Solution: Given that \[ f(z) = \frac{z}{\left( 1 - e^z \right) \sin z} = \frac{g(z)}{h(z)}. \] Recall that $z = z_0$ is a pole of a meromorphic function $f(z) = \frac{g(z)}{h(z)}$ if $h\left( z_0 \right) = 0$. Here \begin{align*} h(z) = 0 \implies \left( 1 - e^z \right) \sin z = 0 & \implies 1 - e^z = 0 \text{ or } \sin z = 0 \\ & \implies z = 2\pi \iota n, \text{ or } z = n\pi, n \in \mathbb{Z} . \end{align*} Take $z = 2\pi \iota n$. Note that \[ h^\prime (z) = -e^z \sin z + \left( 1 - e^z \right) \cos z. \] Now we will determine the order of each of the poles. We have, \[ h^\prime (2 \pi \iota n) \neq 0, \text{ if } n\neq 0. \] therefore, for each $n\in \mathbb{Z}\setminus \{ 0 \} $ the poles $z = 2\pi \iota n$ are simple. On the other hand, \[ h^\prime (0) = 0 \text{ and } h^{\prime\prime} (0) = -2 \] and hence, $z=0$ is a pole of order $2$. Also, for $n\neq 0$, \[ h^\prime (n \pi) \neq 0, \] hence these poles are also simple. Summarizing, the poles are \[ \{ 2n\pi \iota : n \in \mathbb{Z} \} \cup \{ n \pi : n \in \mathbb{Z} \}. \] All poles except $z=0$ are simple.