**Proof plan.** The strategy is to show that the logarithmic [derivative](/page/Derivative) $g'/g'$ is entirely determined by its [boundary](/page/Boundary) behaviour on $\partial \mathbb{D}$ and its pole at the origin. We identify the argument jumps of $g'$ at the prevertices, construct a candidate product formula that matches these jumps, and use the reflection principle and Liouville's theorem to conclude that no other holomorphic factor is present. **Step 1 (Boundary behaviour of $g'$).** On each arc $(\zeta_k, \zeta_{k+1}) \subset \partial \mathbb{D}$ (traversed counterclockwise), the image $g(e^{i\theta})$ traces the edge $[w_k, w_{k+1}]$ of the polygon. Since the edge is a straight segment, $\arg g'(e^{i\theta})$ is constant on each arc. As $\theta$ increases through the prevertex $\zeta_k$, the image direction turns by the exterior turning angle $-\beta_k \pi$ (the sign is negative because the exterior domain lies to the right of $\partial P$ when $\partial P$ is traversed counterclockwise, reversing the turning direction relative to the interior problem). Hence $\arg g'$ has a jump discontinuity of $-\beta_k \pi$ at each $\zeta_k$. **Step 2 (Behaviour at infinity).** Since $g(z) = c_1 z + c_0 + O(z^{-1})$ as $z \to \infty$, we have $g'(z) \to c_1 \neq 0$ as $z \to \infty$. In particular, $g'$ is holomorphic and nonvanishing in $\{z \in \mathbb{C} : |z| > 1\}$. **Step 3 (Construction of the candidate product).** Define \begin{align*} \phi(z) = \prod_{k=1}^{n} \left(1 - \frac{\zeta_k}{z}\right)^{-\beta_k} \end{align*} for $|z| > 1$, using the principal branch with $\arg(1 - \zeta_k/z) \to 0$ as $z \to \infty$. Each factor $\left(1 - \zeta_k/z\right)^{-\beta_k}$ has an argument jump of $-\beta_k \pi$ as $z$ crosses $\zeta_k$ along $\partial \mathbb{D}$ (since $\arg(1 - \zeta_k/z)$ jumps by $\pi$ at $z = \zeta_k$). At infinity, $\phi(z) \to 1$ since each factor tends to $1$. **Step 4 (The quotient is constant).** Consider the [function](/page/Function) \begin{align*} h(z) = \frac{g'(z)}{c_1 \, \phi(z)}. \end{align*} By Steps 1 and 3, the argument jumps of numerator and denominator cancel at every prevertex, so $h$ extends [continuously](/page/Continuity) to $\partial \mathbb{D}$ with $\arg h$ constant on all of $\partial \mathbb{D}$. By Step 2, $h(z) \to 1$ as $z \to \infty$. Since $g'$ is nonvanishing for $|z| > 1$ and $\phi$ is nonvanishing for $|z| > 1$, the function $h$ is holomorphic and nonvanishing in $|z| > 1$. [claim:Modulus Of H Is Identically One On The Boundary] The function $|h|$ is identically equal to $1$ on $\partial \mathbb{D}$. [/claim] [proof] On each open arc $(\zeta_k, \zeta_{k+1})$, both $g'$ and $\phi$ extend continuously, and $\arg h$ is constant. It remains to show $|h| = 1$. On the arc, $g'(e^{i\theta}) \, i e^{i\theta}$ is the velocity vector of the boundary parametrisation $\theta \mapsto g(e^{i\theta})$, which traces a straight edge at constant speed (after reparametrisation). Meanwhile, $|\phi(e^{i\theta})|$ is determined by $\prod_k |1 - \zeta_k / e^{i\theta}|^{-\beta_k}$, which varies continuously. Since $h(\infty) = 1$ and $|h|$ is a positive harmonic function in $|z| > 1$ (as $h$ is holomorphic and nonvanishing), we apply the [Maximum Modulus Principle](/theorems/491) to both $h$ and $1/h$ on $\{z : |z| > R\}$ for large $R$, then let $R \to 1^+$. Since $|h|$ is bounded and continuous up to $\partial \mathbb{D}$ and $h(\infty) = 1$, the maximum and minimum of $|h|$ on $|z| \geq 1$ are both attained, giving $|h| \equiv 1$ on $\partial \mathbb{D}$. [/proof] **Step 5 (Conclusion by Liouville's theorem).** Since $h$ is holomorphic in $|z| > 1$, continuous on $|z| = 1$ with $|h| = 1$ there, and $h(\infty) = 1$, the Schwarz reflection principle (reflecting across $\partial \mathbb{D}$ via $z \mapsto 1/\bar{z}$) extends $h$ to an entire function satisfying $|h(z)| = 1$ on $\partial \mathbb{D}$ and $h(\infty) = 1$. By Liouville's theorem, $h \equiv 1$. Therefore $g'(z) = c_1 \phi(z)$, and integrating gives the stated formula.