I have a function $f(z) = \frac{\cos(z)}{z^6}$. I have to find the singularities and the corresponding residues. I think there is a single pole at $z=0$, which has order 6.
For the residue, I did this: $\text{Res}(f,0) = \frac{1}{5!} \lim_{z\to 0}\frac{d^5}{dz^5}(z^6 \cdot \frac{\cos(z)}{z^6}) = \frac{1}{5!}\lim_{z \to 0} -\sin(x) = 0$
Am I right, or am I missing something?
That is correct, but it is much simpler to say that, since $f$ is an even function, the residue at $0$ is $0$.