= Solution
The function $f$ has a <simple pole> of residue $1/2$ at every $z=k\pi i$, because $(e^{2z}-1)'=2$ there. At $z=0$, the term $1/(2z)$ in $g$ cancels the principal part of $f$. For $1\leq k\leq m$, the summand
$$
\frac z{z^2+\pi^2k^2}
$$
has residue $1/2$ at each of $z=\pm k\pi i$, so it cancels the corresponding pole of $f$. Therefore all possible singularities of $F$ in $|z|<(m+1)\pi$ are <removable singularities>: they are $0$ and $\pm k\pi i$ for $1\leq k\leq m$. \b[After filling them in, $F$ is analytic throughout that disc.]
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