Question

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

Answer 1

$a+ib=\frac{x+iy}{x-iy}$

$a+ib=\frac{{(x+iy)}^2}{x^2+y^2}$

$a+ib=\frac{x^2-y^2}{x^2+y^2}+\frac{2xy}{x^2+y^2}i$

$a=\frac{x^2-y^2}{x^2+y^2}$

$b=\frac{2xy}{x^2+y^2}$

$a^2+b^2=\frac{x^4+y^4+2x^2{y}^2}{{(x^2+y^2)}^2}$

$a^2+b^2=\frac{{(x^2+y^2)}^2}{{(x^2+y^2)}^2}$

$a^2+b^2=1$

$\frac{b}{a}=\frac{\frac{2xy}{x^2+y^2}}{\frac{x^2-y^2}{x^2+y^2}}$

$\frac{b}{a}=\frac{2xy}{x^2-y^2}$