integral of the complex function $1/\cos(1/z)$ 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.
 A: \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.
A: 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 $|z| \leqslant 1$
cos(1/z)==0
