Question

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

Answer 1

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

Taking RHS

$=\frac{(a+ib)(a+ib)}{(a-ib)(a+ib)}=\frac{(a+ib{)}^2}{(a-ib)(a+ib)}=\frac{a^2+(ib{)}^2+2aib}{a^2-(ib{)}^2}=\frac{a^2+i^2{b}^2+2aib}{a^2-\left(i^2\right)\left(b^2\right)}=\frac{a^2+i^2{b}^2+2aib}{a^2-\left(i^2\right)\left(b^2\right)}=\frac{a^2-b^2+2aib}{a^2+b^2}=\frac{a^2-b^2}{a^2+b^2}+i\left(\frac{2ab}{a^2+b^2}\right)$

Now LHS

$x+iy$

Comparing real and imaginary part in LHS and RHS

$x=\frac{a^2-b^2}{a^2+b^2}y=\frac{2ab}{a^2+b^2}\therefore x^2=(\frac{a^2-b^2}{a^2+b^2}{)}^2\to (1)\therefore y^2=(\frac{a^2-b^2}{a^2+b^2}{)}^2\to \left(2\right)$

Adding $1$ and $2$

$x^2+y^2=(\frac{a^2-b^2}{a^2+b^2}{)}^2+\frac{(2ab{)}^2}{(a^2+b^2{)}^2}=\frac{1}{(a^2+b^2{)}^2}((a^2-b^2{)}^2-(2ab{)}^2)=\frac{1}{(a^2+b^2{)}^2}(a^4+b^4-2a^2{b}^2+4a^2{b}^2)=\frac{(a^2{)}^2+(b^2{)}^2+2a^2{b}^2}{(a^2+b^2{)}^2}=\frac{(a^2+b^2{)}^2}{(a^2+b^2{)}^2}=1$

Thus $x^2+y^2=1$

Hence proved.