The function sine z over z, sine z over z has

an isolated singularity of the origin because that's we're dividing by zero.

Otherwise the function is analytic.

Sine z itself has a series representation centered at the origin,

namely z minus z cubed over 3 factorial plus z to the fifth over 5 factorial and

so forth, which you see right here.

If I divide that by Z, I find myself with 1- 1 over 3 factorial,

times z squared plus 1 over 5 factorial, times Z to the fourth, and so forth.

That's the Laurent series representation for sine z over z, which we use to show

that z equals 0 is actually a removable singularity for this function f.

Therefore, the residue of f at 0 is equal to 0.

Next, let's consider residues at simple poles.

Recall that z0 is a simple pole.

If the a- 1 term in the Laurent series representation was non-zero,

but all other k's for k's less than negative one are equal to zero.

So there's exactly one negative power of z- z0 in another Laurent series.

So f has the form of a minus one divided by z- z0.

And then all non negative powers of z minus z0.