 Solution

At a vertex with exterior turning angle ${\beta }_j\pi$, the interior angle is $(1-{\beta }_j)\pi$. A local conformal map from a half-plane therefore behaves as $f(Z)-z_j\sim c_j(Z-X_j{)}^{1-{\beta }_j}$, so its derivative has local exponent $-{\beta }_j$.

On each real boundary interval between prevertices, $\arg f^{\prime }$ is constant because the image is a straight side. Across a prevertex its change is the exterior turning angle. Taking $\arg (Z-X_j)\in (0,\pi )$, the product ${\prod }_j(Z-X_j{)}^{-{\beta }_j}$ has exactly these changes. Dividing $f^{\prime }$ by that product removes the corner singularities and leaves a holomorphic nonzero function with constant boundary argument. Schwarz reflection and the regular behavior at infinity make it constant. This gives the Schwarz-Christoffel formulaThere are three freely normalizable real prevertices, because an automorphism of the upper half-plane has three real parameters. After those are fixed, the remaining $n-3$ prevertex parameters are determined by the polygon's shape through the side-length conditions. They are not further independent normalization choices.

If the last prevertex is at infinity, absorb its constant scale into $C$ and omit its finite factor:For a bounded polygon, ${\sum }_j{\beta }_j=2$, so this derivative behaves like $Z^{{\beta }_n-2}$, consistent with the corner described in the local coordinate $1/Z$.