Given: Re(z2) = 0 and |z| = 2
Let z = x + iy
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