# Need help integrating close contour of $\sec(z)$

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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Remember $\sec(z) = \frac{1}{\cos(z)}$. Check if it has poles in the region $|z-1|=1$. Use Cauchy integral formula if it does. If there not poles, the answer is simple. And do not be afraid to let us know if this is a homework. You will still get help. – Sasha Sep 17 '12 at 13:14
thanks, I will look into that, it's not a homework though, more of a challenge from the teacher! – harinsa Sep 17 '12 at 14:23
Is it possible to apply Cauchy Integral formula to trigonometric equation? If so, how is it done? – harinsa Sep 19 '12 at 16:40

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

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

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

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