Let $z_{1}$ and $z_{2}$ be complex numbers such that $z_{1} \neq z_{2}, |z_{1}| = |z_{2}|$ . If $z_{1}$ has positive real part and $z_{2}$ has negative imaginary part, then $\frac{z_{1} + z_{2}}{z_{1} - z_{2}}$ may be

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real and positive

real and negative

pure imaginary

none of these

answer is C.

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Detailed Solution

$|z_{1}| = |z_{2}| Re (z_{1}) > 0 l m (z_{2}) < 0$

$Re (\frac{z_{1} + z_{2}}{z_{1} - z_{2}}) = \frac{1}{2} (\frac{z_{1} + z_{2}}{z_{1} - z_{2}} + \frac{\bar{z_{1}} + \bar{z_{2}}}{\bar{z_{1}} - \bar{z_{2}}})$

$= \frac{1}{2} (\frac{(z_{1} + z_{2}) (\bar{z_{1}} - \bar{z_{2}}) + (\bar{z_{1}} + \bar{z_{2}}) (z_{1} - z_{2})}{(z_{1} - z_{2}) (\bar{z_{1}} - \bar{z_{2}})})$

$\frac{1}{2} [\frac{z_{1} \bar{z_{1}} - z_{1} \bar{z_{2}} + z_{2} \bar{z_{1}} - z_{2} \bar{z_{2}} + z_{1} \bar{z_{1}} + z_{1} \bar{z_{2}} - z_{2} \bar{z_{1}} - z_{2} \bar{z_{2}}}{(z_{1} - z_{2}) (\bar{z_{1}} - \bar{z_{2}})}]$

$\frac{1}{2} (\frac{2 {|z_{1}|}^{2} - 2 {|z_{2}|}^{2}}{{|z_{1} - z_{2}|}^{2}}) = 0 \Rightarrow (\frac{z_{1} + z_{2}}{z_{1} - z_{2}})$ $is purely imaginary.$