Construct a conformal map from $\operatorname{Im} z >0$ onto $\{z : |z - 2i| < 2, |z - i |>1\}$ Construct a conformal mapping $f$ from the upper-half plane  $\{z : \operatorname{Im} z >0 \}$ onto $ \Omega = \{z : |z - 2i| < 2, |z - i| >1\}$, with the restrictions that $f(0) = 2i$, $f(\pm 1) = 0$
I know that map $z \mapsto \frac{1}{z}$ will take the strip $\{z : -\frac{1}{2} < \operatorname{Im} z < -\frac{1}{4} \}$ bijectively onto $\Omega$. And we can morph the upper-half plane into such a strip by first applying $\operatorname{Log} z$ (principal branch), and then scaling and translating appropriately. However, this course of action does not seem to do the job! The use of $\operatorname{Log} z$ would mean that our $f$ is actually not defined on the negative real axis. But we still need $f(0) = 2i, f(-1) = 0$.
A solution is greatly appreciated. I've been thinking about this one for a while with no luck! 
 A: It is better to construct a map of $\Omega$ onto the half-plane. (In this example, and in general). You can invert it later. 


*

*Starting from $\Omega$, apply $z\mapsto 1/z$ to get $\{-1/2<\operatorname{Im}z<-1/4\}$. 

*Use a linear transformation $z\mapsto 2\pi (z+i/2)$ to get $\{0<\operatorname{Im}z<\pi\}$. 

*Exponential map sends the strip onto the upper half-plane. 

*Where is the point that was at $2i$? It's at $1$ now. We should map this to $0$.

*Where is the point that was at $0$? It got split into $0$ and $\infty$, poor thing. We should map the pieces to $\pm 1$. 

*The transformation $z\mapsto \dfrac{z-1}{z+1}$ takes care of 4 and 5. 


Now invert the steps. The inverse of 3 will be a principal branch of logarithm, by the way. It's just that it's applied after a linear fractional transformation. 


The use of $\mathrm{Log}\, z$ would mean that our $f$ is actually not defined on the negative real axis.

The principal branch of logarithm has continuous boundary values along $(-\infty, 0)$, when it's approached from above. They don't agree with the limits we get by approaching $(-\infty, 0)$ from below, but this does not matter when we stay within the closed upper half-plane. A more  hands-on explanation: the closed upper half-plane is the set $\{re^{i\theta}:r\ge 0, 0\le \theta\le \pi\}$. The map $f(re^{i\theta})=\ln r+ i\theta$ is well-defined and continuous in the closed halfplane, except for $0$. It sends the closed halfplane (minus $0$) onto the closed strip $\{x+iy: 0\le y\le \pi , \ x\in\mathbb R\}$. 
