I am supposed to construct a conformal mapping from $A$ to $B$, where $A$ and $B$ are given by: $A = \{z=x+iy: 0<x<1\ \text{and}\ 0<y<1\},\ B = \{z \in \mathbb{C}: 1<|z|<e^{2\pi}\} \setminus (1,e^{2\pi})$.
Using the map $f_1(z) = z^4$, I can map $A$ to the square $C = \{z=x+iy: -1<x<1\ \text{and}\ -1<y<1\} \setminus [0,1)$ (by taking the usual branch cut of log $z$). Then, I can map $C$ to a bigger square $C' = \{z=x+iy: -e^{2\pi}<x<e^{2\pi}\ \text{and}\ e^{2\pi}<y<e^{2\pi}\} \setminus [0,e^{2\pi})$ by the map $f_2(z) = e^{2\pi z}$. Now, via the map $f_3(z) = \frac{z - i}{z + i}$, we map the positive real axis to the unit disc, and hence we have $C'$ being mapped to $C'' = C' \setminus \overline{D(0;1)}$. I am not sure how to proceed after this, since I cannot find a map that maps a line segment to the semicircle arc in $B$.
Any hint or help will be appreciated!