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?

  • 1
    $\begingroup$ 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 $n = 1$. $\endgroup$ – froggie Nov 27 '12 at 14:15

Suppose $f$ is not injective in $D(0,r)$. Let $z_0,z_1\in D(0,r)$ distinct such that $f(z_0)=f(z_1)=w.$

Let $1>\rho>r$ and denote $\partial D(0,\rho)$ by $\gamma.$ Define $g(z)=f(z)-w$ and notice that $g$ has at least two zeros in $D(0,\rho)$. The winding number $$\begin{align}n(g(\gamma),0)=&\frac{1}{2\pi i}\int_{g(\gamma)}\frac{1}{z}\,dz\\ =&\frac{1}{2\pi i}\int_0^{2\pi}\frac{g'(\gamma(t))}{g(\gamma(t))}\gamma'(t)\,dt\\ =&\frac{1}{2\pi i}\int_{\gamma}\frac{g'(z)}{g(z)}\,dz, \end{align}$$ which by the Argument principle yields that $$\begin{align}n(g(\gamma),0)=&\sum_{a\in D(0,\rho)}\operatorname{ord}_g(a)\cdot n(\gamma,a)\\\geq&\,2. \end{align} $$ That is, $g(\gamma)$ self intersects. But since $\text{Im}(\gamma)\subset U$, and $f$ is injective in $U$, we reach a contradiction.


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