 Solution

The zeros of the denominator satisfy $e^z=\ell \pi$ with nonzero integer $\ell$, since an exponential never vanishes. Equivalently the poles areAt such a point $e^{z_{m,k}}=(-1{)}^{k}m\pi$, and the derivative of the denominator is $e^z\cos (e^z)=(-1{)}^{m+k}m\pi \ne 0$. All these poles are simple, with