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

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

e get

ts to zero, we get

= |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.

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