= Solution
{c}
A function on the open set $D$ is <holomorphic> if at every $a\in D$ the complex limit
$$
f'(a)=\lim_{\substack{z\to a\\z\neq a}}
\frac{f(z)-f(a)}{z-a}
$$
exists. The limit is independent of the direction of approach.
A <pole> of order $n\geq1$ at $a$ means that $f(z)=(z-a)^{-n}h(z)$ in a punctured neighborhood, where $h$ extends holomorphically across $a$ and $h(a)\neq0$. Equivalently $(z-a)^nf(z)$ 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,
$$
\cos z-1=-\tfrac12(z-a)^2+O((z-a)^4).
$$
At $a\neq0$ the numerator does not vanish, so the <pole> has order two. At zero the numerator cancels one order:
$$
\frac z{\cos z-1}=-\frac2z-\frac z6+O(z^3).
$$
Therefore
$$
\boxed{z=0\text{ is a simple pole};\quad
z=2\pi k,\ k\neq0,\text{ are double poles}.}
$$
The numerator's cancellation at zero is why not every denominator zero produces the same <pole> order.
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