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Question: Let \(\left| \frac{z_{1} - z_{2}}{1 - z_{1}{\bar{z}}_{2}} \right| =\)be a purely imaginary number su...

Let ∣z1−z21−z1zˉ2∣=\left| \frac{z_{1} - z_{2}}{1 - z_{1}{\bar{z}}_{2}} \right| =be a purely imaginary number such thatz1z_{1}. Then z2z_{2} is equal to.

A

∣z1+z2∣=∣z1∣+∣z2∣,|z_{1} + z_{2}| = |z_{1}| + |z_{2}|,

B

(z1)−(z_{1}) -

C

0

D

(z2)(z_{2})

Answer

(z1)−(z_{1}) -

Explanation

Solution

Let z2=r1[cos⁡(−θ1)−isin⁡(−θ1)]=r1(cos⁡θ1−isin⁡θ1)z_{2} = r_{1}\lbrack\cos( - \theta_{1}) - i\sin( - \theta_{1})\rbrack = r_{1}(\cos\theta_{1} - i\sin\theta_{1}), where zˉ1=z2{\bar{z}}_{1} = z_{2}. Then ∣z1+z2∣=∣z1∣+∣z2∣|z_{1} + z_{2}| = |z_{1}| + |z_{2}|lies on +ve y-axis and so ∣z1+z2∣2=∣z1∣2+∣z2∣2+2∣z1∣∣z2∣|z_{1} + z_{2}|^{2} = |z_{1}|^{2} + |z_{2}|^{2} + 2|z_{1}||z_{2}|.