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

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

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