Question

Let $z_1$ and $z_2$ be complex numbers such that $z_1 \neq z_2$ and $| z_1| = | z_2 |$ . If Re $(z_1) > 0$ and $Im (z_2) < 0 $ ,then $\frac{z_1+ z_2}{z_1 - z_2}$ is

• A. One
• B. real and positive
• C. real and negative
• D. purely imaginary

Answer 1

Answer: (D)

Let $z_{1}=x_{1}+i y_{1}$ and $z_{2}=x_{2}+i y_{2}$
$Re \left(z_{1}\right)>0 \Rightarrow x_{1}>0$
and $Im \left(z_{2}\right) < 0 $
$ \Rightarrow y_{2} < 0$
Given, $\left|z_{1}\right|=\left|z_{2}\right|$
$\Rightarrow \left|z_{1}\right|^{2}=\left|z_{2}^{2}\right|$
$\Rightarrow z_1\bar{z_1} = z_2\bar{z_2}$
Now, $\left(\frac{z_{1}+z_{2}}{z_{1}-z_{2}}\right)+\left(\frac{\overline{z_1+z_2}}{z_1-z_2}\right)$
$=\left(\frac{z_{1}+z_{2}}{z_{1}-z_{2}}\right)+\left(\frac{\bar{z}_{1}+\bar{z}_{2}}{\bar{z}_{1}-\bar{z}_{2}}\right)$
$=\frac{z_{1} \bar{z}_{1}+z_{2} \bar{z}_{1}-z_{1} \bar{z}_{2}-z_{2} \bar{z}_{2}+z_{1} \bar{z}_{1}+z_{1} \bar{z}_{2}-z_{2} \bar{z}_{1}+z_{2} \bar{z}_{2}}{\left(z_{1}-z_{2}\right)\left(\bar{z}_{1}-\bar{z}_{2}\right)}$
$=\frac{2\left(\left|z_{1}\right|^{2}-\left|z_{2}\right|^{2}\right)}{\left(z_{1}-z_{2}\right)\left(\bar{z}_{1}-\bar{z}_{2}\right)}=0 \left(\because\left|z_{1}\right|^{2}=\left|z_{2}\right|^{2}\right)$
$=\frac{z_{1}+z_{2}}{z_{1}-z_{2}}$ is purely imaginary.