Solution
= Solution
The <Taylor expansions> give
$$
e^{iaz}\sin(bz)=(1+iaz+O(z^2))(bz+O(z^3))=bz+iabz^2+O(z^3),
$$
and hence $f(z)=b/z+iab+O(z)$. Since $b>0$, zero is \b[a <simple pole> with <residue> $b$]. It is not a <removable singularity>, despite the vanishing numerator, because the denominator has a zero of order two.