Mobius transformations mapping to three arbitrary points

While I understand how to use the cross ratio to find a Mobius transformation given three points mapping to $0,1,\infty$, I am confused how to solve a more general case where three points are sent to three other arbitrary points.

Take for example problem 3.21.b which asks us to find $f$ a Mobius transformation which maps to $1\rightarrow 1, -1\rightarrow i,-i\rightarrow -1$

Using the cross ratio, we get two Mobius transformations
\begin{align*}
T_1(z)&=[z,1,-1,-i] &T_2(w)&=[w,1,i,-1]\\
&=\frac{z-1}{z+i}\cdot\frac{-1+i}{-1-1} &&=\frac{w-1}{w+1}\cdot\frac{i+1}{i-1}\\
&=\frac{z(-1+i)+(1-i)}{-2z+(-2i)}&&=\frac{w(1+i)+(-1-i)}{w(1-i)+(-1+i)}.
\end{align*}
However it is at this point that I am confused. It says in the book that $f=T_2^-1\circ T_1$. While it is entirely possible to solve the problem this way, it is extremely messy and tedious. Is this the only way to solve this type of problem or is there a trick I am missing?

@cdunn Good question! In this particular case, I think it helps to note that
$$\frac{i+1}{i-1} = -i.$$

More importantly, while you do need to distribute expressions across the terms $(w+1)$ and $(w-1)$ you don't need to do any other exansion. So, from your second step, I would proceed like so:

$$\begin{align}
\frac{z-1}{z+1}(i-1) &= 2i\frac{w-1}{w+1} \\
(z-1)(i-1)(w+1) &= 2i(z+i)(w-1) \\
((z-1)(i-1) - 2i(z+i))w &= -2i(z+i) - (z-1)(i-1)
\end{align}$$

So that,

$$
w = \frac{-2i(z+i)-(z-1)(i-1)}{(z-1)(i-1)-2i(z+i)}.
$$

That final expression need not be simplified any further.

So is your final expression for $w$ the Mobius transformation?

yea I think it is. The process he used makes sense and checking by plugging in the 3 values gives the correct response. Thanks for asking this question, I also got to the point you did originally and was intimidated by the amount of work to combine the functions

@cdunn That's correct! That's what I meant when I said the "final expression need not be simplified further".