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Source: Oxford Exam $A2 \ 1999$

We want to construct a conformal map $F$ from the unit disc $\mathbb{D}=\{z:|z|<1\}$ to $\mathbb{C} \setminus S$ where $S$ is the half-line $\{x+i:x \in (-\infty,0] \}$ with the additional property that $F(0)=0$.

Here are my thoughts, can someone see if they are correct please:

  1. Use the Möbius transformation $f_1:z \mapsto \frac{1-z}{z+1}$ to send $\mathbb{D}$ to the right half plane.
  2. Use the map $f_2:z \mapsto z+ \alpha$ where $\alpha \in \mathbb{C}$ we will determine later. This simply takes the right half plane to the right half plane.
  3. Use the map $f_3:z \mapsto z^2$ to map the right half plane to $\mathbb{C}$ with a cut along the negative reals.
  4. Use the map $f_4:z \mapsto z+i$ to move the cut up and map to $\mathbb{C} \setminus S$.

Now the composition of these four maps is conformal so we certainly have a conformal map between the sets.

We now need $F(0)=f_4 \circ f_3 \circ f_2 \circ f_1(0)=0$.

$f_1(0)=1$, $f_2(1)=1+\alpha$, $f_3(1+\alpha)=\alpha^2+2\alpha+1$, $f_4(\alpha^2+2\alpha+1)=\alpha^2+2\alpha+1+i$

So we need to solve $\alpha^2+2\alpha+1+i=0$ and this is the $\alpha$ required to ensure $F(0)=0$.

We can find that $\alpha=\sqrt{-i}-1$.

Thus we are done.

Is this correct?

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  • $\begingroup$ It seems to me that 2. will map the right half plane to something different unless the real part of $\alpha$ is zero. $\endgroup$ – MvG May 22 '13 at 7:07
  • $\begingroup$ @MvG Yes you are right; can you suggest how I can ammend my approach to the question? $\endgroup$ – Mathmo May 22 '13 at 9:19
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The map $f_4\circ f_3$ as you constructed it will conformally map the right half plane to the desired $\mathbb C\setminus S$. So it should be possible to express any vaid solution as a combination of a conformal map from the unit disk to the right half plane, concatenated with that function. So every valid $F$ can be written as $F=f_4\circ f_3\circ f_{12}$ for some conformal map $f_{12}: \mathbb D\to\{z\in\mathbb C:\operatorname{Re}(z)>0\}$.

So what ways are there to conformally map the unit disk to the right half plane? According to Riemann's mapping theorem, that map is unique up to Möbius transformation. Since it is possible to perform $f_{12}$ using a Möbius transformation, this tells us that it is enough to consider only Möbius transformations in this step.

A Möbius transformation on $\mathbb C\cup\{\infty\}$ is uniquely defined given three distinct points and their images. Since we know that the boundary of the unit disk must map to $S$, we can choose two points on the unit disk as the preimages of $0$ and $\infty$. Without loss of generality, we can even rotate the unit disk such that one of these points will be the point $1$. We can furthermore require the point $0$ to map to $\sqrt{-i}=\frac{1-i}{\sqrt2}$, so that $f_4\circ f_3$ will take it from there back to $0$ as required. All together, our Möbius transformation $M$ (which will play the role of $f_{12}$) is uniquely defined by

\begin{align*} 0 &\mapsto \sqrt{-i}=\tfrac{1-i}{\sqrt2}=e^{-i\frac\pi4} \\ 1 &\mapsto 0 \\ e^{i\varphi} &\mapsto \infty \end{align*}

for some real parameter $\varphi$. Plugging this into a computer algebra system (sage in my case), you can obtain

$$ M: z\mapsto\frac {e^{i\left(\varphi-\frac\pi4\right)}\cdot (z-1)} {z - e^{i\varphi}} $$

Now plug an arbitrary point from the boundary of $\mathbb D$ into that function. It should end up on the imaginary axis, the boundary of your half plane. So its real part should be zero. Again with help from computer algebra, I got

$$\operatorname{Re}\bigl(M(-1)\bigr) = \frac{\sin\varphi + \cos\varphi + 1}{\sqrt2\,\left(\cos\varphi+1\right)}$$

Note: The choice of the example point here is pretty much arbitrary, but not completely so. In my first attempt, I plugged $-i$ instead of $-1$ into this, which is a special case for which this approach will not work. The reason is that later on we will see that $-i$ is the point which gets mapped to $\infty$, and which therefore has no well-defined real part. If in doubt, one should probably perform this computation with two different points.

In any case, requiring that real part to be zero, one can obtain the solution $\varphi=-\frac\pi2$. With this, the Möbius transformation is fully defined:

$$ M: z\mapsto\frac {e^{-i\frac34\pi}\cdot (z-1)} {z - e^{-i\frac12\pi}} = \frac{(1+i)(1-z)}{\sqrt2(z+i)} $$

This transformation will map the boundary of the unit circle onto the imaginary axis, since it does so for three points and a circle (including lines as a special case) is already uniquely defined by three points. So this will map your unit disk to the right half plane, and put $M(0)$ in the right place. Now you can combine this with $f_3$ and $f_4$ to obtain the complete map $F$:

$$F = f_4\circ f_3\circ M: z\mapsto \frac{2 i \, z^{2} - \left(2 i + 2\right) \, z}{z^{2} + 2 i \, z - 1}$$

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  • $\begingroup$ @mercio has provided an answer above that I think works fine and I have accepted. Thanks for your post though, it was useful nonetheless. $\endgroup$ – Mathmo May 23 '13 at 12:51
  • $\begingroup$ @Mathmo: I found and fixed my mistake. I made a poor choice the first time, using exactly the wrong third point on the circle to deduce $\varphi$. Now it works. The other answer was very useful in finding my mistake, though, so thanks to mercio for that as well. I'll leave my post in place in case it might be useful, now that it is no longer wrong. $\endgroup$ – MvG May 23 '13 at 13:50
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Your choice of $f_2$ is bad. We need a conformal map from the right-half plane to the right-half plane, that sends $f_1(0) = 1$ to $f_3^{-1}\circ f_4^{-1}(0) = f_3^{-1}(-i) = (1-i)/\sqrt 2$.

Your choice of a translation sends the right-half plane to one of its translate to the left, which is no longer the right-half plane, and then the square map messes up everything.

Since the automorphisms of the top-half plane are of the form $z \mapsto (az+b)/(cz+d)$ with $ad-bc > 0$, the conformal map we need is of the form $z \mapsto (az-ib)/(icz+d)$ with $ad-bc > 0$. For example, $f_2(z) = (z-i)/\sqrt 2 $ is one map that works, and it is very easy to see that it does map the right-half plane to itself.

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