Question

If $\left|z\right|=1$, prove that $\frac{z-1}{z+1}(z\ne -1)$, is a pure imaginary number. What will you conclude if $z=1$?

Answer 1

Let $z=x+iy$. Then ${\left|z\right|}^2=x^2+y^2$.
Therefore the condition $\left|z\right|=1$ is equivalent to
$x^2+y^2=1.$
Now $\frac{z-1}{z+1}=\frac{x+iy-1}{x+iy+1}$
$=\frac{(x-1+iy)(x+1-iy)}{(x+1+iy)(x+1-iy)}$
$=\frac{(x^2+y^2-1)+2iy}{{(x+1)}^2+y^2}=\frac{2iy}{{(x+1)}^2+y^2}$ by $\left(1\right)$
Hence $\frac{z-1}{z+1}$ is purely imaginary when $\left|z\right|=1$
provided $z\ne -1.$
When $z=1$, we have $\frac{z-1}{z+1}=0$.
Now recall that according to the definition $2$ given in $\S 2$, $0$ is a pure imaginary number, since the point $0$ which corresponds to $z=0$ lies on both real and imaginary axes.
So in this case also, $\frac{z-1}{z+1}$ is a pure imaginary number.