Question

If $\mid \frac{2i{z}_i-z_i-z_2}{2i{z}_1+z_1+z_2}\mid =\mid \frac{cos\theta +isin\theta }{cos\theta -isin\theta }\mid$,then $\frac{z_1}{z_2}$ is

• A. Purely real
• B. purely imaginary
• C. nothing can be said
• D. none of these

Answer 1

The correct option is A Purely real

The given relation can be written as
$\left|\frac{z_1+i\frac{z_1+z_2}{2}}{z_1-i\frac{z_1+z_2}{2}}\right|=1\Rightarrow \left|\frac{\frac{2z_1}{z_1+z_2}+i}{\frac{2z_1}{z_1+z_2-i}}\right|=1$

Hence $\frac{2z_1}{z_1+z_2}$ is a complex number such that its distance from i and -i is always

equal. Hence this complex number is purely real.$\Rightarrow \frac{2z_1}{z_1+z_2}=\frac{2\overline{z_1}}{\overline{z_1}+\overline{z_2}}$

$\Rightarrow z_1(\overline{z_1}+\overline{z_2})=\overline{z_1}(z_1+z_2)\Rightarrow \frac{z_1}{z_2}=\frac{\overline{z_1}}{\overline{z_2}}\Rightarrow \frac{z_1}{z_2}$ is purely real.