How to find the residues of (tan(z))/z at its singularities

Thanks for the help!!

Clearly z = 0 is a singularity. However it is removable, because
lim(z→0) tan(z)/z = 1 by L'Hopital's Rule. (So, tan(z)/z has a Maclaurin series at z = 0.)

Hence, z = 0 yields residue 0 (since there is no 1/z terms in the Laurent series at z = 0).
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Next, tan z = sin z/cos z has simple poles when cos z = 0
==> z = π/2 + πk for any integer k.

(That these are simple can be checked by noting that (d/dz) (tan(z)/z) {at z = π/2 + πk} is nonzero for any integer k.)

Computing residues:
lim(z→ π/2 + πk) (z - (π/2 + πk)) tan(z)/z
= lim(z→ π/2 + πk) ((z - (π/2 + πk)) / cot z) * lim(z→ π/2 + πk) 1/z
= lim(z→ π/2 + πk) 1 / (-csc^2(z)) * lim(z→ π/2 + πk) 1/z, by L'Hopital's Rule
= lim(z→ π/2 + πk) -sin^2(z) * lim(z→ π/2 + πk) 1/z
= -1 * 1/(π/2 + πk)
= -1/(π/2 + πk).

I hope this helps!