Complex integral

1. Sep 21, 2015

jjr

1. The problem statement, all variables and given/known data
Describe all the singularities of the function $g(z)=\frac{z}{1-\cos{z}}$ inside $C$ and calculate the integral

$\int_C \frac{z}{1-\cos{z}}dz,$

where $C=\{z:|z|=1\}$ and positively oriented.

2. Relevant equations

Residue theorem: Let C be a simple closed contour, described in the positive sense. If a function $f$ is analytic inside and on C except for a finite number of singular points $z_k (k=1,2,...,n)$ inside C, then

$\int_C f(z)dz=2\pi i\sum_{k=1}^{n}Res_{z=z_k}f(z)$

and the series

$\cos{z}=(1-\frac{z^2}{2!}+\frac{z^4}{4!}-...)$

Laurent series representation of a function
$f(z)=\sum_{n=0}^\infty a_n(z-z_0)^n+\frac{b_1}{z-z_0}+\frac{b_2}{(z-z_0)^2}+...$

3. The attempt at a solution

The function has singularities whenever $\cos{z}=1$, i.e. when $z=2\pi n,\hspace{3mm} n=0,\pm 1,\pm 2$. Within the boundary this only happens once, when $n=0$.

The integral itself is easily calculated by the residue theorem, but I have to find the residues first.

I use the series expansion and get

$g(z)=\frac{z}{1-\cos{z}}=\frac{z}{1-(1-\frac{z^2}{2!}+\frac{z^4}{4!}-...)}=\frac{1}{\frac{z}{2!}-\frac{z^3}{4!}+...}$

I am a bit stuck at this point. To be able to find the residues and to classify the singularity I need to know how many of the $b_n$ terms of the Laurent series representation vanish. But I'm not sure how I can get it on the form $...\frac{b_1}{z-z_0}+\frac{b_2}{(z-z_0)^2}+...$

Any suggestions would be very helpful

Sincerely,
J

2. Sep 21, 2015

RUber

An alternate definition of the residue depends on the limit as the contour approaches the singular point.
Letting $z = |z|e ^{i\theta}$
$\lim_{|z|\to 0 } \int_0^{2\pi} \frac{z}{1-\cos z}|z|\, d\theta$?

Hint, since the function is continuous, the integral is bounded, and the limits of integration do not depend on the variable in the limit, you should be able to pass the limit through the integral.

3. Sep 21, 2015

Krylov

On $\{ z : 0 < |z| < 2\pi\}$ the function $g$ has a Laurent expansion of the form
$$g(z) = \gamma_{-1}z^{-1} + \gamma_0 + \gamma_1 z + O(z^2)$$
(Why? In particular, how can you conclude a priori that zero is a simple pole?) You had already derived that
$$g(z) = \frac{1}{\frac{z}{2!} - \frac{z^3}{4!} + O(z^5)}$$
Now combine these two expressions to find the residue $\gamma_{-1}$.