complex analysis - Cauchy's Residue Theorem on a Singularity Outside a Contour

I recently ran into the following exercise:

Evaluate
$$\oint_\Gamma\frac{\cos z}{(z-\pi)^2}dz,$$where $\Gamma$ is a complete circuit of the circle $|z|=1$.

Clearly, the singularity lies outside the contour:

                                    

However, recall Cauchy's Residue Theorem:

If $\Gamma$ is a simple closed positively oriented contour and $f$ is analytic inside and on $\Gamma$ except at the points $z_1$, $z_2$, ..., $z_n$, then
$$\int_\Gamma f(z)dz=2\pi i\sum_{j=1}^{n}\text{Res}(z_j),$$
where
$$\text{Res}(f;z_0)=\lim_{z\to z_0}\frac{1}{(m-1)!}\frac{d^{m-1}}{dz^{m-1}}[(z-z_0)^mf(z)].$$

If I use this to evaluate the integral, I will have that

$$\text{Res}(f;\pi)=\lim_{z\to\pi}\frac{d}{dz}[\cos z]=0,$$and hence the value of the integral will evaluate to $0$, which is the correct answer.

Is this purely coincidental? Because, as I understand it, for Cauchy's theorem to hold, the singularities must lie within the contour (or perhaps not?). Thanks in advance!

Answer

You don't have to do anything to evaluate your integral, because $\cos(z)/(z-\pi)^2$ is holomorphic on the unit disc. So by Cauchy's theorem for a disk (or whichever more general variant you'd prefer), the integral is zero. You can also evaluate this using the residue theorem, but none of the residues are inside the disk so once again you get zero.

The function does indeed have a pole of order $2$ at $\pi$ and the residue is $0$. This does not mean that the function has a removable singularity at the pole, it just means that the function has an analytic antiderivative in some punctured disk about $\pi$.