Find a conformal map from two circle tangent to each other from inside to the upper half plane.

Question

Firstly, I am really sorry that I made the title pretty confusing, so I will hugely appreciate if any one can edit it so that we have a better title.

I worked on a question last week stating as

Conformally map the region $$D=\Big\{z\in\mathbb{C}:|z-1|<1\ \text{and}\ \Big|z-\dfrac{2}{3}\Big|>\dfrac{2}{3}\Big\}.$$ onto the upper half-plane.

I posted this question last week, but deleted as no one responded for a long time except for two comments in the post. I stopped thinking this problem and went to do some other problems over the last weeks, but today I went back to this problem, and thanks to that second comment, I think I've solved it.

Since I've deleted the last post, and I forgot the name of the user who left that comment, and since this question is really interesting, at least to me, I post this question again, and will answer it myself, so that I can:

1) Express my appreciation to that user;

2) Express my apology to that user for me not able to give the credit to him/her;

3) Share this interesting problem and share my solution so that more people can get some ideas about how to solve this type of problem, because I think this problem is pretty classic about how to use Möbius Transformation to map the origin to $\infty$ and to change the geometric interpretation.

Please feel free to point out any mistakes or typos.

Answer 1

The idea of the construction is to use Möbius Transformation to map the origin to infinity, so that we straighten the two circles to be a strip.

Define the Möbius Transformation $$f:\hat{\mathbb{C}}\longrightarrow\hat{\mathbb{C}},\ \text{by}\ z\mapsto f(z):=\dfrac{z-1}{z},$$ which is always a bijective holomorphic function.

Now, let's figure out where $f(z)$ maps the inner and outer circle to. Set the inner circle to be $$S_{1}:=\Big\{z\in\mathbb{C}:\ z=\dfrac{2}{3}e^{i\theta}+\dfrac{2}{3},\ \theta\in[0,2\pi]\Big\}, $$ and similarly set the outer circle to be $$S_{2}:=\Big\{z\in\mathbb{C}:\ z=e^{i\theta}+1,\ \theta\in[0,2\pi]\Big\}.$$

For $z\in S_{1}$, we have \begin{align*} f(z)&=\dfrac{z-1}{z} \\ &=\dfrac{2 e^{i\theta}-1}{2e^{i\theta}+2}\\ &=\dfrac{2+2\cos\theta}{8+8\cos\theta}+\dfrac{6 i\sin\theta}{8+8\cos\theta}\\ &=\dfrac{1}{4}+i\dfrac{3}{4}\tan\Big(\dfrac{\theta}{2}\Big)\\ &:=u+iv. \end{align*}

Thus, $f(z)$ maps $S_{1}$ to $\Big\{w=u+iv:\ u=\dfrac{1}{4},\ v\in\mathbb{R}\ \Big\},$ which is a vertical straight line parallel to the imaginary axis passing through $\dfrac{1}{4}$.

For $z\in S_{2}$, we have \begin{align*} f(z)&=\dfrac{z-1}{z}\\ &=\dfrac{e^{i\theta}}{e^{i\theta}+1}\\ &=\dfrac{1}{2}+i\dfrac{1}{2}\tan\Big(\dfrac{\theta}{2}\Big) \end{align*}

Thus, $f(z)$ also maps $S_{2}$ to $\Big\{w=u+iv:\ u=\dfrac{1}{2},\ v\in\mathbb{R}\ \Big\},$ which is another vertical straight line parallel to the imaginary axis passing through $\dfrac{1}{2}$.

Therefore, totally, $$f:D\longrightarrow D_{1}:=\Big\{w=u+iv:\ u\in\Big[\dfrac{1}{4},\dfrac{1}{2}\Big], v\in\mathbb{R}\Big\}.$$

Now, we want to move $D_{1}$, so that one side of the strip is the imaginary line. Thus, we use this map $$g(z):D_{1}\longrightarrow D_{2}:=\Big\{w=u+iv:\ u\in\Big[0,\dfrac{1}{4}\Big], v\in\mathbb{R}\Big\},\ \text{by}\ z\mapsto g(z):=z-\dfrac{1}{4},$$ which is a commonly-used conformal map.

