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Question: If \|z<sub>1</sub>\| = \|z<sub>2</sub>\| and arg (z<sub>1</sub>) + arg (z<sub>2</sub>) = p/2, then –...

If |z1| = |z2| and arg (z1) + arg (z2) = p/2, then –

A

z1 z2 is purely real

B

z1 z2 is purely imaginary

C

(z1 + z2)2 is purely real.

D

arg (z1–1) + arg (z2–1) = π2\frac{\pi}{2}

Answer

z1 z2 is purely imaginary

Explanation

Solution

Sol. Let |z1| = |z2| = r

Ž z1 = r(cos q + i sin q)

and z2 = r (cos⁡(π2−θ)+isin⁡(π2−θ))\left( \cos\left( \frac{\pi}{2} - \theta \right) + i\sin\left( \frac{\pi}{2} - \theta \right) \right)

Ž z1z2 = r2 i, which is purely imaginary

z1 + z2 = r [(cos q + sin q) + i (cos q + sin q)]

Ž (z1 + z2)2 = 2r2 . (cos q + sin q)2 . i

which is purely imaginary.

Also arg (z1–1) + arg (z2–1) = – π2\frac{\pi}{2}.

Hence (2) is the correct answer.