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I'm having trouble with this problem in Ahlfors' Complex Analysis (page 238):

If a vertex of the polygon is allowed to be at $\infty$, what modification does the formula undergo? If in this context $\beta_k = 1$, what is the polygon like?

The formula he is referring to is (probably) $$F(w) = C \int_0^w \prod_{k=1}^n ( w - w_k)^{-\beta_k}dw + C'$$ which maps the unit disk coformally onto a polygon $\Omega$ with outer angles $\{\beta_k \pi\}$. Here $\{w_k\}$ are points on the unit circle, and $C,C'$ are complex constants.

Since $F(w)$ has no dependence on the vertices of $\Omega$ I can't see what modification he is talking about. Shouldn't the exact same formula hold in all cases?

Also, I believe that letting a vertex $z_k$ tend to $\infty$ should always force the corresponding outer angle $\beta_k \pi$ into $ \pi$ (that is $\beta_k \to 1$). Thus, I can't see how a polygon with an infinite vertex $z_k$ could have the corresponding $\beta_k$ other than 1.

Are my conclusions correct? If not, please help me figure this out.

Thanks!

P.S.

I do understand that having some $\beta_k=1$ means that the polygon admits two infinite parallel line segments (?).

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I can't see how a polygon with an infinite vertex $z_k$ could have the corresponding $\beta_k$ other than $1$.

For example, the quadrant $Q=\{x+iy: x,y>0\}$ has two vertices: $0$ and $\infty$, with angles $\pi/2$ at each, hence both $\beta$ numbers are $1/2$.

If a finite point $w_k$ gets mapped to a finite vertex of interior angle $\phi_k$, the map should locally look like $(w-w_k)^{\phi_k/\pi}$, and thus its derivative should have the factor $(w-w_k)^{\phi_k/\pi - 1} = (w-w_k)^{-\beta_k}$.

If a finite point $w_k$ gets mapped to an infinite vertex of interior angle $\phi_k$, the map should locally look like $(w-w_k)^{-\phi_k/\pi}$, and thus its derivative should have the factor $(w-w_k)^{-\phi_k/\pi - 1}=(w-w_k)^{\beta_k-2}$.

In a nutshell: you'll have $\beta_k-2$ instead of $-\beta_k$ in the exponent corresponding to the vertex at infinity.

Example: infinite strip has two angles of size $0$, both at infinity. The beta numbers are $1, 1$. (By the way, your observation about the geometry of $\beta=1$ is correct.) The Schwarz-Christoffel map is $$\int \frac{1}{(w-a)(w-b)} \,dw $$ which is the logarithm composed with a Möbius transformation.

The Schwarz-Christoffel map substantially simplifies when we take the domain to be the upper half-plane and put one of the points $w_k$, say $w_n$, into infinity. In the case of an infinite vertex, it is natural to let $w_n$ be mapped $\infty$. Then the corresponding term drops out, so we don't have to worry about its exponent.

For example, to map the upper half-plane onto the quadrant $Q$ while sending $\infty$ to $\infty$, we only need to integrate $\int (w-w_1)^{-1/2}$, with the expected result.

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  • $\begingroup$ Thanks. Could you please elaborate on "the map should locally look like..." parts? I can't see why is that so. $\endgroup$ – user1337 Oct 7 '13 at 10:24

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