Let $\bigcup_{j=1}^n [0,a_i] = S$ be a 'star' where $a_i \in \mathbb{C}$ and the $[0,a_i]$ denote the line segments from $0$ to $a_i$ in the plane, all of the $a_i$ here are distinct and nonzero. There is a conformal map $G$ mapping $\mathbb{C} \setminus \overline{\mathbb{D}}$ to $\mathbb{C} \setminus S$ and sending $\infty$ to $\infty$ (conformal here means holomorphic and bijective, sending $\infty$ to $\infty$ means that $\frac{1}{G(\frac{1}{z})}$ has a zero at zero), and given we specify the sign of $G'(\infty)$ this map is in fact unique (but we will not need this for the following question).

It is clear that $G^{-1}$ extends analytically to the half open segments $(0,a_i]$ , and these correspond to some open arcs of the circle, say $(z_i,z_{i+1})$ which are all necessarily disjoint, and the $z_i$ correspond to $0$ (i.e. starting at $0$ and traversing towards $a_i$ 'from the right' of the line segment and then traversing from $a_i$ towards $0$ 'from the left' of the line segment), now put $u = \log |G|$ and note that this is certainly harmonic in $\mathbb{C} \setminus \overline{\mathbb{D}}$, we would like to show it extends to a harmonic function on $\mathbb{C} \setminus \{0, z_1,...,z_n \}$.

To do so we first note that $\partial_{r} u = 0$ on the open arcs $(z_i,z_{i+1})$, because $\arg G$ is constant on these arcs and then the (polar) Cauchy-Riemann equations gives us the result. Apparently, this allows us to conclude by the Schwarz reflection principle that $u$ extends harmonically to the interior of the unit disk, minus the origin.

Question 1:

My first question is why does this hold? I can see intuitively that the normal derivative on those arcs being zero implies $u$'s conjugate is constant on those arcs, so I can apply schwarz reflection, but my concern then is why is zero omitted, moreover I am not sure why we get a harmonic extension that is still harmonic on the other arcs we are not reflecting with respect to, it seems to me that if $u$'s conjugate is constant on one of the arcs, we can use schwarz reflection to extend it to the interior of the disk, but this extension need not be harmonic on the other open arcs.

Question 2:

It is also true that we can show $u = \log |G| = \log \prod_{j=1}^n |z-z_j|^{\frac{2p_j}{n}}$ for some $p_j \geq 0$ such that $\sum_{j=1}^n p_j = n$ by 'local considerations' at the $z_j$, how can we show this?

Thank you for any help in advance.

EDIT: I believe that $u$ having a harmonic extension to $\mathbb{D} \setminus \{0\}$ given $\partial_{r} u = 0$ on $\partial \mathbb{D}$ plus all the open arcs $\bigcup_{j=1}^{n} (z_{j},z_{j+1})$ is true (I have confirmation from a reputable source, but still no proof) , apparently given we already know $\partial_{r} u = 0$ on the boundary, all we need to do is define $u$'s extension as $u(\frac{1}{\overline{z}})$.

Given we take the above fact, I have been able to resolve my second question, namely one can show in the same way as how one might approach the proof to the Schwarz-Christoffel formula (by 'straightening out' a corner with interior angle $\pi \alpha_j$ ($\alpha_j \in (0,2)$) in the image of a conformal map by a suitable branch of $\zeta^{\frac{1}{\alpha_j}}$, then applying the reflection principle, and finally bending the line back to its original angle via a suitable branch of $z \rightarrow z^{\alpha_j}$) that near $z_j$ we have $G(z) = (z-z_j)^{\frac{2 p_j}{n}} h(z)$ where $h(z_j) \neq 0$ is holomorphic in a small disk around $z_j$ (of course this holds only in some small deleted neighbourhood of $z_j$ in the exterior of the unit disk).

