find order of pole

Question

$f(z) = \frac{1}{2-e^z}$

I found that $f$ is holomorphic on $\mathbb{C}-A$, where $A$ is the discrete set $A = \{\log(2) + 2\pi ik, k \in \mathbb{Z} \}$. If I now take $z_k = \log(2) + 2\pi ik$, then $z_k$ is a pole, since $\lim_{z\to z_k} |f(x)| = \infty$.

Trying to find the order of the pole $z_k$, I got stuck in the computation of $\lim_{z\to z_k} \frac{z-z_k}{2-e^z}$.

I know that the solution should be $\lim_{z\to z_k} \frac{z-z_k}{2-e^z} = \lim_{z\to z_k} \frac{1}{-e^z} = \lim_{z\to z_k} -\frac{1}{2}$. So the pole is of order 1. But I don't see how I can get there.

Thanks for any hints.

Answer 1

Hint: The denominator $g(z)=2-e^z$ has only simple zeros because $g'(z)=-e^z$ is never zero.

Answer 2

Just notice that, taken any pole $z_k$, $2 = e^{z_k}$, hence the limit $$ \lim_{z \to z_k} \frac{ e^z - 2}{z - z_k} = \lim_{z \to z_k} \frac{ e^z - e^{z_k}}{z - z_k} = \frac{d}{dz}[e^z](z_k) = e^{z_k} $$