 Solution

A function on the open set $D$ is holomorphic if at every $a\in D$ the complex limitexists. The limit is independent of the direction of approach.

A pole of order $n\ge 1$ at $a$ means that $f(z)=(z-a{)}^{-n}h(z)$ in a punctured neighborhood, where $h$ extends holomorphically across $a$ and $h(a)\ne 0$. Equivalently $(z-a{)}^{n}f(z)$ has a nonzero finite limit at $a$.

For the given quotient, $\cos z=1$ is equivalent, on putting $q=e^{iz}$, to $q+q^{-1}=2$, or $(q-1{)}^2=0$. Thus all possible poles are $a=2\pi k$, $k\in \mathbb{Z}$, with no additional nonreal solutions. At such a point,At $a\ne 0$ the numerator does not vanish, so the pole has order two. At zero the numerator cancels one order:ThereforeThe numerator's cancellation at zero is why not every denominator zero produces the same pole order.