If $b+ic=(1+a)z$ and $a^2+b^2+c^2=1$ then prove that $\dfrac{1+iz}{1-iz}=\dfrac{a+ib}{1+c}$
My Attempt $$ z=\frac{b}{1+a}+i\frac{c}{1+a}\implies iz=\frac{-c}{1+a}+i\frac{b}{1+a}\\ \frac{1+iz}{1-iz}=\frac{\frac{1+a-c}{1+a}+i\frac{b}{1+a}}{\frac{1+a+c}{1+a}-i\frac{b}{1+a}}=\frac{1+a-c+ib}{1+a+c-ib}.\frac{1+a+c+ib}{1+a+c+ib}\\ =\frac{(1+a+ib)^2-c^2}{(1+a+c)^2+b^2}=\frac{1+a^2-b^2+2a+2iab+2ib-c^2}{1+a^2+c^2+2a+2c+2ac+b^2} $$ I don't think its going anywhere with my attempt, how can I solve it easily as it was asked as a multiple choice question ?
\begin{align} \frac{1+iz}{1-iz}&=\frac{(1+a+ib)^2-c^2}{(1+a+c)^2+b^2}\\ &=\frac{1+a^2-b^2-c^2+2a+2iab+2ib}{1+a^2+b^2+c^2+2a+2c+2ac}\\ &=\frac{1+a^2-(1-a^2)+2a+2iab+2ib}{1+1+2a+2c+2ac}\\ &=\frac{2a^2+2a+2iab+2ib}{2+2a+2c+2ac}\\ &=\frac{a^2+a+iab+ib}{1+a+c+ac}\\ &=\frac{a(a+1)+ib(a+1)}{(1+a)+c(1+a)}\\ &=\frac{a+ib}{1+c}\\ \end{align}
Can you find out your mistake now?
If you are interested, see spherical representation of complex numbers.