= Solution
The denominator factors as $z^2+iz+6=(z-2i)(z+3i)$, with no cancellation against the numerator. Thus there are
$$
\boxed{\text{simple poles at }z=2i,-3i;\qquad\text{a simple zero at }z=4.}
$$
To inspect infinity, put $w=1/z$:
$$
f(1/w)=\frac{w-4w^2}{1+iw+6w^2}=w+O(w^2).
$$
Hence infinity is a <removable singularity at infinity>, with limiting value zero. After this natural extension to the <Riemann sphere> it is also a \b[simple zero]. There are no other zeros or singularities.
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