Solution
= Solution
Using the Taylor series of the exponential and series division,
$$
\frac1{e^{2z}-1}
=\frac1{2z}-\frac12+\frac z6-\frac{z^3}{90}+O(z^5).
$$
Thus the first three nonzero Laurent terms are
$$
\boxed{\frac1{2z}-\frac12+\frac z6.}
$$
The singularities occur at $z=k\pi i$. The nearest nonzero ones are $\pm\pi i$, so this <Laurent series> is valid on \b[$0<|z|<\pi$].