Determining possible values of a complex integral

Question

I'm looking at the complex integral:

$ \frac{1}{2 \pi i} \int_{C(0;r)}\frac{cos(z)}{z^{n+1}}dz $

and I'm wondering how I'd go about determining all the possible values of this for $ r>0 $ and $ n \geq 0$

With no similar examples at my disposal I'm wondering how I might approach this.

I'm thinking about using the fact that the integral of $\frac{1}{z}$ is $2 \pi i$

Answer 1

This is the residue of $$ f(z) = \frac{\cos(z)}{z^{n+1}} $$ at $0$ see wikipedia. Since $f$ is analytic in $\mathbb{C}^*=\mathbb{C}\setminus\{0\}$ it doesn't depend on $r>0$. To compute this value note that for $$ f(z) = \sum_{k\in\mathbb{Z}} a_n z^n $$ the residue at $0$ is $a_{-1}$, which is shown using what you mentioned and the fact that this same integral over $\frac{1}{z^n}$ for any $n\not=1$ is zero. Hence to obtain your answer figure out what the series expansion of $f$ is.

Answer 2

By Cauchy's integral formula for higher derivatives we have $$\frac{2\pi i}{n!}f^{(n)}(0)=\int_{C(0;r)} \frac{\cos(z)}{z^{n+1}}dz$$ where $f(z)=\cos(z)$ and $f^{(n)}(0)$ is the $n$th derivative of $f$ at $0$.