Residues at poles of periodic functions I'm trying to compute the residues at all poles of $f(z) = \frac{1}{1-\cos(z)}$ inside $C = \{z\in\mathbb{C}| \: |z|=8\}$. The poles can be found at each $2\pi k, k\in\mathbb{Z}$. Inside the circle, the poles are $-2\pi, 0, 2\pi$. To determine the residue at $0$, I took the limit of $\text{res}_0=\lim_{z\to 0}\frac{d}{dz}\frac{z^2}{1-\cos(z)}$. Expanding cosine into a series and taking the derivative, I ended up with $res_0 f=0$. However, I struggled with the other two.
I looked up the problem and the textbook I found it in simply states that $\text{res}_{2\pi k}f = \text{res}_0f$ for all k. I guess it is somewhat intuitive that this is the case, since I'm looking at a periodic function. On the other hand, this was never clearly stated in our course and I find this rather useful, so is this just accepted as obvious or is there a more detailed explanation to this that I could leverage in an exam to avoid having to calculate several residues? Thanks in advance!
 A: The residue of $f$ at an isolated singularity $z_0$ is the coefficient $a_{-1}$ of the Laurent series of $f$ at $z_0$:
$$
 f(z) = \sum_{n=-\infty}^{\infty} a_n (z-z_0)^n \, .
$$
Now if $f$ is periodic with period $p$ then
$$
 f(z) = f(z-p) = \sum_{n=-\infty}^{\infty} a_n (z-(z_0+p))^n
$$
in a neighbourhood of $z=(z_0+p)$, so that the series on the right is the Laurent series of $f$ at $z_0+p$, and the residue of $f$ at $z_0+p$ is the same value $a_{-1}$.

Another way to compute the residue is
$$
 \operatorname{Res}(f, z_0) = \frac{1}{2\pi i} \int_C f(z)\, dz
$$
where $C$ a circle around $z_0$ (in counter-clockwise direction) with sufficiently small radius. This can also be used to show that
$$
\operatorname{Res}(f, z_0) = \operatorname{Res}(f, z_0+p)
$$
if $f$ is $p$-periodic.

Yet another way is to use the definition of the residue as the unique complex number $R$ such that
$$
 f(z) - \frac{R}{z-z_0}
$$
has a holomorphic antiderivative in a punctured neighbourhood  of $z_0$. If
$$
  f(z) - \frac{R}{z-z_0} = F'(z)
$$
for $0 < |z-z_0| < \delta$ then
$$
  f(z) - \frac{R}{z-(z_0+p)} = f(z-p) - \frac{R}{z-p-z_0}= F'(z-p)
$$
for $0 < |z-(z_0+p)| < \delta$, which again shows $f$ has the same residues at $z_0$ and $z_0+p$.
A: Yes, since the functions are periodic there is a translation invariance.
You could call back to the definition of the residual...
$\text{Rez}_\alpha (f(x)) = R \implies \oint_{|z-\alpha|=\delta} f(z) \ dz = \oint_{|z-\alpha|=\delta} \frac {R}{z-\alpha} \ dz$
As we consider this small path around the poles in your example.
Since the function is periodic $\frac {1}{1-\cos (0+\zeta)} = \frac {1}{1-\cos(2\pi+\zeta)}= \frac {1}{1-\cos(-2\pi+\zeta)}$ for all $\zeta$
$\oint_{|z|=\delta} f(z) \ dz  = \oint_{|z-2\pi|=\delta} f(z) \ dz = \oint_{|z+2\pi|=\delta} f(z) \ dz$ so the residues at each point must be the same.
Or, If you want to use limits, you could say.
$\lim_\limits{|z-2\pi|\to 0} \frac {d}{dz} \frac {(z-2\pi)^2}{1-\cos z}$
Let $w = z+2\pi$
$\lim_\limits{|w|\to 0} \frac {d}{dw}\frac {dw}{dz}\frac {w^2}{1-\cos (w+2\pi)}$
$\frac {dw}{dz} = 1$  and $\cos (w+2\pi) = \cos w$
$\lim_\limits{|w|\to 0} \frac {d}{dw}\frac {w^2}{1-\cos w}$
