**Proof plan.** The proof reduces to the half-plane case via a Möbius transformation. We compose the [conformal map](/page/Conformal%20Maps) $f : \mathbb{D} \to P$ with a biholomorphism $T : \mathbb{H} \to \mathbb{D}$ and apply the [Schwarz–Christoffel Half-Plane Formula](/theorems/683) to the composition $f \circ T$. Computing $f'$ via the chain rule and expressing the prevertices in terms of their images on $\partial \mathbb{D}$ yields the disc formula. Claim 1 carries out the change of variables; Claim 2 identifies the extra factor from the Möbius Jacobian. **Step 1: Reduction to the half-plane formula.** Let $T : \mathbb{H} \to \mathbb{D}$ be any Möbius transformation mapping the upper half-plane biholomorphically onto the disc, for instance \begin{align*} T(z) = \frac{z - i}{z + i}. \end{align*} The composition $F = f \circ T : \mathbb{H} \to P$ is a conformal bijection from $\mathbb{H}$ onto $P$. The prevertices of $F$ on $\mathbb{R} \cup \{\infty\}$ are $s_k = T^{-1}(\zeta_k)$ for $k = 1, \ldots, n$. By the [Schwarz–Christoffel Half-Plane Formula](/theorems/683), \begin{align*} F(z) = A \int_{z_0}^z \prod_{k=1}^{n} (u - s_k)^{\alpha_k - 1} \, du + B \end{align*} for constants $A \in \mathbb{C} \setminus \{0\}$ and $B \in \mathbb{C}$ (absorbing the factor for any $s_k = \infty$ into $A$). **Step 2: Change of variables.** [claim:Change Of Variables To The Disc] Substituting $u = T^{-1}(w)$ in the half-plane [integral](/page/Integral) and expressing the result in terms of the disc prevertices $\zeta_k$ yields \begin{align*} f(z) = c_1 \int_{w_0}^{z} \prod_{k=1}^{n} \left(1 - \frac{w}{\zeta_k}\right)^{\alpha_k - 1} \, dw + c_2 \end{align*} for constants $c_1 \in \mathbb{C} \setminus \{0\}$ and $c_2 \in \mathbb{C}$. [/claim] [proof] The inverse Möbius transformation is $T^{-1}(w) = i(1 + w)/(1 - w)$, with [derivative](/page/Derivative) $(T^{-1})'(w) = 2i/(1 - w)^2$. Substituting $u = T^{-1}(w)$ in the half-plane integral, each factor transforms as \begin{align*} u - s_k = T^{-1}(w) - T^{-1}(\zeta_k) = \frac{-2i(w - \zeta_k)}{(1 - w)(1 - \zeta_k)}. \end{align*} Raising to the power $\alpha_k - 1$ and taking the product: \begin{align*} \prod_{k=1}^n (u - s_k)^{\alpha_k - 1} = \prod_{k=1}^n \left[\frac{-2i}{(1-w)(1-\zeta_k)}\right]^{\alpha_k - 1} \prod_{k=1}^n (w - \zeta_k)^{\alpha_k - 1}. \end{align*} The first product contributes a factor of $(1 - w)^{\sum_k (1 - \alpha_k)} = (1 - w)^2$ (using the angle-sum identity $\sum \alpha_k = n - 2$, so $\sum(1 - \alpha_k) = 2$), multiplied by constants depending on the $\zeta_k$. Combined with the Jacobian $(T^{-1})'(w) = 2i/(1-w)^2$, the $(1-w)^2$ factor cancels: \begin{align*} \prod_{k=1}^n (u - s_k)^{\alpha_k - 1} \cdot (T^{-1})'(w) = C \prod_{k=1}^n (w - \zeta_k)^{\alpha_k - 1}, \end{align*} where $C$ is a nonzero constant absorbing all factors of $-2i$, $(1-\zeta_k)^{1-\alpha_k}$, and $2i$. Factoring out $(-\zeta_k)^{\alpha_k - 1}$ from each term $(w - \zeta_k)^{\alpha_k - 1}$ and absorbing into the constant, we may rewrite the product as $\prod_k (1 - w/\zeta_k)^{\alpha_k - 1}$ up to a multiplicative constant. Setting $c_1 = A \cdot C \cdot \prod_k (-\zeta_k)^{\alpha_k - 1}$ and $c_2 = B$ gives the stated formula. [/proof] **Step 3: Verification of the normalisation.** The resulting formula inherits all properties from the half-plane version: convergence, holomorphicity, local injectivity ($f'(z) = c_1 \prod_k (1 - z/\zeta_k)^{\alpha_k - 1} \neq 0$ for $z \in \mathbb{D}$ since $\zeta_k \in \partial \mathbb{D}$), [boundary](/page/Boundary) correspondence, and global injectivity, by the [Well-Posedness of the Schwarz–Christoffel Map](/theorems/684) applied through the Möbius change of variables.