Then, we rotate $D_{2}$ to a strip that lives in the upper-half plane. We use another known conformal map $$h(z):D_{2}\longrightarrow D_{3}:=\Big\{w=u+iv:\ u\in\mathbb{R}, v\in\Big[0,\dfrac{1}{4}\Big]\Big\},\ \text{by}\ z\mapsto h(z):=iz.$$

We then dilate $D_{3}$ to be a strip betwen $0$ and $\pi i$, i.e. use the conformal map $$d(z):D_{3}\longrightarrow D_{4}:=\Big\{w=u+iv:\ u\in\mathbb{R}, v\in[0,\pi]\Big\},\ \text{by}\ z\mapsto d(z):=4\pi z.$$

Now, everything is set so that we can use our final and also well-known conformal map $$\ell (z):D_{4}\longrightarrow\mathbb{H},\ \text{by}\ z\mapsto \ell(z):=e^{z}.$$

Therefore, our desired conformal mapping is $$F:=\ell\circ d\circ h\circ g\circ f:D\longrightarrow\mathbb{H}, $$ and a quick calculation gives us $$F(z)=-\exp\Big(\dfrac{3z-4}{z}\Big).$$

Answer 2

First of all, good job. I just want to leave some comments here but they are too long for a comment.

(1) When you tried to compute where $S_1,S_2$ went under $f$, you explicitly computed the expression. Although this is good practice, knowing that a Mobius transformation maps circles to circles (a line is a circle with a point at $\infty$) saves you a lot of trouble,because it suffices to find two points' images.

(2) There is a rule for this type of problems. Usually we need to find conformal mappings between disks, strips, sectors, half-planes. Usually we use the following (and also translations, rotations):

$$\text{Mobius transformation}: \\\text{disks}\leftrightarrow\text{half-planes},\\ \text{disks}\leftrightarrow\text{disks},\\ \text{half-planes}\leftrightarrow\text{half-planes}$$ $$e^{\alpha z}:\text{strip}\mapsto\text{sector}$$ $$\log:\text{sector}\mapsto\text{strip}$$

Also, by properly choosing $\alpha$ (scaling the width of the strip), you can decide the angle of a sector that $e^{\alpha z}$ maps a strip to, following which we can map a strip to a half-plane (a sector with an angle of $\pi$).

(3) Your question has an interesting variant. How to map two acentric circles to cocentric circles? For example,

Map the region $\{z\in\mathbb C:|z-1|>1,\ |z|<3\}$ to a ring-shaped region with concentric boundaries.

Answer 3

Mapping the origin to infinity in order to map the circles to lines is a good approach. I would simply start with $f(z) = \frac 1z$.

Determining the image of $C_1 = \{ |z-1|=1 \} $ and $C_2 = \{ |z-\frac 23| = \frac 23 \} $ under $f$ can be simplified by using the fact that conformal mappings preserve angles: Both $C_1$ and $C_2$ intersect the real axis at a right angle, and $f$ maps the (extended) real axis onto itself. It follows that $f(C_1)$ and $f(C_2)$ also intersect the real axis at a right angle. Therefore $f(C_1)$ and $f(C_2)$ are lines containing the points $z=\frac 12$ resp. $z = \frac 34$, and are parallel to the y-axis: $$ f(C_1) = \{ z \mid \operatorname{Re}z = \frac 12 \} \cup \{ \infty \} \\ f(C_2) = \{ z \mid \operatorname{Re}z = \frac 34 \} \cup \{ \infty \} $$ and $D$ is mapped to the strip between those lines: $$ f(D) = \{ z \mid \frac 12 < \operatorname{Re} < \frac 34 \} $$

From here you can continues as you did: Map this vertical strip to the horizontal strip $$ \{ z \mid 0 < \operatorname{Im} < \pi \} $$ and finally to the upper half-plane with the exponential function.