The exercise asks to find zeroes and singularities of $$f(z) = \frac{1-\cos(z)}{e^{2iz}-1}$$ using its Laurent expansion.
The denominator equals zero at every $z_{k} = k\pi$, so the first question is:
$\bullet$ For k odd, the numerator equals 2 so these are definitely singularities. For k even, though, the numerator also vanishes: how do I know if those are singularities too?
About the expansion, I guess I must expand around the generic $z_{k}$ to find what type of singularity it is (pole or essential). The first thing I notice is that
$$e^{2iz} = e^{2i(z-z_{0})}$$
Now I want the term $(z-z_{0})$ to appear at the numerator also, and using the addition formula for $\cos x$ I obtain
$$\cos \bigl((z-z_{0})+z_{0}\bigr) = (-1)^k \cos (z-z_{0})$$
After this I thought it could have been useful replacing $\cos (z-z_{0})$ using the relationship
$$ \cos (z-z_{0}) = \frac{e^{i(z-z_{0})}+e^{-i(z-z_{0})}}{2}$$
but after that and substituting back in the original function I can't proceed. Any help would be very appreciated.
Let's just look at the Taylor expansions at zero:
$$\frac{1 - \cos z}{e^{2iz} - 1} = \frac{1 - (1 - z^2/2 + O(z^4))}{1 + 2iz + O(z^2) - 1} = \frac{z^2/2 + O(z^4)}{2iz + O(z^2)} = \frac 1 {4i} z + O(z^2).$$
after dividing the series. Hence, the singularity is removable. This is the key result: The numerator vanishes to second order at any zero of the denominator, and the denominator only vanishes to first order there. The fact that $f$ is periodic means that we don't have to do too much more work.