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By the Riemann Mapping Theorem we know every region (open, connected subset of $\mathbb{C}$), that isn't the whole plane is conformally equivalent to the unit disk $\mathbb{D}$. By the Schwarz-Christoffel mapping we can conformally map the upper half plane $\mathbb{H} = \lbrace z : \Im(z) \geq 0 \rbrace$ to a polygon $P$, with interior mapping to interior and boundary mapping to boundary. Now the whole idea of Quasiconformal maps is that you can't conformally map a rectangle onto a square. Let $f_1 : \mathbb{H} \to P_1$ be a Schwarz-Christoffel mapping to a rectangle, $f_2: \mathbb{H} \to P_2$ to a square, then wouldn't $f_2^{-1} \circ f_1: P_2 \to P_1$ be a conformal map from square to rectangle? I realize that the boundary would be where we get problems, but is there some proof that this would always be non-conformal? Thank you.

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    $\begingroup$ Likely you can't map the corners to the corners. $\endgroup$ – Chappers May 8 '18 at 18:19

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