Complex curved integral: $\int_{\gamma(t)}\frac{1}{z+2}dz$, with: $\gamma(t)=\cos(3t)+i5\sin(5t)$ where $t\in [-\pi,\pi]$.

Question

I am stuck computing the following complex integral: $$\int_{\gamma(t)}\frac{1}{z+2}dz$$ with: $\gamma(t)=\cos(3t)+i5\sin(5t)$ where $t\in [-\pi,\pi]$.

What is the easiest way? How do you represent the curve?

The result is "$0$" but I do not know how it is done.

Would appreciate if somebody could explain it to me.

Answer 1

Hint. This is a Lissajous curve. Does this "twisted" closed curve (see picture) go around the pole $z=-2$? Note that $|\mbox{Re}(\gamma(t))|=|\cos(3t)|\leq 1$ and then use the Residue Theorem.