Question

If the ratio $\left(\frac{1-z}{1+z}\right)$ is purely imaginary, then find value of $|z|$.

Answer 1

$z=x+cy\Rightarrow \frac{1-z}{1+z}=\left[\frac{1-x-cy}{1+x+cy}\right]=\frac{(1-x-cy)(1+x-cy)}{(1+x+cy)(1+x-cy)}=\frac{1+x-cy-x-x^2+cxy-cy-cxy+x^2{y}^2}{{(1+x)}^2+y^2}=\frac{1-x^2-y^2-2xy}{{(1+x)}^2+y^2}$

So, $=\frac{1-x^2-y^2}{{(1+x)}^2+y^2}=01-x^2-y^2=0x^2+y^2=1\left|z\right|=\sqrt{x^2+y^2}=\sqrt{1}=1$