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I am looking for $\underset{|z|=1}{\oint}\frac{1}{\cos\left(\frac{1}{z}\right)}dz$ I was able to do the following: $$\underset{|z|=1}{\oint}\frac{1}{\cos\left(\frac{1}{z}\right)}dz=\underset{|z|=1}{\oint}\frac{1}{\cos\left(\overline{z}\right)}dz=\underset{|\overline{z}|=1}{\oint}\overline{\frac{1}{\cos\left(z\right)}}dz$$

But got stuck here, any help would be greatly appretiated.

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Does the 'residue theorem' sound familiar? – Berci Jan 28 '13 at 16:46
@Berci: it does, i was still anable to solve it (1/cos(z)) have infinity many residue points – user44874 Jan 28 '13 at 17:22
@levap: it does not bring we anywhere – user44874 Jan 28 '13 at 17:25
Try parametrizing the curve and use a little symmetry argument. – mrf Jan 28 '13 at 17:32
@mrf you mean like $z=e^{i \alpha} $ ? tried, did not get to anything – user44874 Jan 28 '13 at 18:41
up vote 2 down vote accepted

\begin{align} w & = 1/z \\[8pt] dw & = -1/z^2 \, dz \\[8pt] \frac{-dw}{w^2} & = dz \end{align}

$$ \int_\text{circle} \frac{1}{\cos\frac1z}\,dz = -\int_\text{circle} \frac{1}{\cos w} \left(\frac{-dw}{w^2}\right) . $$ (The circle is traversed in the opposite direction; hence the first minus sign.)

The residue of $1/w^2$ at $w=0$ is $0$, and so you should get $2\pi i$ times $1/\cos 0$ times that.

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yeh, except that the $w=0$ is of order 2 and i get $(\frac{1}{cos(w)})'=-\frac{2 sin(w)}{1+cos(2 w)}$ thanks! – user44874 Jan 28 '13 at 19:10

1) Find the poles inside the unit circle

that means you have to find the roots of:

$\cos (\frac{1}{z})=0$ , $|z| \leqslant 1$

2) Find the residue in each pole inside $|z| \leqslant 1$

for simple poles (multiplicity=1) :

$\operatorname{Res}(f,c)=\lim_{z\to c}(z-c)f(z)$

for any pole (multiplicity=n):

$\mathrm{Res}(f,c) = \frac{1}{(n-1)!} \lim_{z \to c} \frac{d^{n-1}}{dz^{n-1}}\left( (z-c)^{n}f(z) \right)$

3) Use $\oint_\gamma f(z)\, dz = 2\pi i \sum \operatorname{Res}( f, a_k )$

If it's not a pole, you will know what kind of singularity is when you do the limit or when you do the Laurent series expansion.

Something tells me you have an essential singularity when z=0 (when you divide by 0).

Laurent series of this function at z=0

You have many poles but not an infinity in $|z| \leqslant 1$


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there are infinity many of these poles and each have different order. the 'residue theorem' seems just not to be the suitable method here ... – user44874 Jan 28 '13 at 18:56

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