Question

lf $\left|\frac{z_1+z_2}{z_1-z_2}\right|=1$ then $\frac{z_1}{z_2}$ is

• A. positive real
• B. negative real
• C. purely imaginary
• D. 0

Answer 1

The correct option is C purely imaginary

$\left|\frac{z_1+z_2}{z_1-z_2}\right|=1$

${\left|\frac{z_1+z_2}{z_1-z_2}\right|}^2=1$

$\Rightarrow \frac{(z_1+z_2)(\overline{z_1}+\overline{z_2})}{(z_1-z_2)(\overline{z_1}-\overline{z_2})}=1$

$\Rightarrow {\left|z_1\right|}^2+z_1\overline{z_2}+z_2\overline{z_1}+{\left|z_2\right|}^2$$={\left|z_1\right|}^2-z_1\overline{z_2}-z_2\overline{z_1}+{\left|{}_2\right|}^2$

$2(z_1\overline{z_2}+z_2\overline{z_1})=0$

$\Rightarrow \frac{z_1}{z_2}+\frac{\overline{z_1}}{\overline{z_2}}=0\left[dividingby{z}_2\overline{z_2}\right][z+\overline{z}=0\Rightarrow zispurelyimag.]$

$\Rightarrow \frac{z_1}{z_2}$ is purely imaginary