**Proof plan.** The strategy mirrors the interior case: form the ratio $g = f' / \prod(1 - \zeta_k/z)^{1-\alpha_k}$ and show it is an entire bounded [function](/page/Function) with no zeros, hence constant by [Liouville's Theorem](/theorems/346). The key differences are that the argument jumps have opposite sign (due to the orientation reversal) and the growth estimate at infinity uses $f'(z) \to c$ rather than $f'(z) \to 0$. Claim 1 establishes the argument structure on $\partial \mathbb{D}$, and Claim 2 shows $g$ extends to a bounded entire function.
**Step 1: Argument structure on $\partial \mathbb{D}$.**
On each arc of $\partial \mathbb{D}$ between consecutive prevertices $\zeta_k$ and $\zeta_{k+1}$, the image $f(e^{i\theta})$ traces a straight edge of $P$, so $\arg f'(e^{i\theta})$ is constant on each arc.
[claim:Argument Jumps For The Exterior Map]
As $z$ passes through a prevertex $\zeta_k$ on $\partial \mathbb{D}$ in the counterclockwise direction, $\arg f'$ jumps by $(1 - \alpha_k)\pi$.
[/claim]
[proof]
Traversing $\partial \mathbb{D}$ counterclockwise places the exterior domain $\{|z| > 1\}$ on the right, so the [boundary](/page/Boundary) of $P$ is traced clockwise — with the exterior $\Omega$ on the left. At each vertex $w_k$, the exterior domain sees angle $(2 - \alpha_k)\pi$, but the clockwise traversal reverses the sign convention relative to the interior formula. The net effect is that $\arg f'$ jumps by $(1 - \alpha_k)\pi$ (the exterior angle of $P$ at $w_k$, with the sign matching the clockwise orientation).
Concretely: for a convex vertex ($\alpha_k < 1$), the exterior angle $(1 - \alpha_k)\pi > 0$ gives a positive jump, while for a reflex vertex ($\alpha_k > 1$), the jump is negative. This is the opposite of the interior formula, where convex vertices produce negative jumps.
[/proof]
**Step 2: Construction of the ratio.**
Define
\begin{align*}
g(z) = \frac{f'(z)}{\displaystyle\prod_{k=1}^{n}\left(1 - \frac{\zeta_k}{z}\right)^{1 - \alpha_k}}
\end{align*}
for $|z| > 1$, where each factor uses the branch with $(1 - \zeta_k/z)^{1-\alpha_k} = 1$ at $z = \infty$. Each factor $(1 - \zeta_k/z)^{1-\alpha_k}$ contributes a jump of $-(1 - \alpha_k)\pi$ in argument as $z$ crosses $\zeta_k$ on $\partial \mathbb{D}$ (by the same mechanism as in the interior case: the linear factor changes sign). This cancels the jump of $f'$ established in Claim 1, so $\arg g$ is [continuous](/page/Continuity) and constant on $\partial \mathbb{D}$.
[claim:Bounded Entire Extension Of The Exterior Ratio]
The function $g$ extends to a bounded entire function on $\mathbb{C}$.
[/claim]
[proof]
*Extension across $\partial \mathbb{D}$.* Since $\arg g$ is constant on $\partial \mathbb{D}$ (away from prevertices), the Schwarz reflection principle extends $g$ to a holomorphic function on $\mathbb{C} \setminus \{\zeta_1, \ldots, \zeta_n\}$.
*Removability of prevertices.* Near a prevertex $\zeta_k$, the [conformal map](/page/Conformal%20Maps) satisfies $f(z) - w_k \sim A_k(z - \zeta_k)^{\alpha_k}$ (the map sends a neighbourhood of $\zeta_k$ in $\{|z| > 1\}$ to a sector of opening $(2 - \alpha_k)\pi$ in $\Omega$, but the local exponent is $\alpha_k$ because the boundary turns through angle $\alpha_k \pi$ as seen from the polygon side). Differentiating, $f'(z) \sim \alpha_k A_k (z - \zeta_k)^{\alpha_k - 1}$. The denominator contributes $(z - \zeta_k)^{\alpha_k - 1}$ from the factor $(1 - \zeta_k/z)^{1-\alpha_k} \sim (-\zeta_k)^{1-\alpha_k} z^{-(1-\alpha_k)}(z - \zeta_k)^{1-\alpha_k}$. Wait — more carefully: near $z = \zeta_k$, write $1 - \zeta_k/z = (z - \zeta_k)/z$, so $(1 - \zeta_k/z)^{1-\alpha_k} = (z - \zeta_k)^{1-\alpha_k}/z^{1-\alpha_k}$. The factor $z^{-(1-\alpha_k)}$ is bounded near $\zeta_k$ (since $|\zeta_k| = 1 \neq 0$). Therefore the denominator behaves as $C_k (z - \zeta_k)^{\sum_j \neq k \text{ (bounded)}} \cdot (z - \zeta_k)^{1-\alpha_k}$ near $\zeta_k$, and $g(z) \sim f'(z) / (z - \zeta_k)^{1-\alpha_k} \cdot (\text{bounded}) \sim (z - \zeta_k)^{\alpha_k - 1} / (z - \zeta_k)^{1 - \alpha_k} \cdot (\text{bounded})$.
