 Solution

Near a prevertex of a Schwarz-Christoffel map $X_j$, a conformal map to an interior angle $(1-{\beta }_j)\pi$ has local formIts derivative therefore has the factor $(Z-X_j{)}^{-{\beta }_j}$. On each real interval between prevertex, the argument of the product of these factors is constant, so its integral traces a straight polygon side. Passing a prevertex of a Schwarz-Christoffel map changes the argument by the prescribed exterior angle ${\beta }_j\pi$. Thushas exactly the required straight sides and corner angles, with the branches chosen in the upper half-plane. Equivalently $f^{\prime \prime }/f^{\prime }$ has simple poles $-{\beta }_j/(Z-X_j)$; removing them leaves the constant normalization determined by behavior at infinity. The additive integration constant translates the polygon and $C$ rotates and scales it. This is the Schwarz-Christoffel formula.

The automorphisms of the upper half-plane are real Möbius transformations with positive determinant, modulo a common nonzero scalar. They have three real degrees of freedom, so three prevertex of a Schwarz-Christoffel map positions may be assigned independently. The remaining positions are constrained by the side-length ratios.

If $X_n=\infty$, omit its finite factor:Since ${\sum }_j{\beta }_j=2$, the derivative behaves like $Z^{{\beta }_n-2}$ at infinity; in the coordinate $U=1/Z$, this is exactly the required corner exponent there. Thus the missing angle is encoded by behavior at infinity.