- If \(P\) is a simply connected region bounded by a polygon with vertices \(a _ { 1 } , \ldots , a _ { n }\) and angles \(\alpha _ { 1 } \pi , \ldots , \alpha _ { n } \pi\) , and \(F\) is a conformal map of the disc D to \(P ,\) then there exist complex numbers \(B _ { 1 } , \ldots , B _ { n }\) on the unit circle, and constants \(c _ { 1 }\) and \(c _ { 2 }\) so that
\[ F (z) = c _ {1} \int_ {1} ^ {z} \frac {d \zeta}{(\zeta - B _ {1}) ^ {\beta_ {1}} \cdots (\zeta - B _ {n}) ^ {\beta_ {n}}} + c _ {2}. \]
[Hint: This follows from the standard correspondence between H and D and an argument similar to that used in the proof of Theorem 4.7.]
Solution
Put \(\beta_j=1-\alpha_j\). Choose a Möbius map \[ \phi:\mathbb H\longrightarrow\mathbb D \] whose inverse sends real points \(x_j\) to the boundary points \(B_j\in\partial\mathbb D\) that correspond under \(F\) to the polygon vertices \(a_j\).
Then \(G=F\circ\phi\) maps \(\mathbb H\) conformally onto \(P\). By the Schwarz–Christoffel formula, \[ G'(w)=C\prod_{j=1}^n(w-x_j)^{-\beta_j}. \tag{1} \] Now write \(w=\phi^{-1}(z)=(az+b)/(cz+d)\). For each \(j\), \[ w-x_j=\frac{(a-cx_j)(z-B_j)}{cz+d}, \qquad \frac{dw}{dz}=\frac{ad-bc}{(cz+d)^2}. \] Because the angles of a polygon satisfy \[ \sum_{j=1}^n\alpha_j=n-2, \] we have \[ \sum_{j=1}^n\beta_j =n-(n-2)=2. \] Therefore all powers of \(cz+d\) cancel when (1) is transformed back to the disk, and \[ F'(z)=C_1\prod_{j=1}^n(z-B_j)^{-\beta_j} \] for a nonzero constant \(C_1\). Integrating from any fixed base point (in particular from \(1\), interpreted with a path in the disk when \(1\) is not itself a prevertex) gives \[ F(z)=c_1\int_1^z \frac{d\zeta}{(\zeta-B_1)^{\beta_1}\cdots(\zeta-B_n)^{\beta_n}} +c_2, \] with suitable branch choices and constants \(c_1\ne0,c_2\in\mathbb C\).