Question

If $z$ is a complex number such that $|z|=1$, prove that $\frac{z-1}{z+1}$ is purely imaginary. What will be your conclusion, if $z=1$?

Answer 1

Let $\frac{z-1}{z+1}=t\Rightarrow z=\frac{t+1}{1-t}$

$\left|z\right|=\frac{|t+1|}{|1-t|}=1$

$|1+t|=|1-t|$

Now, put $t=x+iy$ and square on both sides

${|1+x+iy|}^2={|1-x-iy|}^2$

${(1+x)}^2={(1-x)}^2$

${(1+x)}^2-{(1-x)}^2=0$

$A^2-B^2=(A+B)(A-B)$

$(1+x-1+x)(1+x+1-x)=0$

$2x-2=0$

$4x=0$

$x=0$

$t=yi=$purely imaginary

If $z=1$

$\frac{z-1}{z+1}=\frac{1-1}{1+1}=0$ which is purely real