**Proof plan.** The strategy is to show that the [derivative](/page/Derivative) $f'$ must be a constant multiple of the product $\prod (z - x_k)^{\alpha_k - 1}$. We do this by forming the ratio $g = f' / \prod (z - x_k)^{\alpha_k - 1}$ and proving that $g$ is an entire bounded [function](/page/Function), hence constant by [Liouville's Theorem](/theorems/346). The key idea is that $\arg f'$ is piecewise constant on $\mathbb{R}$ with prescribed jumps at the prevertices, and each factor $(z - x_k)^{\alpha_k - 1}$ contributes exactly the matching jump in argument, so the ratio has continuous (and in fact constant) argument on $\mathbb{R}$. The proof uses three claims: Claim 1 identifies the argument jumps of $f'$; Claim 2 shows the ratio $g$ has constant argument on $\mathbb{R}$; and Claim 3 establishes that $g$ extends to a bounded entire function. We treat the case where all prevertices are finite. If $x_m = \infty$, the argument is identical with the factor $(z - x_m)^{\alpha_m - 1}$ removed and a compensating adjustment to the growth estimate at infinity. **Step 1: Piecewise constancy of $\arg f'$ on $\mathbb{R}$.** On each open interval $(x_k, x_{k+1}) \subset \mathbb{R}$ between consecutive prevertices, the image $f(t)$ traces a straight edge of $P$ (since $f$ maps each such interval homeomorphically onto the edge from $w_k$ to $w_{k+1}$). Along a straight edge, the tangent direction is constant, so $f'(t)$ points in a fixed direction for all $t \in (x_k, x_{k+1})$. [claim:Argument Jump Of The Derivative] For each $k \in \{1, \ldots, n\}$, the one-sided [limits](/page/Limit) of $\arg f'(t)$ at the prevertex $x_k$ satisfy \begin{align*} \arg f'(x_k^+) - \arg f'(x_k^-) = (\alpha_k - 1)\pi. \end{align*} [/claim] [proof] As $t$ increases through $x_k$, the image $f(t)$ transitions from the edge $w_{k-1}w_k$ to the edge $w_k w_{k+1}$. The direction of the tangent vector rotates by the exterior angle at $w_k$. The exterior angle at a vertex with interior angle $\alpha_k \pi$ is $(1 - \alpha_k)\pi$, measured as the supplement of the interior angle. The signed change in $\arg f'$, with the counterclockwise (interior-on-left) convention, is $-(1 - \alpha_k)\pi = (\alpha_k - 1)\pi$. Concretely, for a convex vertex ($\alpha_k < 1$), the image turns clockwise and $\arg f'$ decreases; for a reflex vertex ($\alpha_k > 1$), the image turns counterclockwise and $\arg f'$ increases. [/proof] **Step 2: Construction of the ratio $g$.** Define \begin{align*} g : \mathbb{H} \to \mathbb{C}, \qquad g(z) = \frac{f'(z)}{\displaystyle\prod_{k=1}^{n}(z - x_k)^{\alpha_k - 1}}, \end{align*} where each factor uses the branch with $\arg(z - x_k) \in (0, \pi)$ for $z \in \mathbb{H}$, extending [continuously](/page/Continuity) to $\mathbb{R} \setminus \{x_1, \ldots, x_n\}$ by taking $\arg(t - x_k) = 0$ for $t > x_k$ and $\arg(t - x_k) = \pi$ for $t < x_k$. Since $f$ is conformal on $\mathbb{H}$, we have $f'(z) \neq 0$ for all $z \in \mathbb{H}$, and the denominator is nonzero on $\mathbb{H}$ (as $x_k \in \mathbb{R}$). Therefore $g$ is holomorphic and nonvanishing on $\mathbb{H}$. [claim:Constant Argument On The Real Axis] The function $g$ extends continuously to $\mathbb{R} \setminus \{x_1, \ldots, x_n\}$, and $\arg g(t)$ is constant for all $t \in \mathbb{R} \setminus \{x_1, \ldots, x_n\}$. [/claim] [proof] On each interval $(x_k, x_{k+1})$, $\arg f'(t)$ is constant (Step 1). It remains to show that the denominator contributes matching constant arguments on each interval, with jumps that cancel those of $f'$. For $t \in (x_k, x_{k+1})$, the argument of the denominator is \begin{align*} \sum_{j=1}^n (\alpha_j - 1) \arg(t - x_j) = \sum_{j=1}^{k} (\alpha_j - 1)\cdot 0 + \sum_{j=k+1}^{n}(\alpha_j - 1) \cdot \pi = \pi \sum_{j=k+1}^{n}(\alpha_j - 1), \end{align*} since $t > x_j$ for $j \leq k$ (giving $\arg(t - x_j) = 0$) and $t < x_j$ for $j \geq k + 1$ (giving $\arg(t - x_j) = \pi$). This quantity depends on $k$ but not on $t$, so the argument of the denominator is constant on each interval. As $t$ crosses $x_k$ from left to right, the factor $(t - x_k)^{\alpha_k - 1}$ transitions from $\arg = (\alpha_k - 1)\pi$ (when $t < x_k$) to $\arg = 0$ (when $t > x_k$). The jump in argument of the denominator is therefore $-(\alpha_k - 1)\pi$, which cancels the jump $(\alpha_k - 1)\pi$ in $\arg f'$ established in Claim 1. Since both jumps cancel at every prevertex, $\arg g(t)$ has the same value on every interval, i.e., it is globally constant on $\mathbb{R} \setminus \{x_1, \ldots, x_n\}$. [/proof] **Step 3: Extension to an entire bounded function.** [claim:Bounded Entire Extension] The function $g$ extends to a bounded entire function on $\mathbb{C}$. [/claim] [proof] *Extension across the real axis.* Since $\arg g(t) = \theta_0$ for some constant $\theta_0$ on $\mathbb{R} \setminus \{x_1, \ldots, x_n\}$, the function $e^{-i\theta_0} g$ is real and positive on $\mathbb{R}$ away from the prevertices. By the Schwarz reflection principle, $g$ extends to a holomorphic function on $\mathbb{C} \setminus \{x_1, \ldots, x_n\}$ via $g(\bar{z}) = e^{2i\theta_0}\overline{g(z)}$. *Removability of the prevertices.* Near a prevertex $x_k$, the [conformal map](/page/Conformal%20Maps) $f$ satisfies $f(z) - w_k \sim A_k (z - x_k)^{\alpha_k}$ for some $A_k \neq 0$ (since $f$ maps a half-plane neighbourhood of $x_k$ to a sector of opening angle $\alpha_k \pi$). Differentiating, $f'(z) \sim \alpha_k A_k (z - x_k)^{\alpha_k - 1}$. The denominator behaves as $(z - x_k)^{\alpha_k - 1} \cdot h_k(z)$ where $h_k$ is continuous and nonzero near $x_k$. Therefore $g(z) \to \alpha_k A_k / h_k(x_k) \neq 0$ as $z \to x_k$, so each singularity is removable. After removing all singularities, $g$ is entire. *Boundedness at infinity.* The angle-sum identity gives $\sum_{k=1}^n (\alpha_k - 1) = -2$. For $|z| \to \infty$, each factor $(z - x_k)^{\alpha_k - 1} \sim z^{\alpha_k - 1}$, so the denominator satisfies \begin{align*} \prod_{k=1}^n (z - x_k)^{\alpha_k - 1} \sim z^{\sum(\alpha_k - 1)} = z^{-2}. \end{align*} Meanwhile, $f'(z) = O(|z|^{-2})$ as $|z| \to \infty$ in $\mathbb{H}$, because $f$ maps a neighbourhood of $\infty$ in $\overline{\mathbb{H}}$ to a bounded portion of $\partial P$ (the image of the interval $(x_n, \infty) \cup \{-\infty, \infty\} \cup (-\infty, x_1)$ is a finite edge of $P$), so $f(z) \to w_\infty$ and the decay $f'(z) = O(|z|^{-2})$ follows from the Schwarz–Pick lemma or direct estimation. Therefore $g(z) = f'(z) / \prod (z - x_k)^{\alpha_k - 1}$ remains bounded as $|z| \to \infty$. Combining: $g$ is entire and bounded. [/proof] **Step 4: Conclusion via Liouville's theorem.** By [Liouville's Theorem](/theorems/346), $g$ is a constant $c_1 \in \mathbb{C} \setminus \{0\}$ (nonzero because $f$ is conformal). Therefore \begin{align*} f'(z) = c_1 \prod_{k=1}^{n} (z - x_k)^{\alpha_k - 1}. \end{align*} Integrating from the base point $z_0$ to $z$ along any path in $\mathbb{H}$ — the [integral](/page/Integral) is path-independent by [Cauchy's Theorem for Simply Connected Domains](/theorems/344) since the integrand is holomorphic on $\mathbb{H}$ — gives \begin{align*} f(z) = c_1 \int_{z_0}^z \prod_{k=1}^{n} (\zeta - x_k)^{\alpha_k - 1} \, d\zeta + f(z_0). \end{align*} Setting $c_2 = f(z_0)$ completes the proof.