Given that

$a^2 + b^2 +c^2 = 1 \& b +ic = \left(1 + a\right)z$

$\Rightarrow z = \frac{\left(b+ic\right)}{1+a}$

$\Rightarrow iz = \frac{-c+ib}{1+a}$

By componendo and dividendo

$\Rightarrow \frac{1+iz}{1-iz}=\frac{\left(1+a-c\right)+ib}{\left(1+a+c\right)-ib}$

$\Rightarrow \frac{1+iz}{1-iz}=\frac{\left(1+a-c\right)+ib}{\left(1+a+c\right)-ib}\times \frac{\left(1+a+c\right)+ib}{\left(1+a+c\right)+ib}$

$\Rightarrow \frac{1+iz}{1-iz}=\frac{{\left(1+a+ib\right)}^2-c^2}{{\left(1+a+c\right)}^2-{\left(ib\right)}^2}$

$\Rightarrow \frac{1+iz}{1-iz}=\frac{1+a^2-b^2+2a+2ib+2iab-c^2}{1+a^2+c^2+2ac+2a+2c+b^2}$

We know that $a^2+b^2+c^2=1$

$\Rightarrow \frac{1+iz}{1-iz}=\frac{1+a^2-1+a^2+2a+2ib(1+a)}{2+2a+2c\left(1+a\right)}$

$\Rightarrow \frac{1+iz}{1-iz}=\frac{\left(1+a\right)\left(2a+2ib\right)}{\left(2+2c\right)\left(1+a\right)}=\frac{a+ib}{1+c}.$