Question

Let $z=x+iy$. If $\frac{z-1}{z+1}$ is purely imaginary then prove that $|z|=1$.

Answer 1

Now,

$\frac{z-1}{z+1}$

$=\frac{x-1+iy}{x+1+iy}$

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

$=\frac{x^2-1+y^2+i\left\{y\right(x+1)-y(x-1\left)\right\}}{(x+1{)}^2+y^2}$

Now if this complex number is purely imaginary then we must have,

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

or, $x^2+y^2=1$

or, $|z|=1$.