Question

If $z$ and $\alpha$ complex numbers such that $|z|=|\alpha |=r,r>0$ and $\omega =\frac{z-\overline{\alpha }}{r^2+z\overline{\alpha }}$, find $Re\left(\omega \right)$.

Answer 1

$|z=|\alpha |=r$

$z=a_1+i{b}_1\Rightarrow z=\sqrt{a_1^2+b_1^2}=r\Rightarrow a_1^2+b_1^2=a_2^2+b_2^2$

$x=a_2+i{b}_2\Rightarrow x=\sqrt{a_2^2+b_2^2}=r{a}_1^2-a_2^2=b_2^2-b_1^2$

$w=\frac{a_1+i{b}_1-a_2-i{b}_2}{r^2+a_1{a}_2-b_1{b}_2}\Rightarrow \frac{a_1-a_2+i{b}_1-i{b}_2}{r^2+a_1{a}_2-b_1{b}_2}$

$Re\left(w\right)=\frac{a_1-a_2}{r^2+a_1{a}_2-b_1{b}_2}$

$a_1^2+b_1^2$ or $a_2^2+b_2^2$