Question

Let $z_1$ and $z_2$ be complex numbers such that $z_1\ne z_2$ and $\left|z_1\right|=\left|z_2\right|$. If $Re\left(z_1\right)>0$ and $lm\left(z_2\right)<0$, 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

The correct option is D purely imaginary

$z_1=x_1+i{y}_1;z_2=x_2+i{y}_2$
$R_e\left(z_1\right)>0\Rightarrow x_1>0$ and $I_m\left(z_2\right)<0\Rightarrow y_2<0$
$\left|z_1\right|=\left|z_2\right|\Rightarrow {\left|z_1\right|}^2={\left|z_2\right|}^2$
$\Rightarrow z_1\overline{z_1}=z_2\overline{z_2}$
Now $\left(\frac{z_1+z_2}{z_1-z_2}\right)+\left(\frac{\overline{z_1+z_2}}{z_1-z_2}\right)$
$\left(\frac{z_1+z_2}{z_1-z_2}\right)+\left(\frac{\overline{z_1}+\overline{z_2}}{\overline{z_1}-\overline{z_2}}\right)=\frac{z_1\overline{z_1}+z_2\overline{z_1}-z_2\overline{z_2}+z_1\overline{z_2}-z_2\overline{z_1}-z_2\overline{z_1}}{(z_1-z_2)(\overline{z_1}-\overline{z_2})}$
$=\frac{2({\left|z_1\right|}^2={\left|z_2\right|}^2)}{(z_1-z_2)(\overline{z_1}-\overline{z_2})}=0(\because {\left|z_1\right|}^2={\left|z_2\right|}^2)$
$\Rightarrow \frac{z_1+z_2}{z_1-z_2}$ is purely imaginary