Find the bilinear transformation which maps $z=(1, i, -1)$ respectively into $w =(i, 0, -i)$
My try: Here, $w_1=i$, $w_2=0$, $w_3=-i$, $z_1=1$, $z_2=i$, $z_3=-1$ $\text{As the formula states,}$ $$\begin{align}\\ &{\begin{aligned}\\ \frac{(w-w_1)(w_2-w_3)}{(w-w_3)(w_2-w_1)}&=\frac{(z-z_1)(z_2-z_3)}{(z-z_3)(z_2-z_1)}\\ \end{aligned}\\}\\ &\implies\frac{(w-i)(0+i)}{(w+i)(0-i)}=\frac{(z-1)(i+1)}{(z+1)(i-1)}\\ &\implies\frac{(w-i)i}{(w+i)(-i)}=\frac{(z-1)(i+1)^2}{(z+1)(-2)}\\ &\implies-\frac{w-i}{w+i}=\frac{(z-1)2i}{(z+1)(-2)}\\ &\implies\frac{w-i}{w+i}=\frac{(z-1)i}{z+1}\\ &{\begin{aligned}\\ \implies\require{cancel}\frac{w-\cancel{i}+w+\cancel{i}}{\cancel{w}-i-\cancel{w}-i}&=\frac{i(z-1)+z+1}{i(z-1)-z-1}\\ \end{aligned}\\}\\ &\implies w=i\frac{(z-1)i+z+1}{(z-1)i-(z+1)}\\ &\implies w=\frac{i(z+1)-z+1}{i(z-1)-z-1}\\ \end{align}\\ $$ Now, I can't understand how to shape this equation like $w=\frac{az+b}{cz+d}$. Please help.
You get $\omega=\dfrac{(i-1)z+i+1}{(i-1)z-(1+i)}$.
It is important to remember that here $a,b,c,d$ are complex numbers.