Then we get $\log|G| = \log|z-z_j|^{\frac{2p_j}{n}} + C_j + \epsilon(z)$ in a punctured disk at $z_j$ where $\epsilon(z)$ is harmonic in this punctured disk and vanishes as $z \rightarrow z_j$ and $C_j$ is some non-zero constant. Moreover if we normalize $G$ so that at $\infty$ its laurent expansion is of the form $z + a_2z^2 + ...$, then we know that $\log|G| = \log|z| + \epsilon(z)$ as $z \rightarrow \infty$, where the $\epsilon$ term also vanishes as $z \rightarrow \infty $ and is harmonic in a neighbourhood of $\infty$, then since near $0$ $\log|G| = \log|G|(\frac{1}{\overline{z}}) = -\log|\overline{z}| + \epsilon(\frac{1}{\overline{z}}) = -\log|z| + \epsilon(z)$ (again the $\epsilon$ term vanishes as $z \rightarrow 0$), this suggests us to consider the function $\sum_{j=1}^{n} \log|(z-z_j)(\frac{1}{z} - \overline{z_j})|^{\frac{p_j}{n}} $, locally near the $z_j$ it looks like $\log|G|$ up to some constant summand and term vanishing as $z \rightarrow z_j$, and at $\infty$ and near $0$ its difference with $\log|G|$ goes to zero, therefore its difference with $\log|G|$ is a harmonic function on the entire Riemann sphere, and in particular must be bounded, but then by Liouville's theorem it is a constant, and this constant is necessarily zero since this difference is zero at $\infty$ and $0$.

This proves that $\log|G| = \log \prod_{j=1}^n |(z-z_j)(\frac{1}{z} - \overline{z_j})|^{\frac{p_j}{n}}$

Therefore what remains to prove is the first point, namely that the condition $\partial_{r} u = 0$ on the boundary enables us to extend our $u$ to the unit disk , as well as all the open arcs on the boundary.

Edit 2: Originally I asked this question because I wanted to show that all such conformal maps have a specific form. I have since been able to show that they have the form I desired through other means (despite this, I am still interested in a solution to my Question 1).

I will prove that if $f : \mathbb{D} \rightarrow \mathbb{C}$ is a conformal map from the interior of the unit disk to the exterior of the star, sending $f(0) = \infty$ (so basically the map I want, except the map I want is $G(z) = f(\frac{1}{z})$), and the segments $[0,a_j]$ and $[0,a_{j+1}]$ are at an angle $\frac{2 \pi p_j}{n}$ from each other and I impose the condition that $f$ has a simple pole at $0$ with residue $1$ there, then $f(z) = \frac{1}{z} \prod_{j=1}^n (z-z_j)^{\frac{2\pi p_j}{n}}$.

The crux of the proof is the already mentioned straightening of corners, first let $M$ be some mobius map from $\mathbb{H}^{+}$ to $\mathbb{D}$ chosen carefully so that $\infty$ is not sent to any prevertex $z_j$ or point on the circle corresponding to a maximum $a_i$, now tracing through the usual proof of the Schwarz-Christoffel mapping from $\mathbb{H}^{+}$ to some simply connected region with polygonal boundary, we see that after an even number of reflections, (say $2$) , our new branch of the conformal map is the same as our original up to a rotation (this is because reflecting with respect to a line through the origin and then another line through the origin is some rotation $w \rightarrow \exp(i \phi)w$), so the same holds with respect to the circular arcs $(z_j,z_{j+1})$ (the image is not changed by the mobius map, or more explicitly after reflecting twice for the upper half plane transported map, we get $\exp(i \theta) f \circ M$, so precomposing with $M^{-1}$ the new branch of $f$ in the unit disk is $\exp(i \theta) f \circ M \circ M^{-1} = \exp(i \theta) f$), after iterating this procedure we have continued $\frac{f'}{f}$ to $\mathbb{C} \setminus \{0,z_1,...,z_n\}$ (since after an even number of reflections $\frac{f'}{f}$ returns back to its original branch).

