Given: Re(z²) = 0 and |z| = 2

Let z = x + iy

[given]

Squaring both the sides, we get

x² + y² = 4 …(i)

Since, z = x + iy

⇒ z² = (x + iy)²

⇒ z² = x² + i²y² + 2ixy

⇒ z² = x² + (-1)y² + 2ixy

⇒ z² = x² – y² + 2ixy

It is given that Re(z²) = 0

⇒ x² – y² = 0 …(ii)

Adding eq. (i) and (ii), we get

x² + y² + x² – y² = 4 + 0

⇒ 2x² = 4

⇒ x² = 2

⇒ x = ±√2

Putting the value of x² = 2 in eq. (i), we get

2 + y² = 4

⇒ y² = 2

⇒ y = ±√2

Hence, z = √2 ± i√2, -√2 ± i√2