We have $\frac{wi+1-i}{-w+2i} = \frac{-z+1}{z-1}$
We need to get this in the form of $w = \frac{az+b}{cz+d}$
This is what I tried,
$=\frac{wi+1-i}{-w+2i} = \frac{-z+1}{z-1}$
$=\frac{wi-\frac{1}2i+1}{-w} = -z+1/z-1$
$=\frac{wi-\frac{1}2i}{-w} = \frac{-z+1}{z-1}-1$
$=wi-\frac{1}2i = -w\frac{-z+1}{z-1}-1$
$=wi = -w\frac{-z+1}{z-1}-1+\frac{1}2i$
$=wi(z-1) = -w-1+\frac{1}2i$
$=wiz-wi = -w-1+\frac{1}2i$
$=w+wiz-wi = -1+\frac{1}2i$
$=w+wi-wi = -1+\frac{1}2i * z$
$=w = -1+\frac{1}2i * z$
Does anyone have any ideas of how to get to $w = \frac{az+b}{cz+d}$