Question

If $a^2+b^2+c^2=1,b+ic=(1+a)z$, prove that $\frac{a+ib}{1+c}=\frac{1+iz}{1-iz}$ ,
Where $a,b,c$ are real numbers and $z$ is a complex number.

Answer 1

$\frac{b+ic}{1+a}=z$ $\therefore \frac{b-ic}{1+a}=\overline{z}$
$z+\overline{z}=\frac{2b}{1+a}$, $z-\overline{z}=\frac{2ic}{1+a}$
and $z\overline{z}=\frac{b^2+c^2}{{(1+a)}^2}=\frac{{1-a}^2}{{(1+a)}^2}=\frac{1-a}{1+a}$
$\frac{1+iz}{1-iz}=\frac{1+iz}{1-iz}\times \frac{1+i\overline{z}}{1+i\overline{z}}=\frac{1+i(z+\overline{z})-z\overline{z}}{1-i(z-\overline{z})+z\overline{z}}$
$\frac{1+iz}{1-iz}=[1+i\frac{2b}{1+a}-\frac{1-a}{1+a}]÷[1+\frac{2c}{1+a}+\frac{1-a}{1+a}]$
$=\frac{2(a+ib)}{2(1+c)}=\frac{(a+ib)}{(1+c)}$