# Order of pole of function

Question Let $f(z)$ be meromorphic function given by $f(z)=\frac{z}{(1-e^z)\sin{z}}$ then which of the following are correct?

(1) $z=0$ is pole of order $2$

(2) for every $k$ in $\mathbb{Z}$, $z=2kπi$ is simple pole.

(3) for every $k$ in $\mathbb{Z}-\{0\}$, $z=kπ$ is a simple pole.

(4) $z=π+2πi$ is a pole.

My attempt: I solved and saw that, (1) is false, (2) is true, and (4) is false.But I am not sure about (3). Here is my work for (3).

Consider, $$\lim_{z\to kπ} (z-kπ)\frac{z}{(1-e^z)(\sin{z})}$$

$$=\lim_{z\to kπ}\frac{z^2-zkπ}{\sin{z}-e^z\sin{z}}$$

Which is ($0/0$ form) so using LH rule we get,

$$=\lim_{z\to kπ}\frac{2z-kπ}{\cos{z}-(e^z\cos{z}+e^z\sin{z})}$$

$$=\frac{kπ}{\cos{kπ}-(e^{kπ}\cos{kπ}-0)}$$

$$=\frac{kπ}{\cos{kπ}-(\cos{kπ}+i\sin{kπ})\cos{kπ}}$$

$$=\frac{kπ}{\cos{kπ}-(\cos{kπ})^2}$$

But this not always nonzero and finite(as we required by definition of pole) ( further can we check this by Laurent series too?)

So how can be (3) is true? (In key it was given (3) is true!!)

You've made an error when you went from $e^{k\pi}$ to $\cos{k\pi}+i\sin{k\pi}$: that would be $e^{ik\pi}$. You should end up with $$\frac{k\pi}{(1-e^{k\pi})\cos{k\pi}},$$ which is finite and nonzero for $k \in \mathbb{Z} \setminus \{ 0\}$.