|
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. |
|||||||||||
|
|
\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. |
||||
|
|
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 $|z| \leqslant 1$ |
||||
|
