(Solution check) Find a linear fractional transformation that maps the points $1, i,$ and $-1$ onto the points $-1, 0,$ and $1$ on the real axis.

Question

I am trying to understand the solution given in the textbook for an example:

Find a linear fractional transformation that maps the points $1, i,$ and $-1$ onto the points $-1, 0,$ and $1$ on the real axis.

Solution. By a theorem involving cross-ratios, the desired mapping satisfies $$\frac{z-1}{z+1} \cdot \frac{i+1}{i-1} = \frac{w+1}{w-1} \cdot \frac{-1}{1}. \qquad{(1)}$$

After solving for $w$ and simplifying we get $$w = \frac{z-i}{iz-1}.$$

This last part that says "After solving for $w$ and simplifying" is where I get stuck. When I tried to do the calculation in (1) by cross multiplying I ended up getting $$zwi - zi - wi + i = zw + z + w + 1 \Rightarrow w = \frac{z+1+(z-1)i}{-z-1+(z-1)i}$$ which is a complete mess from the answer. Is there an elegant way to get the solution from the solution I got? I can't seem to see where I messed up my arithmetic, if I did.

Answer 1

\begin{eqnarray*} \frac{z+1+(z-1)i}{-z-1+(z-1)i}=\frac{z(1+i)+(1-i)}{z(-1+i)-(1+i)} \color{blue}{ \frac{1-i}{1-i}} =\cdots =\color{red}{ \frac{z-i}{iz-1}} . \end{eqnarray*}