A question about Argument Principle/Rouche's Theorem I am trying to solve my friend's homework assignment, but I got stuck. The question is:
Let f be a meromorphic function on a domain D, let E be a domain such that $\overline {E}\subseteq D$. If $|f(z)|<1$ for all $z \in \partial E$ and if $\partial E$ is a simple closed curve, then the function f(z)-1 has as many roots in E as it has poles in E.
This is the last question in a homework assignment and the previous two questions are standard "Rouche's Theorem" questions(the zeros of a polynomial in a disc/annulus); however, we have a meromorphic function here, so I think Rouche's Theorem is not relevant for that question so maybe we can use Argument Principle, but I have no idea about how to proceed.
Thanks in advance for any help.  
 A: Rouché's theorem, or rather its proof, is relevant.
For $\lambda \in [0,1]$, consider the function $g_\lambda(z) = \lambda f(z) - 1$. Since $\lvert f(z)\rvert < 1$ on $\partial E$, none of the $g_\lambda$ has zeros or poles on $\partial E$. The argument principle says
$$N(\lambda) = \frac{1}{2\pi i}\int_{\partial E} \frac{g_\lambda'(z)}{g_\lambda(z)}\,dz$$
is the number of zeros of $g_\lambda$ in $E$ minus the number of poles of $g_\lambda$ in $E$, both counted according to their multiplicity.
You want to show $N(1) = 0$.
A: There is a more general version of Rouché's theorem that works for meromorphic functions (taken from here):

Theorem (Rouché) Let $D \subset \mathbb{C}$ be a simply connected domain, $f$ and $g$ two meromorphic functions on $D$ with a finite set of zeros and poles $F$. Let $\gamma$ be a positively oriented simple closed curve which avoids $F$ and bounds a compact set $K$. If $|f - g| < |g|$ along $\gamma$, then
  $$ N_f - P_f = N_g - P_g$$
  where $N_f$ (resp. $P_f$) is the  number of zeros (resp. poles) of $f$ in $K$, counted with multiplicity (same for $g$).

I'm sure you'll find what to choose for $f$ and $g$ to solve your problem. Note that it's important that $D$ is simply connected!
The proof is kind of easy (much nicer than the one on the English Wikipedia article for the usual version of Rouché's theorem, if you ask me), so I'll reproduce it here:
Let $h= f/g$, it is a meromophic function on $D$ that has neither zeros nor poles along $\gamma$. The assumption on $f$ and $g$ ensures that $|h - 1|<1$ along $\gamma$. In other words the image of $\gamma$ by $h$ is sent in $D(1,1)$. This implies that $\int_\gamma \frac{h'(z)}{h(z)}dz = 0$ 1. But on the other hand, $ \frac{h'(z)}{h(z)} =  \frac{f'(z)}{f(z)} +  \frac{g'(z)}{g(z)}$. Conclude with the argument principle.
1 I'm not sure what the best argument for that is (I don't find the French Wikipedia article very convincing here), but here's one: $\frac{h'}{h}$ is holomorphic on a neighborhood of $\gamma$ and has a primitive there: $L \circ h$ where $L$ is a branch of the logarithm on $D(1,1)$.
