Map $\{x+iy \mid x^2+y^2<1 \text{ and } x^2 + (y-1)^2<2\}$ conformally to UHP

Question

From an old qualifying exam:

Let $D$ be the domain $$D :=\{x+iy \mid x^2+y^2<1 \text{ and } x^2 + (y-1)^2<2\}.$$

• Map the domain onto the upper half-plane.
• Obtain a function $f(z)$ analytic in the domain $D' := D \cap \{x+iy \mid x>0\}$ and which takes on the boundary values $\text{Re}f(z) = -1$ on the segment of the imaginary axis $-1<y<1$, and $2$ on the bounding circular arcs in $D'$, excluding the points $z=i$ and $z=i-\sqrt{2}i$. Further $\text{Im}f(0)=0$. Is $f(z)$ unique?

(For the first, I have to assume they want a conformal mapping, but they did not indicate it...go figure. For the second, one of the arcs is not "circular", but I guess I know what they mean.)

Let's focus on the first part for now; maybe if I get that I can get the second part. I have struggled to come up with some ideas for the first one. I don't know of any results about conformal mapping and parabolae. Maybe I should look at the polar form of a parabola?

Answer 1

Yes, in the context of a complex analysis exam you can assume that "map $D$ onto $D'$" means to map conformally (biholomorphically). Appealing to the fact that both domains have the same cardinality, and thus a bijection exists is not sufficient. :)

one of the arcs is not "circular"

Yes, and the question is correctly worded: the boundary value should be $-1$ on the vertical line segment (non-circular) and $2$ on the circular arcs.

When you map $D$ onto the UHP following Lubin's method, the points $\pm 1$ end up at $0$ and $\infty$. Where does the vertical line segment $[i, i-i\sqrt{2}]$ go? Since $\pm 1$ are symmetric about it, $0$ and $\infty$ will be symmetric about its image under the Möbius transformation used by Lubin. Hence, the Möbius transformation sends this segment to a circular arc centered at $0$. The subsequent power map keeps the circular shape of this arc. You'll find the radius by looking at where $0$ goes, for example.

Basically, you will find that the image of $D'$ is a half-disk. You need a holomorphic function with real values $-1$ and $2$ on two parts of its boundary. Idea: use another Möbius map to transform the half-disk to a sector in the plane, in which a multiple of $i\log z$ does the job. (The real part of $i\log z$ is $-\arg z$, which takes on two different constants on the half-line bounding the sector).

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Concerning uniqueness of $f$, it is tempting to argue as follows: if $f_1$ and $f_2$ are two such functions, then $h:=\operatorname{Re}(f_1-f_2)$ is a harmonic function vanishing on the boundary. By the maximum principle $h$ is identically zero. It follows that $f_1-f_2$ is a purely imaginary constant, which is zero because $\operatorname{Im}(f_1-f_2)$ vanishes at $0$. (There's a nonzero chance that this is what the problem author had in mind.)

Problem is, we can't apply the maximum principle because we don't know anything about the behavior of $h$ at two boundary points. On any simply connected domain there is a nonzero harmonic function that has zero boundary values at every point except one. Indeed, on the unit disk the function $\operatorname{Re}\frac{1+z}{1-z}$ has this property (the Poisson kernel, basically). By composition with a conformal map, we can transplant this example to other domains.

Thus, there are infinitely many distinct harmonic functions which attain the values $-1$ and $2$ on the indicated arcs. Each of them can be written (uniquely) as the real part of a holomorphic function whose imaginary part vanishes at $0$. All these are different holomorphic functions that satisfy the given requirements.

One can restore uniqueness by imposing the additional condition that $\operatorname{Re} f$ is bounded on $D'$. Then a different version of the maximum principle applies: if $h$ is a harmonic function in a bounded domain $\Omega$, $h$ is bounded from above in $\Omega$, and $\limsup_{z\to \zeta}h(z)\le M$ for all $\zeta\in\partial \Omega$ except for some finite set, then $h\le M$ in $\Omega$. In particular, a bounded harmonic function whose boundary values on $\partial \Omega$ are zero at all but finitely many points must be identically zero.

Answer 2

First from the definition it seems $D := \{|x^2+y^2|<1\}$ or the unit disc itself. If that is not a mistake, then the map $z \rightarrow i \dfrac{1+z}{1-z}$ sends the unit disc to the upper half plane conformally.

However, a more interesting problem would be to map $D:= \{|x^2+y^2| >1 \text{ and } x^2 + (y-1)^2 < 2\}$ to the UHP. Here D would be the simply-connected region bounded between the two circles that are tangent to each other at $z = -i$. Since this is a special point, it would help to map this point to infinity using $z \rightarrow \dfrac{1}{z+i}$. Then the unit circle gets mapped to the line $z = - i/2$ and the larger circle gets mapped to the line $z = -i/4$. So we get an infinite strip parallel to the real axis. This is because due to conformality the circles which were tangent to each other at $-i$ get mapped to generalized circles (in this case straight lines) which are tangent to each other at infinity, which means two parallel straight lines. From here the steps are easy.

1. Translate the strip vertically up using $z \rightarrow z+ i/2$.
2. Dilate the strip using $z \rightarrow 4\pi z$.
3. Finally use $z \rightarrow e^z$ to send the above domain to the UHP.

Answer 3

The region is the intersection of the unit disk (centered at the origin) and a disk of radius $\sqrt2$ centered at $(0,1)$. The points on the boundaries of both are $\pm1$, and we want to send one of these to zero, the other to infinity. If you take $$f_1(z)=\frac{z-1}{i(z+1)}\,,$$ then the upper arc (unit circle) goes to the positive real axis, and the lower arc goes from $0$ to $\infty$ by way of $1-i$. The interior point $0$ of the domain goes to $i$, so you see that the image is the wedge of angle $3\pi/4$ at the origin. Just take the $4/3$-power now to expand it to the UHP.