Solve the system of equations Re (z2) = 0,

Solution:

According to the question,

We have,

Re (z2) = 0, |z| = 2

Let z = x + iy.

Then, |z| = √(x2 + y2)

Given in the question,

√(x2 + y2)  = 2

⇒ x2 + y2 = 4 … (i)

z2 = x2 + 2ixy – y2

= (x2 – y2) + 2ixy

Now, Re (z2) = 0

⇒ x2 – y2 = 0 … (ii)

Equating (i) and (ii), we get

⇒ x2 = y2 = 2

⇒ x = y = ±√2

Hence, z = x + iy

= ±√2 ± i√2

= √2 + i√2, √2 – i√2, –√2 + i√2 and –√2 – i√2

Hence, we have four complex numbers.