Question

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

Answer 1

Given : $Re\left(z^2\right)=0,|z|=2$

Let $z=x+iy$



Now,

$z^2=(x+iy)(x+iy)$

$\Rightarrow z^2=x^2+(iy{)}^2+2ixy$

$\Rightarrow z^2=x^2-y^2+2ixy$ $[\because i^2=-1]$

We know, $Re\left(z^2\right)=0$

$x^2-y^2=0$

$\Rightarrow x^2=y^2$ ⋯(1)

Now,

$|z|=2$

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



Using equation (1), we get

$\Rightarrow \sqrt{x^2+x^2}=2$

$\Rightarrow \sqrt{2x^2}=2$

$\Rightarrow 2x^2=4$

$\Rightarrow x^2=2$

$\Rightarrow x=\pm \sqrt{2}$


From equation (1),

$y=\pm \sqrt{2}$

Therefore, the possible complex numbers satisfying the given equations are

$z=x+iy=\sqrt{2}+i\sqrt{2},\sqrt{2}-i\sqrt{2},-\sqrt{2}+i\sqrt{2},-\sqrt{2}-i\sqrt{2}$

$z=\pm \sqrt{2}\pm i\sqrt{2}$