Question

Find the number of solutions of $Re\left(z^2\right)=0$ and $\left|z\right|=a$, where $z$ is a complex number and $a>0$

Answer 1

Let $z=x+iy$

Then

$z^2=x^2+2ixy-y^2$

and

$Real\left(z^2\right)=x^2-y^2=\mathrm{0.......}\left(1\right)$

and

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

here , $|z|=a$

Hence,

$\sqrt{x^2+y^2}=a$

$x^2+y^2=a^2.......\left(2\right)$

Adding (1) and (2) we get

$2x^2=a^2$

$x^2=\frac{a^2}{2}$

$x=\pm \frac{a}{\sqrt{2}}$

$y=\pm \frac{a}{\sqrt{2}}$