# Mapping unit disk to polygon

Let $$D$$ be the open unit disk.

I already know the Poisson integral formula:

"Let the complex-valued function $$g(e^{i \theta})$$ be piecewise continuous and bounded for $$\theta$$ in $$[0,2\pi]$$. Then the function $$f(z)$$ defined by

$$f(z)=\frac{1}{2\pi} \int_{0}^{2\pi} \frac{(1-|z|^2)}{(|e^{it}-z|^2)}g(e^{it})dt$$ is the harmonic extension of $$g$$ in the unit disk"

"Let $$\Omega$$ be a subset of the complex plane that is a bounded convex domain whose boundary is a Jordan curve $$\Gamma$$. Let $$g$$ map the unit circle continuously onto $$\Gamma$$ and suppose that $$g(e^{it})$$ runs once around $$\Gamma$$ monotonically as $$e^{it}$$ runs around the unit circle. Then the harmonic extension given in the poisson integral formula is univalent in $$D$$ and defines a harmonic mapping of $$D$$ onto $$\Omega$$."

But now I have to prove the following statement:

"Let $$f(z)$$ be the harmonic extension (i.e. harmonic extension from the poisson integral formula, see above) of a stepwise constant function $$g$$ on the unit circle, i.e., if $$\{e^{it_k}\}$$ is a partition of the unit circle with $$t_0<\cdots then $$g(e^{it})=v_k$$ for $$t_{k-1}. Suppose the vertices $$v1; v2; ... ; vn$$, when traversed in order, define a convex polygon whose interior is denoted by $$\Omega$$. Then $$f(z)$$ is univalent in $$D$$ and defines a harmonic mapping from $$D$$ onto $$\Omega$$."

Anyone an idea how to prove this last statement? You can use the previous two statements (I already proved them so that's okay).

A big thank you if you can help me!

Kind regards.

• Latex for "f(z)=1/(2pi) integral_{0}^{2pi} (1-|z|^2)/(|e^{it}-z|^2)g(e^{it})dt" is $f(z)=\frac{1}{2\pi} \int_{0}^{2\pi} \frac{(1-|z|^2)}{(|e^{it}-z|^2)}g(e^{it})dt$. I have replaced it into your question. Mar 18 at 10:15
• Besides, I think that "harmonic extension of f " should be "harmonic extension of g". Mar 18 at 10:18
• I think you need a first mapping from the unit disk onto the upper half plane using $Z=i \frac{z+1}{z-1}$, then use the so-called Schwarz-Christoffel transformation Mar 18 at 10:23
• Yeah that indeed works. But I am now looking for another method than Schwarz christoffel . Mar 18 at 10:35
• Thank you, Jean Marie, but I think that's not an explanation why the poisson integral delivers a mapping from te unit disk to the polygon? My purpose was to find, besides schwarz christoffel, another method to map onto a polygon. And that should be the poisson integral. So I need to prove the statement in my question. Mar 18 at 10:37