I am trying to find the fractional linear transformation that sends $0$ to $1$ and maps $-1$ to $1+i$.
My approach is to setup the equation $w(z) = \frac{az+b}{cz+d}$ and plug in the values $0$ and then $-1$ and place this equations equal to $1$ and $1+i$ respectively. However, the equations I get are $\frac{b}{d}=1$ and $\frac{-a+b}{-c+d}=1+i$. From this point, I haven't been able to get what the values for a, b, c and d should be, any suggestions on how to attack this problem?
First, here's the hard way:
Once you've got $\dfrac b d=1$, you conclude $b=d$, and then you can put $b$ in place of $d$ and the equation $\dfrac{-a+b}{-c+d}=1+i$ becomes $\dfrac{-a+b}{-c+b} = 1+i$. Multiplying both sides by the denominator, we get $$ -a+b = (1+i)(-c+b) = (1+i)(-c) + (1+i)b. \tag 1 $$ The variable $b$ appears twice, so in order to get it to appear only once we can subtract $b$ from both sides, getting $-a = (1+i)(-c) + ib$, or $$ a = (1+i)c -ib. \tag 2 $$ If you let $b$ and $c$ be anything you like, that determines $a$., except that you can't allow $b=c$ because then you'd have a $0$ in the denominator in $\dfrac{-a+b}{-c+b}$. Then $(2)$ would imply that you get $z\mapsto\dfrac{z+1}{z+1}$, which certainly maps $0$ to $1$ as desired, but is indeterinate when $z=-1$. If we let $b=2$ and $c=1$, we get $a=1-i$. Etc.
Now the easy way: Since you have only two constraints, $0\mapsto 1$ and $-1\mapsto1+i$, a straight line will do it, so you can set $c=0$ and $d=1$ and you have $\dfrac{az+b}{cz+d} = az+b$. Since you concluded that $b/d=1$, we should have $b=1$ and you have $az+1$. This satisfies the requirement $0\mapsto1$. Now plug in $-1$: $$ a(-1)+1 = 1+i. $$ Thus $$ -a = i, $$ and finally our function is seen to be $$ z\mapsto -iz+1. $$
As suggested when we did this the hard way, this is not the only possibility, but it seems to be the simplest.