Integrating $\frac{z}{1-\cos(z)}$ over the unit circle

Question

I want to evaluate $ \int_C f(z)\,dz $ where $f(z)$ is :

$$f(z)=\frac{z}{1-\cos(z)}$$

and $C$ is the unit circle, counterclockwise.

I kept having problems with it. If someone can help, it would really be appreciated .

Answer 1

The function $f$ is holomorphic except at points where $\cos z = 1$, i.e. at points where $z=2\pi k$, $k \in \mathbb{Z}$. You can check that $1-\cos z$ has a double zero at $z=0$, so $f$ has a simple pole there. By the residue theorem $$ \int_{C} \frac{z}{1-\cos z}\,dz = 2\pi i \operatorname{Res}\limits_{z=0} \frac{z}{1-\cos z} = 2\pi i \lim_{z\to 0} \frac{z^2}{1-\cos z} = 4\pi i $$ (since $1-\cos z = \dfrac{z^2}{2} + O(z^3)$).

Answer 2

By residue theorem $$\int_C f(z)dz=2i\pi Res(f,0)=$$

and $Res(f,0)$ is the coefficient $a_{-1}$ in the Laurent series of $\frac{z}{1-\cos z}=\sum_{-\infty}^\infty a_n z^n$ so we find $Res(f,0)=2$ and then $$\int_C f(z)dz=4i\pi $$