Question: Find the bilinear transformation which carries the circle $|z|=3$ into $|z-1|=1$, the point $3+3i$ into $1$ and the point $3$ into $0$.
My Attempt: First, I've done problems like this before, but I ran into something at the beginning, so I want to first write my proof, then mention what I ran into and hope to have my solution verified and to see if I can get an explanation on why I had that problem and what it means. So here goes:
Let $f(z)=w$ we such a transformation. We have that $f(3+3i)=1$ and $f(3)=0$. Now, we see that, using $z^*=\frac{R^2}{\bar z-\bar a}+a$, that $3+3i$ is symmetric to $3-3i$ with respect to $|z|=1$ and that $1$ is symmetric to $\infty$ with respect to $|z-1|=1$. Thus, we have $f(3-3i)=\infty$. So, using the cross ratio, we have $(w,1,0,\infty)=(z,3+3i,3,3-3i)$, and so we have $\frac{w-0}{w-\infty}\frac{1-\infty}{1-0}=\frac{z-3}{z-(3-3i)}\frac{3+3i-(3-3i)}{3+3i-3}$. Going through the calculation, we get that $w=\frac{(z-3)2i}{z-3+3i}$.
So, first, does this solution look correct? Next, when I was trying to find that "last point", I was running into an issue using $f(3)=0$. I kept getting $3$ is symmetric to $3$ with respect to $|z|=3$ and $0$ is symmetric to $0$ with respect to $|z-1|=1$, so I was just getting a transformation value that I already knew. Why? Or, should I, of course, have used the point with the nonzero imaginary part? Any help is greatly apprecaited! Thank you.
The transformation required is $f(z)=\frac{(z-3)(1-i)}{2z-3-3i}$
One can find it by noting that $g(w)=f(3/w)-1, |w| \le 1$ is a unit disc automorphism that carries $\frac{1-i}{2} \to 0$ and $1 \to -1$ hence it is $g(w)=\alpha \frac{w-\frac{1-i}{2}}{1-\frac{1+i}{2}w}, |\alpha|=1$ and putting $w=1$ gives $\alpha=i$, so giving $g(w)= i \frac{w-\frac{1-i}{2}}{1-\frac{1+i}{2}w}$ and $f(z)=g(3/z)+1$ which when explicited leads to the answer above.
(an easy check shows that the required values are attained and then $f(3i)=1-i$ and $f(-3)=\frac{2-4i}{5}$ which are clearly on $|z-1|=1$ together with $f(3)=0$ and since a circle is uniquely determined by three points and $f$ sends circles to circles, shows that $f(|z|=3)$ is indeed $|z-1|=1$)