Möbius transformation given 1 $\rightarrow$ 1, -1 $\rightarrow$ i, -i $\rightarrow$ -1

Question

I'm struggling a bit with Möbius transformations. Im supposed to find a Möbius transformation that maps

1 $\rightarrow$ 1

-1 $\rightarrow$ i

-i $\rightarrow$ -1

I know that Möbius transformation has the form $f(z)=\frac{az+b}{cz+d}$

which gives me

$f(1)=\frac{a+b}{c+d}=1$ $\Rightarrow$ $a+b=c+d$

$f(-1)=\frac{-a+b}{-c+d}=i$ $\Rightarrow$ $-a+b=i(-c+d)$

$f(-i)=\frac{-ai+b}{-ci+d}=-1$ $\Rightarrow$ $-ai+b=ci-d$

From here i dont really know where to go.

I guess i can set $a=1$ since we got one free variable, and doing equationssystems, but that doesnt give me the right answer.

It gives me $f(z)=\frac{5z-5+2i}{(1-2i)z+3+4i}$

which does not map $1 \rightarrow 1$

Would anyone help me out to solve this in the best possible way?

Answer 1

There is, in a manner of speaking, a "simple" way that doesn't involve solving a system of equations (though of course it is just hiding that detail). I'll detail it simply because it's a neat standard tool to have in one's pocket: cross ratios.

Here's the idea: given three complex numbers $\alpha, \beta, \gamma$, the unique Möbius transformation sending $\alpha \mapsto 1$, $\beta \mapsto 0$, and $\gamma \mapsto \infty$ is given by the cross ratio $$S(z) = (z, \alpha; \beta, \gamma) := \frac{(z - \beta)(\alpha - \gamma)}{(z - \gamma)(\alpha - \beta)}.$$ (There is a fact hidden here, which is interesting to think about, namely that a Möbius transformation is uniquely determined by its values on any three points on the Riemann sphere.)

The point here is that if I know how to make a Möbius transformation $S$ sending $1 \mapsto 1$, $-1 \mapsto 0$, and $-i \mapsto \infty$, then I can also find one, call it $T$, sending $1 \mapsto 1$, $i \mapsto 0$, and $-1 \mapsto \infty$. Then the inverse of $T$ sends $1 \mapsto 1$, $0 \mapsto i$, and $\infty \mapsto -1$, meaning that the composition $$ T^{-1} \circ S $$ sends $1 \mapsto 1 \mapsto 1$, $-1 \mapsto 0 \mapsto i$, and $-i \mapsto \infty \mapsto -1$.

Now, inverses of Möbius transformations are not very hard to compute, but this is where the cute cross ratios come in: essentially what the above says is that the transformation we need is precisely $$ (z, 1; -1, -i) = (w, 1; i, -1), $$ where we think of $z$ as the input and $w$ as the output. Hence, setting the cross ratios up, we get $$ \frac{(z + 1)(1 + i)}{(z + i)(1 + 1)} = \frac{(w - i)(1 + 1)}{(w + 1)(1 - i)} $$ and what remains is to simplify this and rearrange it to have $w$ alone on one side.