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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.

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1 Answers 1

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$\cos z=0\Longleftrightarrow z=\frac{(2n+1)\pi}{2}\,\,,\,\,n\in\Bbb Z$

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

$\oint_{|z-1|=1}\sec z\,dz=2\pi i(-1)=-2\pi i$