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.
Solution:
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