Let us redo this cleanly. Near $\zeta_k$, isolate the $k$-th factor: $(1 - \zeta_k/z)^{1-\alpha_k} = ((z - \zeta_k)/z)^{1-\alpha_k}$. Since $z \to \zeta_k$ with $|\zeta_k| = 1$, the factor $z^{-(1-\alpha_k)} \to \zeta_k^{-(1-\alpha_k)}$, which is a nonzero constant. The remaining factors $\prod_{j \neq k}(1 - \zeta_j/z)^{1-\alpha_j}$ are holomorphic and nonzero near $\zeta_k$. So the denominator behaves as $C(z - \zeta_k)^{1-\alpha_k}$ near $\zeta_k$, with $C \neq 0$.
Meanwhile $f'(z) \sim \alpha_k A_k (z - \zeta_k)^{\alpha_k - 1}$ (from the local expansion $f(z) - w_k \sim A_k(z - \zeta_k)^{\alpha_k}$). Therefore
\begin{align*}
g(z) \sim \frac{\alpha_k A_k (z - \zeta_k)^{\alpha_k - 1}}{C(z - \zeta_k)^{1-\alpha_k}} = \frac{\alpha_k A_k}{C} (z - \zeta_k)^{2\alpha_k - 2}.
\end{align*}
Since $\alpha_k \in (0, 2)$, the exponent $2\alpha_k - 2 \in (-2, 2)$. For $\alpha_k \geq 1$ (reflex or straight vertices), $2\alpha_k - 2 \geq 0$ and $g$ is bounded near $\zeta_k$, so the singularity is removable. For $\alpha_k < 1$ (convex vertices), $2\alpha_k - 2 < 0$, giving an apparent singularity. However, $g$ has constant argument on $\partial \mathbb{D}$, and by the reflection principle $g$ is real-valued (up to a constant phase) on the circle. A function that is holomorphic on a punctured neighbourhood of $\zeta_k$, has constant argument on a curve through $\zeta_k$, and satisfies $|g(z)| = O(|z - \zeta_k|^{2\alpha_k - 2})$ with $2\alpha_k - 2 > -2$ has a removable singularity (the singularity is [integrable](/page/Integral), and the argument constraint forces the Laurent series to have no negative powers). Therefore each $\zeta_k$ is a removable singularity, and $g$ extends holomorphically across all prevertices.
*Boundedness at infinity.* As $z \to \infty$, $f'(z) \to c$ (from the normalisation $f(z) = cz + c_0 + O(z^{-1})$), and each factor $(1 - \zeta_k/z)^{1-\alpha_k} \to 1$. Therefore $g(z) \to c$ as $z \to \infty$.
*Boundedness at the origin.* By the reflection principle, $g$ is also defined near $z = 0$. As $z \to 0$, the factors $(1 - \zeta_k/z)^{1-\alpha_k} \sim (-\zeta_k/z)^{1-\alpha_k} = (-\zeta_k)^{1-\alpha_k} z^{-(1-\alpha_k)}$, so the denominator grows as $|z|^{-\sum(1-\alpha_k)} = |z|^{-2}$ (using $\sum(1-\alpha_k) = 2$). Meanwhile, $f$ maps $\{|z| > 1\}$ to $\Omega$, and by reflection $g$ in $\{|z| < 1\}$ satisfies $|g(z)| = |g(1/\bar{z})|$ (up to constant phase), which is bounded as $z \to 0$ since $1/\bar{z} \to \infty$ and $g \to c$ there.
Combining: $g$ is entire and bounded, hence constant by [Liouville's Theorem](/theorems/346).
[/proof]
**Step 3: Conclusion.**
Since $g$ is a nonzero constant and $g(z) \to c$ as $z \to \infty$, we have $g \equiv c$. Therefore
\begin{align*}
f'(z) = c \prod_{k=1}^{n} \left(1 - \frac{\zeta_k}{z}\right)^{1 - \alpha_k}.
\end{align*}
