Let $f(z)$ be analytic on the unit disc, and suppose that there is an annulus $U = \{z ∈ C \mid r < |z| < 1\}$ such that $f(z)$ restricted to the annulus $U$ is injective. Show that $f$ is injective on the unit disc.
I tried to show that $g(z) = f(z) - f(z_0)$ has only one zero in the unit disk, but how can I use the hypothesis of the annulus?
Let $w \notin f(r\partial\mathbb{D})$, the number of zeros of the function $f(z)-w$ in $r\mathbb{D}$ are equal to $$|\frac{1}{2\pi i}\int_{r\partial\mathbb{D}}\frac{f'(j)}{f(j)-w}dj|=|\frac{1}{2\pi i}\int_{f(r\partial\mathbb{D})}\frac{1}{z-w}dz|=|ind_{f(r\partial\mathbb{D})}(w)|. $$So $f(z)-w$ has $0$ or $1$ zeros in $r\mathbb{D}$ ,since $f(r\partial\mathbb{D})$ is a Jordan curve.\\ Since $w,r$ was arbitrary the proof follows.
This is not an answer, but it might be helpful. If not, I will trash it later.
Note that $$\phi:z\mapsto \frac{1}{|z|}\Big(1-(1-r)|z|\Big) \cdot z,\quad 0\mapsto 0$$ maps any point of the open unit disk $D$ into the anulus $A$ - geometrically, it's just a circular inversion mapping $|z|\in(0,1)$ to $|\phi(z)|\in(r,1)$. More importantly, it's a bijection.
Fix $z_0\not=0$ and let
$g(z)=f(\phi(z))-f(\phi(z_0))$
Then $$g(z)=0\iff f(\phi(z))=f(\phi(z_0))$$ but $\phi(D-\{0\})=A$, so if $z\not=0$ then $\phi(z)=\phi(z_0)$ and so $z=z_0$.
We just need to see what happens if $z=0$. $f$ is analytic so $$f(z)=f(0)+\sum_{n=1}^\infty a_n z^n$$ Now $f(0)=f(\phi(z_0))$ if and only if $$f(0)=f(0)+\sum_{n=1}^\infty a_n \phi(z_0)^n$$ that is $\phi(z_0)=0$, and so $z_0=0$.
Therefore, $g(z)=0$ if and only if $z=z_0$ on the whole open unit disk.
Can you use this info somehow?
