I'm reading a proof of Arnold's theorem about analytic linearization of analytic circle diffeomorphisms. The following result is used in the proof and I don't see why it should be true.


Let $S\_\sigma= \{ z:|\Im z|<\sigma \}$. We have a function $h$ that is holomorphic on $S\_{\sigma-\delta}$, and is real-valued on the real-axis. We are also give that for $z \in S\_{\sigma-\delta}$ we have $|h(z)|<\delta$, and for $z \in S\_{\sigma-2\delta}$ we have that $|h'(z)|<1$. Let $H$ be defined by $H(z)=z+h(z)$, so $H$ sends any point $z \in S\_{\sigma-\delta}$ less than $\delta$ far away from itself.


I want to show that $H(S\_{\sigma-2\delta})$ contains $S\_{\sigma-3\delta}$. We have that $h$ sends the real axis to itself, so $H$ does also. As well, because $h(z)<\delta$ for $z \in S\_{\sigma-\delta}$, for $\Im z=\sigma-2\delta$ we have $\Im H(z) > \sigma-3\delta$, and 
for $\Im z = -(\sigma-2\delta)$ we have $\Im H(z)< -(\sigma-3\delta)$. The text seems to conclude from this that $H(S\_{\sigma-2\delta})$ contains $S\_{\sigma-3\delta}$. This would follow from the above facts if we knew that $H(S\_{\sigma-2\delta})$ had to be simply connected, but why should this be the case? (I hope I said enough to make it clear why this WOULD follow if we knew that the image were simply connected.)



Holomorphic image of a strip in complex plane
I think the easiest argument is using Rouché's theorem, which is itself proved using the argument principle. Let $a \in S\_{\sigma-3\delta}$, let $K \subset S\_{\sigma-2\delta}$ be the circle about $a$ of radius $\delta$, and let $f(z)=z-a$. Then for $z \in K$ we have $|h(z)|<\delta=|f(z)|$, so by Rouché's theorem $f$ and $f+h$ have the same number of zeros inside $K$. Of course $f$ has one zero inside $K$ so too $f+h=H-a$ has one zero inside $K$. Thus $a \in H(S\_{\sigma-2\delta})$, and we conclude $S\_{\sigma-3\delta} \subseteq H(S\_{\sigma-2\delta})$.


