I need help integrating this:
$$\oint_{|z-1|=1} \sec(z) \, dz $$
From the context of my course, I assume that what I'm suppose to do is expand $\sec(z)$ centered at $z_0=1$ into a Laurent series then find the residue, then use the formula to solve it. But expanding sec(z) centered at 1 seems too tedious, so I'm wondering if there is a better way. Also, I can't see any terms in the expansion that will contain $\frac1z$ so is the answer to the problem $0$? or am I missing something.
$$\cos z=0\Longleftrightarrow z=\frac{(2n+1)\pi}{2}\,\,,\,\,n\in\Bbb Z$$
and from here the only pole of $\,f(z):=\sec z\,$ in $\,|z-1|\leq 1\,$ is $\,z=\pi/2\,$, and
$$\operatorname{Res}_{z=\pi/2}(f)=\lim_{z\to\pi/2}\frac{z-\pi/2}{\cos z}\stackrel{\text{L'Hospital}}=\frac{1}{-\sin \pi/2}=-1$$
so that
$$\oint_{|z-1|=1}\sec z\,dz=2\pi i(-1)=-2\pi i$$