I am working on the following statement.
Let $Q = Q(x,y): \mathbb R^2 \to \mathbb R$ be a rational function, which is continuous on the unit circle $S_1(0)$. Let furthermore $f: \mathbb C \to \mathbb C$ defined by
\begin{align}
f(z) := \frac{Q\left(\frac{1}{2}\left(z+\frac{1}{z}\right), \frac{1}{2i} \left(z-\frac{1}{z}\right)\right)}{iz},
\end{align}
where $z = e^{i\varphi}$ and $\varphi \in [0,2\pi)$. Let $\mathcal S = \{z_1, \ldots, z_m\}$ be the set of poles of $f$ inside of the unit circle $S_1(0)$. Then
\begin{align}
\int_{0}^{2\pi} Q(\cos(\varphi),\sin(\varphi))\, d\varphi = 2\pi i\sum_{z_k \in S} \operatorname{Res}(f,z_k).
\end{align}
Although I already have the proof in my lecture notes, I wanted to recreate it on my own and had some trouble. But firstly, here is how far I got:
Proof. We have that
\begin{align*}
\cos(\varphi) = \frac{1}{2}\left(e^{i\varphi} + e^{-i\varphi}\right) = \frac{1}{2}(z+z^{-1}), \quad \sin(\varphi) = \frac{1}{2i} \left(e^{i\varphi} – e^{-i\varphi}\right) = \frac{1}{2i}(z-z^{-1}),
\end{align*}
with $z = e^{i\varphi}$ and $\varphi \in [0,2\pi)$, hence
\begin{align*}
Q(\cos(\varphi),\sin(\varphi)) &= Q\left(\frac{1}{2}(z+z^{-1}), \frac{1}{2i}(z-z^{-1})\right).
\end{align*}
It follows that
\begin{align*}
Q(\cos(\varphi),\sin(\varphi)) = f(z)iz = f(e^{i\varphi})i e^{i \varphi}
\end{align*}
and therefore
\begin{align*}
\int_0^{2\pi} Q(\cos(\varphi), \sin(\varphi)) \, d\varphi = \int_{S_1^+(0)} f(z) \, dz.
\end{align*}
$\mathbb C$ is obviously simply connected and $\mathcal S$ a finite subset of $\mathbb C$. $f$ is a rational function with no poles on $S_1(0)$, since the only poles are in $\mathcal S$ and inside of $S_1(0)$. Therefore $f: \mathbb C \setminus \mathcal S \to \mathbb C$ is analytic and of course $S_1^+(0)$ is a closed contour which maps $[0,2\pi)$ into $\mathbb C \setminus \mathcal S$. Thus by the Residue Theorem
\begin{align*}
\int_0^{2\pi} Q(\cos(\varphi), \sin(\varphi)) \, d\varphi
= 2\pi i \sum_{z_k \in \mathcal S} \underbrace{W(S_1^+(0), z_k)}_{= 1} \operatorname{Res}(f,z_k)
= 2\pi i\sum_{z_k \in \mathcal S} \operatorname{Res}(f,z_k).
\end{align*}
In order to use the Residue Theorem I wanted to make sure that all needed assumptions are satisfied, that is why I wrote it out so precisely in the last sentences. But I don't understand why the only poles are inside the unit circle. Why is it not possible that there are also ones outside? Maybe this is obvious but at the moment I can't see it.
Edit: We formulated the Residue Theorem in the following way:
Let $D \subset \mathbb C$ a simply-connected domain, and $\mathcal S = \{z_1, \ldots, z_m\} \subset D$ be a finite subset of $D$. Let $f: D \setminus \mathcal S \to \mathbb C$ be analytic and $\gamma : [\alpha, \beta] \to D \setminus \mathcal S$ be a closed contour. Then
\begin{align*}
\int_\gamma f \, dz = 2\pi i\sum_{z_k \in \mathcal S} W(\gamma,z_k)\operatorname{Res}(f,z_k),
\end{align*}
where $W$ is the winding number of $\gamma$.
If someone could make things clear with that theorem, my question would be answered totally, thanks.
Best Answer
It is possible, and usually there will be poles outside the unit disk.
But we are not interested in these poles, since the contour of integration - the unit circle - does not wind around them, hence they contribute nothing to the integral. In other words,
$$W(S_1^+(0), p) = 0$$
for all poles $p \in \mathbb{C}\setminus \overline{D_1(0)}$, so let $\mathcal{P}$ be the set of all poles of $f$ in $\mathbb{C}$, then we have
$$\sum_{z \in \mathcal{P}} W(S_1^+(0),z)\operatorname{Res}(f,z) = \sum_{z\in \mathcal{S}} \operatorname{Res} (f,z).$$