= Solution
{c}
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, $\arg f'$ 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 $f'$ 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 formula>
$$
\boxed{f'(Z)=C\prod_{j=1}^n(Z-X_j)^{-\beta_j}.}
$$
There are \b[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:
$$
\boxed{f'(Z)=C\prod_{j=1}^{n-1}(Z-X_j)^{-\beta_j}.}
$$
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 $1/Z$.
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