Question

If $x+iy=\frac{a+ib}{a-ib}$, prove that $x^2+y^2=1$

Answer 1

$x+iy=\frac{a+ib}{a-ib}$
$|x+iy|=\left|\frac{a+ib}{a-ib}\right|$ (Taking modulus on both sides)
$|x+iy|=\frac{|a+ib|}{|a-ib|}[\left|\frac{z_1}{z_2}\right|=\frac{|z_1|}{|z_2|}]$
$\sqrt{x^2+y^2}=\frac{\sqrt{a^2+b^2}}{\sqrt{a^2+(-b{)}^2}}$
$x^2+y^2=\frac{(a^2+b^2)}{(a^2+b^2)}$
$x^2+y^2=1$
Hence Proved