This is a question from an old exam. After looking through the suggested solution, I have a few questions.
The problem is to find an analytic function $f$ mapping the open unit disc $\mathbb{D}$ one to one onto $D = \{z \in \mathbb{C}:|z-1+3i|<2\}$ such that $f(0) = 2-2i$ and $f'(0) > 0$.
The map $g(z) = 2z + 1 - 3i$ maps $\mathbb{D}$ one-to-one onto $D$ but does not satisfy the desired conditions. But since $g((1+i)/2) = 2 - 2i,$ composing it with the conformal self-map $\varphi_{(1+i)/2}$ so that $h(z) = g \circ \varphi_{(1+i)/2}$, satisfies $h(0) = 2-2i.$
Now, composing with a rotation also satisfies the condition, so each of $f_{e^{i \theta}} = g \circ \varphi_{(1+i)/2}(e^{i \theta}0)$, where $e^{i \theta}\in \partial \mathbb{D},$ satisfy $f_{e^{i \theta}}(0) = 2-2i.$
The following is where they lose me:
Differentiating, we have $f'_{e^{i \theta}}(0) = g'((1+i)/2) \varphi'_{(1+i)/2}(0)e^{i\theta} = -e^{i \theta}.$
So $g'((1+i)/2) = 2,$ and $\varphi'_{(1+i)/2}(0) = -1/2,$ and then they multiply by $e^{i \theta}$ to get $-e^{i \theta}$. But I would have thought it should be $\varphi_{(1+i)/2}(0e^{i \theta}) = \varphi'_{(1+i)/2}(0) = -1/2,$ so that the result would be $-1$, not $-e^{i\theta}...$
And in the end I'm even more confused, as they state that we should now set $f(z) = f_{-1}(z) = \frac{4-4i+(2-4i)z}{2+(1-i)z},$ and that this is the final answer to the problem. Where did that come from and how do I reach that result?