Question about how to get the residue for infinite amount of poles

Question

So I am asked to find the residue for each pole such as $$ f(z) = \frac{z}{1-\cos(2z)} $$ I understand pole of order 2 with $z= 2\pi k$ excluding zero.

I also understand that residue equals to cauchy integral over the curve with integrand f(z)

I am not sure how to apply it because I am not given the radius and

only used to using short cut formula for residue and simple poles.

From my understanding I know the generalized cauchy formula with denominator of integrand

of power of $(k+1)$ derives $f^{(k)}(z_0)$ and I also know that when there is pole of order two you

can get the residue right away by evaluating the holomorphic function inside the integrand

and multiply $\pi/2$ but this is infinite amount of poles so I am not sure how I could

do this. Can someone explain me why integral over the curve of f(z) is residue?

I mean integral over entirely analytic function on domain D is 0 and residue therefore

equals 0 but this is all the knowledge I know.

Answer 1

I am not sure how to apply it because I am not given the radius and

It doesn't matter as long as there's only one singularity enclosed by the curve.

Can someone explain me why integral over the curve of f(z) is residue?

I don't think I understand this question. This is just a definition.

There are some easier ways to calculate residues when the singularity is a pole. If $z_0$ is a pole of order $n$ of $f$, then: $$ \operatorname{res}_{z_0}f = \frac{1}{(n-1)!} \lim_{z\to z_0}\left(\frac{d}{dz}\right)^{n-1} ((z-z_0)^nf(z)) $$

Answer 2

The poles are all multiples of $\pi$, not $2\pi$, and $0$ is included.

Hint: if $z = \pi k + t$, $1 - \cos(2z) = 2 t^2 + \ldots$.

You almost never compute a residue by actually integrating.

Answer 3

We want to calculate:

\begin{align} \mathop{\mathrm{Res}}_{z = n\pi}\left(\frac{z}{1-\cos(2z)}\right) \end{align}

For the case $n=0$, we use the Maclaurin series of $\cos(x)$: \begin{align} \frac{z}{1-\cos(2z)} &= \frac{z}{1-(1-z^2/2!+z^4/4!-z^6/6!\ldots)}\\ &= \frac{z}{z^2/2!-z^4/4!+z^6/6!\ldots}\\ &= \frac{z}{2z^2(1-2z^2/4!+2z^4/6!+\ldots)}\\ &= \frac{1}{2}\frac{1}{z(1-2z^2/4!+2z^4/6!+\ldots)} \end{align}

At this point you can use Cauchy's integral formula with $ f=\frac{1}{1-2z^2/4!+2z^4/6!+\ldots}$

\begin{align} \mathop{\mathrm{Res}}_{z = 0}\left(\frac{z}{1-\cos(2z)}\right) = \frac{1}{2} \end{align}

For the case $n\neq0$, we use the subsitution $w=z-n\pi$ and take the residue at $w=0$ (i.e. $z=n\pi$):

\begin{align} \frac{z}{1-\cos(2z)} &= \frac{w+n\pi}{1-\cos(2w+2n\pi)}\\ &= \frac{w+n\pi}{1-\cos(2w)}\\ &= \frac{w}{1-\cos(2w)} + \frac{n\pi}{1-\cos(2w)}\\\\ \end{align} Here we can use again Maclaurin series and Cauchy's integral formula, but using our previous result:
The first term is (as before) $1$.
The second term has a pole of 2nd order at $w=0$, hence its residue equals $0$.

Finally we have:

\begin{align} \mathop{\mathrm{Res}}_{z = n\pi}\left(\frac{z}{1-\cos(2z)}\right) &= \frac{1}{2}\quad \end{align}

I think this is an easy way to calculate the residues.
Otherwise, you should calculate the Laurent series at every $z=n\pi$.

About the radius of convergence, the Laurent series at $z=n\pi$ have a radius of convergence $r<\pi$, so that it "won't touch the next pole".