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This problem is taken from Section VIII.4 of Theodore Gamelin's Complex Analysis:

Let $f(z)$ be an analytic function on the open unit disk $\mathbb{D}=\{|z|<1\}$. Suppose there is an annulus $U = \{r<|z|<1\}$ such that the restriction of $f(z)$ to $U$ is one-to-one. Show that $f(z)$ is one-to-one on $\mathbb{D}$.

Any hints?

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Hint: First, use the argument principle to show that there is an $n\geq 1$ such that $f$ is an $n$-to-$1$ function onto its image. That is, every point $a$ in the image of $f$ has $n$ preimages in $\mathbb{D}$, at least when counted with multiplicity. The point of the problem is to show $n = 1$. Next, show that there is a point $a$ in the image of $f$ such that every preimage of $a$ lies in your annulus (you can do this with open mapping, or equivalently maximum modulus). Finally, use the assumption to conclude $a$ has only 1 preimage, thus $n = 1$. – froggie Nov 27 '12 at 14:15

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