= Solution
{c}
Near a <prevertex of a Schwarz-Christoffel map> $X_j$, a <conformal map> to an interior angle $(1-\beta_j)\pi$ has local form
$$
f(Z)-f(X_j)\sim c_j(Z-X_j)^{1-\beta_j}.
$$
Its <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 $\beta_j\pi$. Thus
$$
\boxed{f'(Z)=C\prod_{j=1}^n(Z-X_j)^{-\beta_j}}
$$
has exactly the required straight sides and corner angles, with the branches chosen in the <upper half-plane>. Equivalently $f''/f'$ 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 \b[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:
$$
\boxed{f'(Z)=C\prod_{j<n}(Z-X_j)^{-\beta_j}.}
$$
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.
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