Residue of $\frac{\cos (\pi z)}{z^2-1}$ I want to find the residue $\displaystyle\frac{\cos (\pi z)}{z^2-1}$ at $z=1$.
The function has two poles, at $-1,1$, both of order $1$. The series expansion for the function gives $\displaystyle\frac{1}{z^2-1}\left(1- \frac{z^2\pi^2}{2!}+O(z^4z\pi^4)  \right)$.
(1) Is then Laurent series the following?
$$\frac{1}{z^2-1}- \frac{z^2\pi^2}{2!(z^2-1)}+\frac{z^4}{4!(z^2-1)}+\dots$$
(2) How do I find the pole of order $1$ from the Laurent series?
(3) I know that the residue should be given by $\displaystyle\lim_{z \to 1}(z-1)\frac{\cos (\pi z)}{z^2-1}=\frac{1}{z+1}- \frac{z^2\pi^2}{2!(z+1)}+\frac{z^4\pi^4}{4!(z+1)}+\dots=\frac{1}{2}-\frac{\pi^2}{2\cdot2!}+\frac{\pi^4}{2\cdot4!}+\dots$ 
Is the residue $1/2$? I am not sure of what term in the Laurent series I should pick up to be the residue. 
 A: You have the function $f(z) = \frac{g(z)}{z^2-1}$, where $g(z)= \cos(\pi z)$ and therefore holomorf. The residue at $z=1$:
Using limits.
\begin{align}
\text{Res}_{z=1} f(z) =\lim_{z\to 1} (z-1)\frac{g(z)}{(z-1)(z+1)} = \frac{g(1)}{2} = \frac{-1}{2}
\end{align}
Using Laurent series. You must have powers of $z-1$ in your laurent series:
\begin{align}
\frac{\cos(\pi z)}{z^2-1} &= \frac{\cos(\pi(z-1) + \pi)}{(z-1)(z+1)} \\&
= -\frac{\cos(\pi( z-1))}{(z-1)((z-1)+2)}\\
& =  -\frac{1}{2}\frac{\cos(\pi( z-1))}{(z-1)((z-1)/2+1)} \\
&= -\frac{1}{2}\cdot \frac{\cos(\pi(z-1))}{(z-1)} \cdot \sum_{k=0}^\infty (-(z-1))^k \\
&= -\frac{1}{2} (1 + O((z-1)^2) \sum_{k=0}^\infty (-1)^k(z-1)^{k-1}\\
&=-\frac{1}{2} (1 + O((z-1)^2) \cdot ((z-1)^{-1}+O(1))\\
&= -\frac{1}{2} (z-1)^{-1} + O(1)
\end{align}
We used the geometric series. Now you can see the coefficient of $(z-1)^{-1}$ is $-\frac 1 2 $ and therefore the residue at $z=1$ is $-\frac 1 2 $.
A: Use the next result: if $g(z)$ is holomorphic at $z_0$ and $h(z)$ has a simple pole at $z_0$, then $Res(gh,z_0)=g(z_0)Res(h,z_0)$. Take 
$$g(z)=\frac{cos(\pi z)}{z+1}$$
$$h(z)=\frac{1}{z-1}$$
The residue is:
$$Res(gh,1)=g(1)Res(h,1)=-\frac{1}{2}$$
