Past exam of the perturbation methods and applied complex variables course of the University of Oxford / 2010 / B / 4

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 $1/Z$.

Use the branch of $\sqrt{1-Z^2}$ that is positive for $Z=iY$, $Y>0$. The right-half-plane slit mapis holomorphic in the upper half-plane, with derivative $Z(1-Z^2{)}^{-3/2}$, the required Schwarz-Christoffel formula form. Its derivative has no interior zero.

To verify the image and global injectivity, factor the construction. Squaring maps the upper half-plane bijectively onto $\mathbb{C}\setminus [0,\infty )$. Thus $1-Z^2$ ranges over $\mathbb{C}\setminus (-\infty ,1]$. The positive-real-part square root maps this onto the right half-plane with the interval $(0,1]$ removed; reciprocation maps it onto the right half-plane slit along $[1,\infty )$. This also identifies the slit tip as a boundary point; the strict inequality in the printed slit description must be understood with that boundary convention.

Approaching the real axis from above givesIn particular $Z=-1$ maps to the lower infinite prime end $A$, $Z=0$ to the tip $B=1$, and $Z=1$ to the upper infinite prime end $C$. The real-axis point at infinity maps to zero on the vertical wall.

Let $g=f^{-1}$, so $g(z)=\sqrt{1-z^{-2}}$, choosing the analytic branch with $\mathrm{Im}g>0$; for example $g(1/2)=i\sqrt{3}$. Let $\zeta \in \Omega$ be the source point. For the convention $-\Delta G={\delta }_{\zeta }$, the Dirichlet Green function in the upper half-plane is obtained by an opposite-sign reflected source:It vanishes for real $Z$, has the logarithmic singularity $-(2\pi {)}^{-1}\log |Z-W|$, and is harmonic elsewhere. Conformal invariance of two-dimensional harmonic functions now givesThe branch specified above maps every boundary bank to the real axis, so the Green function vanishes on both slit banks and the vertical wall. Local conformality preserves the coefficient of the logarithmic singularity. If the convention is instead $\Delta G={\delta }_{\zeta }$, negate this expression.