $\text{y}=\pm\sqrt{2}$
$\therefore\ \text{z}=\text{x}+\text{iy}=\pm\sqrt{2}\pm\text{i}\sqrt{2}$
Hence, we have four complex numbers.

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