Question

If $z_1$ and $z_2$ are two complex numbers such that $z_1\ne z_2$ and $|z_1|=|z_2|$, then $\frac{z_1+z_2}{z_1-z_2}$ may be

• A. purely imaginary
• B. real and positive
• C. real and negative
• D. none of these

Answer 1

The correct option is A purely imaginary

Given, $\left|z_1\right|=\left|z_2\right|$
Let $z_1=r(\cos A+i\sin A)\&z_2=r(\cos B+i\sin B)$
$w=\frac{z_1+z_2}{z_1{-z}_2}=\frac{\cos A+i\sin A+\cos B+i\sin B}{\cos A+i\sin A-\cos B-i\sin B}$
$\Rightarrow w=\frac{\cos A+\cos B+i(\sin A+\sin B)}{\cos A-\cos B+i(\sin A-\sin B)}$
$\Rightarrow w=\frac{2\cos \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right)+2i\sin \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right)}{-2\sin \left(\frac{A+B}{2}\right)\sin \left(\frac{A-B}{2}\right)+2i\sin \left(\frac{A-B}{2}\right)\cos \left(\frac{A+B}{2}\right)}$
$w=\cot \left(\frac{A-B}{2}\right)\left(\frac{\cos \left(\frac{A+B}{2}\right)+i\sin \left(\frac{A+B}{2}\right)}{-\sin \left(\frac{A+B}{2}\right)+i\cos \left(\frac{A+B}{2}\right)}\right)$

$=i\cot \left(\frac{A-B}{2}\right)\left(\frac{i\cos \left(\frac{A+B}{2}\right)-\sin \left(\frac{A+B}{2}\right)}{-\sin \left(\frac{A+B}{2}\right)+i\cos \left(\frac{A+B}{2}\right)}\right)$
$\Rightarrow w=i\cot \left(\frac{A-B}{2}\right)$
Therefore $z$ is purely imaginary.