For a function
$$f(z) = \frac{1}{1-e^{iz}}$$
Calculate the integral using the residue theorem
$$ \oint_{\gamma}\frac{1}{1-e^{iz}}\cdot \mathbb{d}z $$
where $\; \gamma : |z+\pi|=\frac{3}{2}\pi$
The function has a singularity for $z=0$, but since the complex exponential funcion is periodic of $\;T=2\pi i$, the function has infinite singularities given by $z=0 + 2\pi in$ where $n$ is an arbitrary integer.
However, there is only one singularity inside the closed contour $\gamma$.
- For $n=-1, z=-2\pi i\quad$ outside contour
- For $n=0, z=0\quad$ inside contour
- For $n=1, z=2\pi i\quad$ outside contour
My question is, how can I get the residue of the function for the singularity $z=0$?
Thanks in advance.
The function has a pole of order $1$ at $0$ and the residue is $\lim_{z \to 0} \frac {z-0} {1-e^{iz}}=\frac 1 {-i}=i$ (as seen by L'Hopital's Rule or by power series expansion of $e^{iz}$)