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If $f$ is a conformal self map from an annulus to itself, I know it has to map its boundary to its boundary. Suppose $f$ maps a boundary point to a interior point, then apply maximum value principle on $f^{-1}$, we see that $f^{-1}$ is a constant function.

But how could I show that $f$ maps the outer circle to the outer circle, and similarly for the inner circle.

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It doesn't have to do this, I don't think. Consider the annulus which is given by the circles centered at $0$ with radii $1/2$ and $2$. Then the map $f(z) = z^{-1}$ should switch the circles, and since the region doesn't contain $0$, this map is holomorphic with non-zero derivative, and hence is a conformal self-map of the annulus.

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  • $\begingroup$ Thank you, I guess it is a typo in the book then. $\endgroup$
    – Xiao
    Apr 7, 2017 at 2:25

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