Find the most general bilinear transformation that maps $|z-1|=1$ to $\Re(f(z)) = 1$.
Attempt
Assuming that $$ 2 \mapsto 1 \\ 0 \mapsto \infty \\ 1+i \mapsto 1+i $$ we have the bilinear transformation $$ f(z) = 2 \left( \frac{z-1}{z} \right). $$ But how do we find the form of all such mappings? My only guess is to compose $f$ with a $g$ that always maps $\Re(w)$ to itself. I would think that $$ T(z) = (g \circ f)(z) = 2 \left( \frac{z-1}{z} \right) + i\alpha, \quad \forall \alpha \in \mathbb{R} $$ gives the most general form of all such transformations. But how can I prove to myself that this is right (if it is)?