Find the poles of $f$ at $z=2πik$?

Question

Let $f(z)= \dfrac{z}{ (1-e^{z} ) \sin z}$ then the poles of $f$ at $z=2πik$ where $k \in \mathbb{Z}$ $k \neq 0$are of order $1$ or $2$?

\begin{align} \lim_{z \to 2πik} (z-2πik)^2 \dfrac{z}{ (1-e^{z}) \sin z } &= \lim_{z \to 2πik} \dfrac{z}{ \dfrac{1-e^{z}}{z-2πik} \dfrac{\sin z}{z-2πik}} \\ &= 2πik \neq 0 \end{align}

so it is a pole of order $2$ . Is it correct?

Answer 1

Here , $\forall k \in \mathbb Z$, $\quad e^z-1=0\implies e^z=1=e^{2k\pi i}\implies z=2k\pi i\quad$ are poles.

Now $\lim_{z \to 2k\pi i}(z-2k\pi i)f(z)$

$=\lim_{z \to 2k\pi i} \frac{z(z - 2k\pi i)}{(1-e^z)\sin z}$ $\quad (\frac{0}{0}\text{form})$

$=\lim_{z \to 2k\pi i}\frac{2z - 2k\pi i}{(1-e^z)\cos z -e^z \sin z}$, which gives a finite value.

$\implies z=2k\pi i$ is a simple pole.

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A simple pole of an analytic function $f(z)$ is a pole of order one. That is, $(z-z_0)f(z)$ is an analytic function at the pole $z=z_0$.

In other word, for $f(z)$ has pole at $z=z_1 \in \mathbb C$, $$\lim_{z \to z_1} f(z)= \infty$$

For simple pole at $z=z_1 \in \mathbb C$, $$\lim_{z \to z_1} (z-z_1)f(z) \qquad \text{is finite.}$$