I have two quick questions.
Identify the pole in the following function and find the res of said function at it's pole.
$(1)$\ $G(z)=\frac{\cos(z)}{\sin(z)}$
Here, $\frac{\cos(z)}{\sin(z)}=(a_0+a_1z+\frac{a_2z^2}{2!}+\frac{a_3z^3}{3!}+...)$
$\Rightarrow (1-\frac{z^2}{2!}+\frac{z^4}{4!}-...)=(a_0+a_1z+\frac{a_2z^2}{2!}+\frac{a_3z^3}{3!}+...)(z-\frac{z^3}{3!}+\frac{z^5}{5!}-...)$
I'm a little confused how to determine the pole and the residue since there appears to be no negative powers of $(z-z_0)$, where $z_0$ is my undetermined singular point. I'm also stuck because $\sin(z)=0$ \, $\forall$ $z=\pi k$.
$(2)$ I want to evaluate this integral. $\int_{0}^{2\pi}\frac{1}{2+\cos(\theta)}d\theta$. I know how to do this evaluation, but why can we say,
$\int_{0}^{2\pi}\frac{1}{2+\cos(\theta)}d\theta = \int_{|z|=1}f(z) \ dz$, where $f(z)=\frac{-2i}{z^2+4z+1}$. I don't understand why we can take the contour to be a circle of radius 1.
1)
$$\frac{\cos{z}}{\sin{z}} = \frac{\displaystyle 1-\frac{z^2}{2} + \frac{z^4}{24}-\cdots}{\displaystyle z-\frac{z^3}{6}+\frac{z^5}{120}-\cdots} = \frac{1}{z} \frac{\displaystyle 1-\frac{z^2}{2} + \frac{z^4}{24}-\cdots}{\displaystyle 1-\frac{z^2}{6}+\frac{z^4}{120}-\cdots}$$
Then we may write that
$$\frac{\cos{z}}{\sin{z}} = \frac{1}{z} F(z)$$
where $F$ is analytic at $z=0$.
2) To get the rational function in $z$, you substituted $z=e^{i \theta}$, which is the parametric form of $|z|=1$, which is a circle of radius $1$ centered at the origin.