Conformal Mappings dealing with Slits

Question

The goal is to find a conformal mapping of the domain $$U=\lbrace z: \vert z \vert <1, z\not\in [1/2,1)\rbrace$$ to the unit disc $D$. I would like to learn how to deal with the slit; I imagine in one direction the slit has to be introduced while in the other direction it has to be erased (and those are both techniques I'd like to be able to use). I have no experience with slits and examples that I have seen haven't included enough detail or images, so I don't have the right intuition when dealing with these yet. Can I get a detailed explanation (preferably with some diagrams) of how to deal with the slit in one of the directions?

In going from $D$ to $U$, I would probably start by mapping onto a half-plane, but usually the slit is introduced shortly after and that's where I'm lost. I'm not sure what it means to introduce a slit.

Answer 1

This seems complicated, but drawing pictures will help.

Let's start with $U=\mathbb D\setminus [1/2,1)$. Let's first change the strip by mapping $1/2$ to $0$. For this purpose you can use a Mobius transformation of the unit disk that maps $1/2$ to $0$. One such transformation is $\frac{z-1/2}{1-z/2}$. Notice that this maps $1$ to $1$ (otherwise you could use a rotation). Since Mobius transformations map lines to lines, the strip $[1/2,1)$ is mapped 1-1 to the strip $[0,1)$.

Our domain now is $U_1=\mathbb D\setminus [0,1)$, and we wish to map it to the upper half plane. We know that the map $\frac{z-i}{z+i}$ maps the upper half plane $\mathbb H$ onto the unit disk, therefore its inverse maps the unit disk onto the upper half plane, and in particular the slit $[0,1)$ is mapped to $[i,i\infty)$ on the imaginary axis. Hence, our domain now is $U_2=\mathbb H\setminus [i,i\infty)$.

Next, use the function $z^2$ which will map the domain $U_2$ onto $\mathbb C$ minus two rays on the real axis: $(-\infty,-1]$ ( which is the image of $[i,i\infty)$) and $[0,\infty)$. It looks like you have two slits now, but it is actually only one slit since it is a "line segment" from $0$ to $-1$, passing through $\infty$.

Let's convert this "line segment" to a ray in the following way: Use a Mobius transformation to map $-1 $ to $\infty$. For example we can use $\frac{z}{z+1}$. Notice that this maps the real line onto itself, it also maps $0$ to $0$ and it is easy to check that the previous "line segment from $0$ to $-1$ through $\infty$" is now mapped onto the desired ray $[0,\infty)$.

Summarizing, so far we have $\mathbb C\setminus [0,\infty)$. Now, we can unfold this using a branch of $\sqrt{z}$, and map onto the upper half plane. Finally, go back to the unit disk, using $\frac{z-i}{z+i}$.

Answer 2

I think this usually involves some potential theory in one form or another. If you start out with the function $$ w(z) = c - \int\ln (t-z)\, d\mu(t) , $$ where $\mu$ is the equilibrium measure of $[-2,2]$ and $c\in\mathbb R$ is chosen so that $\textrm{Re}\, w$ is the equilibrium potential of $[-2,2]$ (so $c=0$), then $w$ maps $\mathbb C^+$ conformally onto the strip $S =\{ x+iy : x<0,0<y<\pi \}$. This is immediately plausible (and not hard to show rigorously) if you keep track of what $w(z)$ does as you increase $z\in\mathbb R$ from $-\infty$ to $\infty$. Notice in particular that the vertical piece of the boundary of $S$ is $w([-2,2])$.

Now if you take a more general compact subset of $E\subset [-2,2]$ and its equilibrium potential (plus its harmonic conjugate), then on each gap $(a,b)\subset [-2,2]$, $(a,b)\cap E=\emptyset$, you have that $\textrm{Im}\, w(x)$ is constant, while $\textrm{Re}\, w(x)$ increases from zero and then decreases again, so you map to a horizontal slit.

You can now use the conformal map $\varphi(z)=-z-1/z$ from the upper semi-disk onto the upper half plane, and $F(z)=e^{w(\varphi(z))}$ maps the unit disk back to a slit unit disk (you also need to holomorphically continue to the whole disk by reflection). The location of the slits will of course depend on your set $E$.

For the slit you want, you need $E=[-2,a]$, with $a<2$.