Residue $f(z)=\frac{e^z}{e^z-1}$

Question

Find the residue at each pole of the function $$f(z)=\dfrac{e^z}{e^z-1}$$

I wonder that $z=0$ is a pole of $f(z)$ ? And is it a simple pole? Can I use the formula when $z_0$ is a simple pole of $h$ then $$Res\Bigg(\dfrac{g(z)}{h(z)},z_0\Bigg)= \dfrac{g(z_0)}{h'(z_0)}$$ So $\text{Res}(f,0)=1$?

Answer 1

Yes. When the analytic function $\varphi$ that you are dealing with can be written as $\dfrac fg$, with $f$ and $g$ analytic, and:

• $g(z_0)=0$,
• $f(z_0)\neq0$,
• $g'(z_0)\neq0$,

then $z_0$ is a simple pole of $\varphi$ and $\operatorname{res}_{z=z_0}\varphi(z)=\dfrac{f(z_0)}{g'(z_0)}$.

In particular, the residue of $\dfrac{e^z}{e^z-1}$ at every $2\pi in$ ($n\in\mathbb Z$) is $1$.

Answer 2

Yes! it is a simple pole, since $$f(z)=\frac{\phi(z)}{z}$$ where $$\phi(z)=\frac{e^z}{1+\frac{z}{2!}+\frac{z^2}{3!}+\cdots}$$ with $\phi(z)$ is analytic $\phi(0) \neq0$ .Hence $$\text{Res}_{z=0} f(z)=\phi(0)=1$$