Solution
= Solution
The denominator factors as $(z+2-i)(z+2+i)$. Neither root cancels against the numerator, and both are simple, so the finite singularities are
$$
\boxed{z=-2+i,\quad z=-2-i\quad\text{(simple poles).}}
$$
This is a <rational function>, so it has no <branch points>. At infinity, putting $w=1/z$ gives $w/(1+4w+5w^2)$, analytic at zero; infinity is therefore a <removable singularity> with value zero.