So I have these four transformations given by: $$-1+i \mapsto -1 $$
$$0 \mapsto -i $$
$$1-i \mapsto 1$$
And I need to combine them into a single transformation. But every way I tried seems to get a different answer to the solution given. I've tried using the cross ratio method, but it seems that my algebra goes wrong every time even though I'm being extremely careful about it
The solution given was
$$w \mapsto \frac{iz+1-i}{z+1+i}$$
If $\displaystyle \phi(z)=\frac{z-z_2}{z-z_3}\frac{z_1-z_3}{z_1-z_2}$, then $\phi(z_1)=1$, $\phi(z_2)=0$ and $\phi(z_3)=\infty$
Let $\displaystyle f(z)=\frac{z-0}{z-(1-i)}\frac{(-1+i)-(1-i)}{(-1+i)-0}=\frac{2z}{z-1+i}$. Then $f(-1+i)=1$, $f(0)=0$ and $f(1-i)=\infty$.
Let $\displaystyle g(z)=\frac{z-(-i)}{z-1}\frac{-1-1}{(-1)-(-i)}=\frac{-2z-2i}{(-1+i)z+1-i}$. Then $g(-1)=1$, $g(-i)=0$ and $g(1)=\infty$.
So $g^{-1}\circ f$ is the required function.
Note that $\displaystyle g^{-1}(z)=\frac{z-1+i}{z-1-i}$.
\begin{align*} g^{-1}\circ f(z)&=\frac{\displaystyle \left(\frac{2z}{z-1+i}\right)-1+i}{\displaystyle \left(\frac{2z}{z-1+i}\right)-1-i}\\ &=\frac{2z+(-1+i)(z-1+i)}{2z+(-1-i)(z-1+i)}\\ &=\frac{(1+i)z-2i}{(1-i)z+2}\\ &=\frac{iz+1-i}{z+1+i}\\ \end{align*}