Question

If z and w are two complex numbers such that |zw| = 1 and arg (z) – arg (w) = π/2, then show that z̅w = – i.

Answer 1

Let z = |z| (cos θ₁ + I sin θ₁) and w = |w| (cos θ₂ + I sin θ₂)

Given |zw| = |z| |w| = 1

Also arg (z) – arg (w) = π/2

⇒ θ₁ – θ₂ = π/2

Now, z̅ w = |z| (cos θ₁ – I sin θ₁) |w| (cos θ₂ + I sin θ₂)g | = 1

= |z| |w| (cos (-θ₁) + I sin (-θ₁)) (cos θ₂ + I sin θ₂)

= 1 [cos (θ₂ – θ₁) + I sin (θ₂ – θ₁)]

= [cos (-π/2) + I sin (-π/2)]

= 1 [0 – i]

= – i

Hence proved