Proving derivative tends to zero using 'Bloch's Theorem'.

Question

I have been given the following version of the 'Bloch-Landau Theorem' in lectures:

Let $E=\lbrace z\in \mathbb{C}:|z|\leq 1\rbrace$. If $f:E\rightarrow \mathbb{C}$ is analytic and $f'(0)=1$, then $f(B(0,1))$ contains an open ball of radius $1/32$, where $B(0,1):=\lbrace z\in \mathbb{C}:|z|< 1\rbrace$.

I am trying to prove the following using the Bloch-Landau Theorem as given above:

Let $D=\lbrace z\in \mathbb{C}:|z|> 1\rbrace$ and $h:D\rightarrow B(0,1)$ be analytic. Then $h'(z)\rightarrow 0$ as $z\rightarrow \infty$.

I am rather stumped by how the Bloch-Landau Theorem can be used to prove this, since the domains $E$ and $D$ are complements of each other, and there appears to be no guarantee that h is has an inverse (even locally).

Any hints as to how to proceed would be much appreciated.

Answer 1

I think this is easier than Bloch-Landau.

Hint: Define $g:\{0<|z|<1\} \to \{|z|<1\}$ by setting $g(z)=h(1/z).$ Then $g$ is bounded in $\{0<|z|<1\}.$ Hence $g$ has a removable singularity at $0 \dots$