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Let $z_{1}$ and $z_{2}$ be complex numbers such that $z_{1} \neq z_{2}$ and $\left|z_{1}\right|=\left|z_{2}\right| .$ If $\operatorname{Re}\left(z_{1}\right)>0$ and
$\operatorname{Im}\left(z_{2}\right) < 0,$ then $\frac{z_{1}+z_{2}}{z_{1}-z_{2}}$ is
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$\operatorname{Im}\left(z_{2}\right) < 0,$ then $\frac{z_{1}+z_{2}}{z_{1}-z_{2}}$ is
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Verified Answer
The correct answer is:
purely imaginary
Let $z_{1}=x_{1}+i y_{1}$ and $z_{2}=x_{2}+i y_{2}$
$\operatorname{Re}\left(z_{1}\right)>0 \Rightarrow x_{1}>0$
$\begin{array}{lr}\text { and } & \operatorname{Im}\left(z_{2}\right) < 0 \\ \Rightarrow & y_{2} < 0 \\ \text { Given, } & \left|z_{1}\right|=\left|z_{2}\right|\end{array}$
$\Rightarrow \quad\left|z_{1}\right|^{2}=\left|z_{2}^{2}\right|$
$\Rightarrow \quad z_{1} \bar{z}_{1}=z_{2} \bar{z}_{2}$
Now, $\left(\frac{z_{1}+z_{2}}{z_{1}-z_{2}}\right)+\left(\frac{z_{1}+z_{2}}{z_{1}-z_{2}}\right)$
$=\left(\frac{z_{1}+z_{2}}{z_{1}-z_{2}}\right)+\left(\frac{\bar{z}_{1}+\bar{z}_{2}}{\bar{z}_{1}-\bar{z}_{2}}\right)$
$=\frac{z_{1} \bar{z}_{1}+z_{2} \bar{z}_{1}-z_{1} \bar{z}_{2}-z_{2} \bar{z}_{2}+z_{1} \bar{z}_{1}+z_{1} \bar{z}_{2}-z_{2} \bar{z}_{1}+z_{2} \bar{z}_{2}}{\left(z_{1}-z_{2}\right)\left(\bar{z}_{1}-\bar{z}_{2}\right)}$
$=\frac{2\left(\left|z_{1}\right|^{2}-\left|z_{2}\right|^{2}\right)}{\left(z_{1}-z_{2}\right)\left(\bar{z}_{1}-\bar{z}_{2}\right)}=0 \quad\left(\because\left|z_{1}\right|^{2}=\left|z_{2}\right|^{2}\right)$
$=\frac{z_{1}+z_{2}}{z_{1}-z_{2}}$ is purely imaginary.
$\operatorname{Re}\left(z_{1}\right)>0 \Rightarrow x_{1}>0$
$\begin{array}{lr}\text { and } & \operatorname{Im}\left(z_{2}\right) < 0 \\ \Rightarrow & y_{2} < 0 \\ \text { Given, } & \left|z_{1}\right|=\left|z_{2}\right|\end{array}$
$\Rightarrow \quad\left|z_{1}\right|^{2}=\left|z_{2}^{2}\right|$
$\Rightarrow \quad z_{1} \bar{z}_{1}=z_{2} \bar{z}_{2}$
Now, $\left(\frac{z_{1}+z_{2}}{z_{1}-z_{2}}\right)+\left(\frac{z_{1}+z_{2}}{z_{1}-z_{2}}\right)$
$=\left(\frac{z_{1}+z_{2}}{z_{1}-z_{2}}\right)+\left(\frac{\bar{z}_{1}+\bar{z}_{2}}{\bar{z}_{1}-\bar{z}_{2}}\right)$
$=\frac{z_{1} \bar{z}_{1}+z_{2} \bar{z}_{1}-z_{1} \bar{z}_{2}-z_{2} \bar{z}_{2}+z_{1} \bar{z}_{1}+z_{1} \bar{z}_{2}-z_{2} \bar{z}_{1}+z_{2} \bar{z}_{2}}{\left(z_{1}-z_{2}\right)\left(\bar{z}_{1}-\bar{z}_{2}\right)}$
$=\frac{2\left(\left|z_{1}\right|^{2}-\left|z_{2}\right|^{2}\right)}{\left(z_{1}-z_{2}\right)\left(\bar{z}_{1}-\bar{z}_{2}\right)}=0 \quad\left(\because\left|z_{1}\right|^{2}=\left|z_{2}\right|^{2}\right)$
$=\frac{z_{1}+z_{2}}{z_{1}-z_{2}}$ is purely imaginary.
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