Let z_{1} and z_{2} be complex numbers such t | vedclass

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(D) Let $w = \frac{z_{1}+z_{2}}{z_{1}-z_{2}}$.To check if $w$ is purely imaginary, we evaluate $w + \bar{w}$.$w + \bar{w} = \frac{z_{1}+z_{2}}{z_{1}-z

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purely imaginary

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(D) Let $w = \frac{z_{1}+z_{2}}{z_{1}-z_{2}}$.To check if $w$ is purely imaginary, we evaluate $w + \bar{w}$.$w + \bar{w} = \frac{z_{1}+z_{2}}{z_{1}-z_{2}} + \frac{\bar{z}_{1}+\bar{z}_{2}}{\bar{z}_{1}-\bar{z}_{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{(z_{1}\bar{z}_{1} - z_{1}\bar{z}_{2} + z_{2}\bar{z}_{1} - z_{2}\bar{z}_{2}) + (\bar{z}_{1}z_{1} - \bar{z}_{1}z_{2} + \bar{z}_{2}z_{1} - \bar{z}_{2}z_{2})}{|z_{1}-z_{2}|^2}$Since $|z_{1}| = |z_{2}|$, we have $z_{1}\bar{z}_{1} = z_{2}\bar{z}_{2} = |z_{1}|^2 = |z_{2}|^2$.Substituting this, the numerator becomes:$|z_{1}|^2 - z_{1}\bar{z}_{2} + z_{2}\bar{z}_{1} - |z_{2}|^2 + |z_{1}|^2 - \bar{z}_{1}z_{2} + \bar{z}_{2}z_{1} - |z_{2}|^2$$= 2|z_{1}|^2 - 2|z_{2}|^2 = 0$.Since $w + \bar{w} = 0$, $w$ must be purely imaginary.

Let $z_{1}$ and $z_{2}$ be complex numbers such that $z_{1} \neq z_{2}$ and $|z_{1}|=|z_{2}|$. If $\operatorname{Re}(z_{1}) > 0$ and $\operatorname{Im}(z_{2}) < 0$, then $\frac{z_{1}+z_{2}}{z_{1}-z_{2}}$ is

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