Complex integral with $\int_{+\partial D}\frac{\sin\left(\frac{1}{z}\right)\cos\left(\frac{1}{z-2}\right)}{z-5}\mathrm{dz}$

Question

Hi guys in this integral

$$\int_{+\partial D}\dfrac{\sin\left(\dfrac{1}{z}\right)\cos\left(\dfrac{1}{z-2}\right)}{z-5}\mathrm{dz}$$

where $D=\{z\in\mathbb{C}:|z|<3\}$, is $z=5$ a pole, and are $z=2$ and $z=0$ essential singularities ?

If the domain is $|z|<3$, is the integral zero?

Answer 1

Hint You have two options,

1. To rotate one way and consider residues on inside of circle, then you will get two essential singularities which in general can be nasty to evaluate (but not always!).
2. Or you can flip direction of contour and instead consider outside of circle. Then you will get the "nice" pole $z=5$ inside of your area, and potentially a pole at complex infinity will also be "inside" since you turned the circle "inside out" so to speak.

It was a while ago I did this, so I don't remember all the details you will need to refer to to be allowed to do nr 2.