Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
up vote 1 down vote accepted

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

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

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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.