I was trying to find the residue of the function
$$f(z) = \frac{z^2 + \sin z}{\cos z - 1}.$$
Here is the my attempt:
The given function has a pole of order two at $z = 2n\pi$. So, we use the following formula to calculate residue of a function when it has a pole of order m at $z=z_0$.
$$\mathrm{Res}(f(z))_{z=z_0}=\frac{1}{(m-1)!}\lim_{z\to z_0}\left[\frac{d^{m-1}}{dz^{m-1}}(z-z_0)^m f(z)\right]$$
But I am not able to apply this formula as I am getting zero in the denominator while I am taking limit $z\to 2n\pi$. Any help or suggestions will be very helpful for me.
Thanks
Consider the function $f(z)=p(z)/q(z)$, where $q(z)=(z-z_0)^2 r(z)$ and $r$ is analytic and nonzero at $z=z_0$. Then the residue of $f$ at $z=z_0$ is
$$\operatorname*{Res}_{z=z_0} f(z) = \left [\frac{d}{dz} \frac{p(z)}{r(z)} \right ]_{z=z_0} = \frac{p'(z_0)}{r(z_0)} - \frac{p(z_0) r'(z_0)}{r(z_0)^2}$$
Now note that, because $q'(z) = 2 (z-z_0) r(z) + (z-z_0)^2 r'(z)$, $q''(z) = 2 r(z) + 4 (z-z_0) r'(z) + (z-z_0)^2 r''(z) $, and $q'''(z) = 6 r'(z)+6 (z-z_0) r''(z)+(z-z_0)^2 r'''(z)$, then $2 r(z_0)=q''(z_0)$ and $6 r'(z_0)=q'''(z_0)$, and we have
$$\operatorname*{Res}_{z=z_0} f(z) = \frac{2 p'(z_0)}{q''(z_0)} - \frac{2 p(z_0) q'''(z_0)}{3 q''(z_0)^2}$$
Note that $p(z) = z^2+\sin{z}$, $q(z) = \cos{z}-1$, and $z=2 n \pi$, so the residue there is
$$\operatorname*{Res}_{z=2 n \pi} \frac{z^2+\sin{z}}{\cos{z}-1} = \frac{8 n \pi + 2}{-1}+ 0 = -(2+8 n \pi)$$