Holomorphic image of a strip in complex plane 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.)
 A: I assume that you are assuming $\sigma > 3\delta$, otherwise, the problem doesn't make sense.  
Can't you just apply the Argument Principle?  Fix a $z_0\in S_{\sigma-3\delta}$ and an $R>>0$ so big that the box $B_1$ given by $|\Re(z)|\le R$ and $|\Im(z)|\le \sigma-2\delta$ contains $z_0$  well inside in the interior.  Now note that, by the properties you've already derived, if $R$ is big enough, it is easily seen that $H$ carries the boundary of $B_1$ into a curve that winds once around (and does not touch) a box $B_2$ given by $|\Re(z)|\le R-2\delta$ and $|\Im(z)|\le \sigma-3\delta$.  Now, by the Argument Principle, the number of times that $H$ assumes the value $z_0$ in $B_1$ is equal to
$$
\frac{1}{2\pi i}\int_{\partial B_1} \frac{H'(z)\ dz}{H(z) - z_0} = 1
$$
Thus, $z_0$ is in $H(S_{\sigma-2\delta})$.
Alternative Argument:  There's an even easier proof if you use fewer hypotheses:  For notational comparison, let me set $\tau = \sigma-\delta$, and suppose that $0<2\delta < \tau$.  
Suppose that we have a holomorphic function $h$ on $S_\tau$ that satisfies $|h(z)|<\delta$ on $S_{\tau-\delta}$, and set $H(z) = z + h(z)$.  Then $S_{\tau-2\delta}\subset H(S_{\tau-\delta})$.  
To show this, suppose that $a\in S_{\tau-2\delta}$ is given, and consider the circle $C_\delta(a)$ consisting of those $z$ satisfying $|z-a| = \delta$.  Then $C_\delta(a)\subset S_{\tau-\delta}$ and so, for $z\in C_\delta(a)$, we have 
$$
\bigl|H(z) - z\bigr| = |h(z)| < \delta.
$$
In particular, the closed curve $\gamma_1(t) = H(a+\delta e^{2\pi i t})$ ($0\le t\le 1$) is homotopic to the closed curve $\gamma_0(t) = a+\delta e^{2\pi i t}$ in the plane minus the point $a$.  Thus, they have the same winding number about $a$, which, by inspection of $\gamma_0$, is equal to $1$.  By the Argument Principle applied to $\gamma_1$, it follows that $H$ must assume the value $a$ somewhere in the disk whose boundary is $C_\delta(a)$.  Thus, $a$ lies in $H(S_{\tau-\delta})$.  Since $a$ was arbitrary in $S_{\tau-2\delta}$, it follows that $S_{\tau-2\delta}\subset H(S_{\tau-\delta})$, as desired.
A: 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 $S_{\sigma-3\delta} \subseteq H(S_{\sigma-2\delta})$.
