# The poles and residues of $f(z)=\frac{\sin z}{z^2}- \frac{\cos z}{z}$

Let $f(z)=\frac{\sin z}{z^2}- \frac{\cos z}{z}$ then

1. $f$ has a pole of order $2$ at $0$,

2. $f$ has a simple pole at $0$,

3. $\int_{|z|=1}f(z)=0$ anticlockwise,

4. residue of $f$ at $0$ is $-2\pi i$,

I am not able to figure out which of the statement is correct or false. I think only 4 is correct by residue theorem.

-
What are your thoughts on the matter so far? Also, if this is homework, please add the appropriate tag. – Brett Frankel Jun 4 '12 at 21:09
Not a home work, well I think only 4 is correct by residue theorem. – La Belle Noiseuse Jun 4 '12 at 21:15

It all becomes clear once you obtain the Laurent expansion $$\frac{\sin z}{z^2}=\frac{1}{z^2} \left(z-\frac{z^3}{3!}+O\left(z^5\right)\right)=\frac{1}{z}-\frac{z}{6}+O\left(z^3\right)$$ $$\frac{\cos z}{z}=\frac{1}{z} \left(1-\frac{z^2}{2!}+O\left(z^4\right)\right)=\frac{1}{z}-\frac{z}{2}+O\left(z^3\right)$$ $$f(z)=\frac{z}{3}+O\left(z^3\right)$$ Hence $f(z)$ is analytic and (3) is true.