Question

If $z_1,z_2$ are two complex numbers satisfying the relation
$\left|\frac{z_1+z_2}{z_1-z_2}\right|=1$
then

• A. $\frac{z_1}{z_2}$ is purely imaginary
• B. $\frac{z_1}{z_2}>0$
• C. $z_1=ik{z}_2$
• D. $\angle z_{1}O{z}_2=\pi /2$

Answer 1

Answer: (A), (C), (D)

$\frac{z_1}{z_2}$ is purely imaginary
$z_1=ik{z}_2$
$\angle z_{1}O{z}_2=\pi /2$

$\left|\frac{z_1+z_2}{z_1-z_2}\right|=1$
$\Rightarrow \frac{z_1+z_2}{z_1-z_2}=\cos A+i\sin A$
$\Rightarrow \frac{z_1}{z_2}=-\frac{1+\cos A+i\sin A}{1-\cos A-i\sin A}$
$\Rightarrow \frac{z_1}{z_2}=i^2\left[\frac{2{\cos }^2\frac{A}{2}+2i\sin \frac{A}{2}\cos \frac{A}{2}}{2{\sin }^2\frac{A}{2}-2i\sin \frac{A}{2}\cos \frac{A}{2}}\right]$
$\Rightarrow \frac{z_1}{z_2}=i\left(\frac{i\cos \frac{A}{2}+i^2\sin \frac{A}{2}}{\sin \frac{A}{2}-i\cos \frac{A}{2}}\right)=-i$
Therefore, $\frac{z_1}{z_2}$ is purely imaginary.
$\Rightarrow arg\left(\frac{z_1}{z_2}\right)=\frac{\pi }{2}$
$\Rightarrow \angle z_{1}O{z}_2=\frac{\pi }{2}$
Ans: A,C,D