Question

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

Answer 1

Given that: Re(z²) = 0, z = 2

Let z = x + yi

∴ |z| = `sqrt(x^2 + y^2)`

⇒ `sqrt(x^2 + y^2)` = 2

⇒ x² + y² = 4  .....(i)

Since, z = x + yi

z² = x² + y² i² + 2xyi

⇒ z² = x² – y² + 2xyi

∴ Re(z²) = x² – y²

⇒ x² – y² = 0  ....(ii)

From equation (i) and (ii), we get

x² + y² + x² − y² = 4 + 0

⇒ 2x² = 4

⇒ x² = 2

⇒ x = `+-  sqrt(2)` and y = `+-  sqrt(2)`

Hence, z = `sqrt(2) +- isqrt(2), -sqrt(2) +- isqrt(2)`.