I came across the following question:
Find a conformal map from $$H=\left\{z\in\mathbb{C}:\operatorname{Re}(z)>0\right\}$$ onto $$A=\left\{z\in\mathbb{C}:|z-2|<3,|z|>1\right\}.$$ You may leave your answer as a composition of conformal mappings.
Here's my solution which I'm not sure if it's correct. Would like to seek assistance on this. Thanks!
Solution: We find a conformal map from $A$ to $H$ first.
Let $\phi_1(z)=\dfrac{1}{z+1}$. Let us figure out what $A$ gets mapped to via $\phi_1$. So, if we let $w=\dfrac{1}{z+1}$, we have $z=\dfrac{1}{w}-1$. Consider the annulus $|z-2|<3$, so after the transformation, we have $w>\dfrac{1}{6}$. For the region $|z|>1$, we have $\dfrac{1}{w}>2$. So, $\phi_1$ maps $A$ to $A_1$, where $A_1=\left\{z\in\mathbb{C}:1/6<\operatorname{Re}(z)<1/2\right\}$.
Let $\phi_2(z)=z-\dfrac{1}{6}$. So, $\phi_2$ maps $A_1$ to $A_2=\left\{z\in\mathbb{C}:0<\operatorname{Re}(z)<1/3\right\}$.
Let $\phi_3(z)=\operatorname{tan}\left(\dfrac{3\pi z}{2}\right)$, which maps $A_2$ to $H$. It's a similar trick to showing that $(-\infty,\infty)=(0,1)$, or rather, something along those lines.
As such, the required conformal map from $H$ to $A$ is $\phi_3^{-1}\circ \phi_2^{-1}\circ \phi_1^{-1}$.