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According to the Poincaré-Koebe theorem, it is known that the unit disk $\mathbb D$ and the complex plane $\mathbb C$ aren't conformally equivalent.

My question is maybe naive, but I was wondering if this statement is still true in the quasiconformal sense. More precisely, does there exist a quasiconformal map from $\mathbb D$ onto $\mathbb C$ ?

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No, $\mathbb{D}$ and $\mathbb{C}$ are not quasiconformally equivalent. If $f:\mathbb{D} \to \mathbb{C}$ was a quasiconformal map, and if $A$ is the annulus $1/2 < |z|<1$, then $f(A)$ contains some infinite annulus $R<|z|<\infty$, so the conformal modulus of $A$ is finite, but the conformal modulus of $f(A)$ is infinite. However, quasiconformal maps can only change conformal modulus by a bounded factor, so such an $f$ can not exist.

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