Question

If $z\left(\ne -1\right)$ is a complex number such that $\frac{z-1}{z+1}$ is purely imaginary, then find $\left|z\right|$

Answer 1

$\Rightarrow$ $z=x+iy$

$\Rightarrow$ $\frac{z-1}{z+1}$ is purely imaginary.

$\Rightarrow$ $Re\left(\frac{z-1}{z+1}\right)=0$

$\Rightarrow$ $Re\left(\frac{x+iy-1}{x+iy+1}\right)=0$

$\Rightarrow$ $Re\left[\frac{(x-1)+iy}{(x+1)+iy}\right]=0$

$\Rightarrow$ $Re[\frac{(x-1)+iy}{(x+1)+iy}\times \frac{(x+1)-iy}{(x+1)-iy}]=0$

$\Rightarrow$ $Re\left[\frac{(x-1)(x+1)-i(x-1)y+iy(x+1)+y^2}{(x+1{)}^2+y^2}\right]=0$

$\Rightarrow$ $Re\left[\frac{(x^2-1)+y^2+i\left\{y\right(x+1)-(x-1\left)u\right\}}{(x+1{)}^2+y^2}\right]=0$

$\Rightarrow$ $Re\left[\frac{x^2+y^2-1+i(xy+y-xy+y)}{(x+1{)}^2+y^2}\right]=0$

$\Rightarrow$ $Re[\frac{x^2+y^2-1}{(x+1{)}^2+y^2}+i\frac{2y}{(x+1{)}^2+y^2}]=0$

$\Rightarrow$ $\frac{x^2+y^2-1}{(x+1{)}^2+y^2}=0$

$\Rightarrow$ $x^2+y^2-1=0$

$\Rightarrow$ $x^2+y^2=1$

$\Rightarrow$ $|z|=1$ [ As $z=x+iy\to |z|=\sqrt{x^2+y^2}$ ]