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Let $f$ be a conformal map from rectangle $R$ to unit disk. Prove that $f$ analytic at all vertices of $R$ (say $z_k, k=1,2,3,4$), and near $z_k$, $$f(z)=f(z_k)+(z-z_k)^2 h(z)$$ where $h(z)$ is analytic at $z_k$ and non zero.

My attempt: I tried to use Schwarz-Christoffel formula to get a map from rectangle to upper half plane and use Möbius to unit disk, but the first map can't guarantee the analytic of vertices. Thanks for any help.

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You don't need the explicit form of $f$. The Schwarz reflection principle is enough. Let's say $R=(0,a)\times (0,b)$ and we study the behavior at $0$. Consider the reflected rectangles $R_1=(-a,0)\times (0,b)$, $R_2=(-a,0)\times (-b,0)$ and $R_3 =(0,a)\times (-b,0)$.

Let $\gamma_1=\partial R \cap \partial R_1$, the vertical line segment from $0$ to $ib$. Its image under $f$ is an arc of the unit circle, call it $\Gamma_1$. By reflection across $\gamma_1$, $f$ extends to a holomorphic map from $R\cup R_1\cup \gamma_1$ to the domain obtained by taking the union of the unit disk, its exterior, and the arc $\Gamma_1$.

Reflecting two more times, we extend $f$ to a holomorphic function $F$ in $(-a,a)\times (-b,b)$. This shows the analyticity at $0$. Also, keeping track of what the reflection was, you'll see that $F(-z)=F(z)$. Indeed, going from $z$ to $-z$ means reflecting in two sides of a rectangle, which in the image plane involves going $w \mapsto 1/\bar w \mapsto 1/\overline{(1/\bar w)} = w$.

Since $F$ is even, its expansion at $0$ has the form $F(z) = \sum_{k=0}^\infty c_{2k} z^{2k} $ which meets the stated requirement (and more).

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