Bilinear transformation which maps $z=(-1,i,−1) $ and $w=(1,i,−1)$

Question

I have formed three equations $$\left\{\begin{array}{ccc} -a+b+c-d & = & 0 & (1)\\ ai+b+c-di & = & 0 & (2)\\ -a+b-c+d & = & 0 & (3) \end{array}\right.$$

I also have found out a $4$th equation by adding equation $(1)$ and $(3)$: $$-a+b=0\tag{4}\label{4}$$

Therefore $ b=a$

I am not sure what to do next.

EDIT- Should I add equation \eqref{4} and $(2)$ ?

EDIT 1- So i managed to solve it but am unsure if i have the right solution. My answer is $ w= \frac{z-1}{ k(z-2)}$.

Answer 1

Here's a general method that will work for all of these problems:

Given distinct $z_1,z_2,z_3$, the transformation $$ T_{z_1,z_2,z_3}(z)=\frac{(z-z_1)(z_2-z_3)}{(z-z_3)(z_2-z_1)} $$ maps the triple $(z_1,z_2,z_3)$ to $(0,1,\infty)$.

Then if $w_1,w_2,w_3$ are distinct, the map $T$ sending $(z_1,z_2,z_3)$ to $(w_1,w_2,w_3)$ is given by $$ T=T_{w_1,w_2,w_3}^{-1}T_{z_1,z_2,z_3}$$