Since at the points $z_j$ we already know $f(z) = (z-z_j)^{\frac{2 p_j}{n}} h(z)$ with $h(z_j) \neq 0$ and $h$ holomorphic in a neighbourhood of $0$, we know that $\frac{f'}{f}$ has simple poles at the $z_j$ with residues $\frac{2 p_j}{n}$ and since $f$ has a simple pole at $0$, $\frac{f'}{f}$ has residue equal to $ord_{f}(0) = -1$ at $0$ (and has a simple pole there), thus we have that $\frac{f'}{f} - \sum_{j=1}^n \frac{\frac{2 p_j}{n}}{z-z_j} + \frac{1}{z}$ is entire and bounded, and therefore must be constant by Liouville's theorem. As $z \rightarrow \infty$ this expression tends to $0$ ( to see this one notes that after reflecting an odd number of times the form of the corresponding branch of $f$ is $A \overline{f}(\frac{1}{\overline{z}})$ (in fact $|A| = 1$), then since near $0$ , $f(z) = g(z) + \frac{1}{z}$ where $g$ is its holomorphic part , we get that the odd branch (for large $z$) is $g(\frac{1}{z}) + z$, then manually computing $\lim_{z \rightarrow \infty} \frac{f'}{f} = 0$ gives the result), so it must be identically zero. Finally we get $f(z) = \alpha \frac{1}{z} \prod_{j=1}^n (z-z_j)^{\frac{2 p_j}{n}}$ where $\alpha$ is some integration constant which is in fact necessarily exactly $\frac{1}{\prod_{j=1}^n (-z_j)^{\frac{2p_j}{n}}}$ because we normalized $f$ to have residue $1$ at $0$, so we can rewrite this further as $f(z) = \frac{1}{z} \prod_{j=1}^n (1 - \frac{z}{z_j})^{\frac{2p_j}{n}}$.


1 Answer 1


Actually it turns out that the answer is more or less embedded in my second edit, the key point is that the $\partial_{r}u = 0$ condition translates to $u$'s conjugate being constant on the open arc $(z_j,z_{j+1})$, let $f = u+iv$ be holomorphic and locally defined on some ball centered at a point $z \in (z_j,z_{j+1})$, let $M$ be a mobius transformation from the upper half plane to the exterior of the unit disk, denote $z \rightarrow \bar{z}$ by $s_{\mathbb{R}}$ and $z \rightarrow \frac{1}{\bar{z}}$ by $s_{\mathbb{D}}$, and note that $M \circ s_{\mathbb{R}} \circ M^{-1} = s_{\mathbb{D}}$ then $f \circ M$ has constant imaginary part on some open interval of $\mathbb{R}$ mapping to some open arc containing $z$, so $f \circ M - c$ for some constant $c = i \Im{f}$ ($\Im{f}$ being the constant imaginary part on that open interval) has zero imaginary part on this open interval and extends to the lower half plane (more precisely the reflection of the portion of the upper half plane it was defiend on ) and on that open arc, to $s_{\mathbb{R}} \circ [f \circ M - c] \circ s_{\mathbb{R}}$, thus $f \circ M$ extends to the lower half plane and this open interval to $s_{\mathbb{R}} \circ [f \circ M - c] \circ s_{\mathbb{R}} + c = s_{\mathbb{R}} \circ f \circ M \circ s_{\mathbb{R}}$, and $f$ extends to $s_{\mathbb{R}} \circ f \circ M \circ s_{\mathbb{R}} \circ M^{-1} = s_{\mathbb{R}} \circ f \circ s_{\mathbb{D}} = \bar{f(\frac{1}{\bar{z}})}$ in the interior of the unit disk plus the open arc minus the origin, and this things real part is $u(\frac{1}{\bar{z}})$. We can clearly do this for all arcs, which gives the result.


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