Questions about poles

Question

Find order of pole for $$f(z)=\frac{1}{e^{z}-1}$$ at $z=0$.

Now I turned the function into this: $$\frac{1}{\sum_{1}^{\infty}x^k/k!}$$

I think the pole has order $1$ but $\lim(z(f(z)))$ seems to be $1$, this is why I was confused. That limit should be $0$ according to our definition of order of a pole.

Answer 1

You can come up with a Laurent expansion by performing polynomial long division:

$$\frac{1}{z+\frac{z^2}{2!}+\frac{z^3}{3!}+\cdots}=\frac{1}{z}-\frac{z}{12}-\frac{z^3}{720}+\cdots$$

The order of the pole one...

Answer 2

$e^z - 1$ is of the form $zh(z)$ with $h(0) = 1$. You have $$f(z) = \frac{1}{h(z)} \frac{1}{z} \sim \frac{1}{z} \quad \text{ as $z \to 0$}.$$ Therefore the pole has order $1$.

Answer 3

Hint

$$\lim_{z \to 0} \frac{z}{e^z-1}=\frac{1}{\lim_{z \to 0} \frac{e^z-e^0}{z-0}}$$