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

Given:

⇒ Re(z²)=0 and |z|=2

Let us assume Z=x+iy

⇒ Re(z²)=0

⇒ Re((x+iy)²)=0

⇒ Re(x²+(iy)²+2(x)(iy))=0

⇒ Re(x²+i²y²+i(2xy))=0

We know that i²=-1

⇒ Re(x²-y²+i(2xy))=0

⇒ x²-y²=0----------------------(1)

⇒ |z|=2

⇒

⇒ (x²+y²)=2²

⇒ (x²+y²)=4-------------------(2)

Solving (1) and (2) we get

⇒ x= and y=.

∴ .