solve the system of equation

Re(z²)=0,|z|=2

Let z = x + iy
Then z² =  x² + 2ixy - y²
And real(z²) = x² - y²  = 0  (1)

And |z| = $\sqrt{x^2+y^2}$
|z| = 2
Hence $\sqrt{x^2+y^2}$ = 2  (2) 
Or x² +  y² = 4 (3)

Solving (1) and (3) , we get x = $\pm \sqrt{2} and y = \pm \sqrt{2}$