Question

If $arg\left(\frac{z_1}{z_2}\right)=\frac{\pi }{2}$, then find the value of $\left|\frac{z_1+z_2}{z_1-z_2}\right|$ $.$

Answer 1

arg$\left(\frac{z_1}{z_2}\right)=\frac{\pi }{2}$

We know if $z=x+iy$ is a complex number then arg$\left(z\right)={\tan }^{-1}\left(\frac{y}{x}\right)$

So, arg$\left(\frac{z_1}{z_2}\right)=\frac{\pi }{2}$

$\Rightarrow \frac{z_1}{z_2}=0$

Let $\frac{z_1}{z_2}=ia$

$\Rightarrow z_1=ia.z_2$ .........$\left(1\right)$

Now, $\frac{|z_1+z_2|}{|z_1-z_2|}=\frac{|ia.z_2+z_2|}{|ia.z_2-z_2|}$

$=\frac{|ia+1|}{|ia-1|}$

$=\frac{\sqrt{a^2+1}}{\sqrt{a^2+1}}=1$

Hence the value of $\frac{|z_1+z_2|}{|z_1-z_2|}=1$