Computing $\int_{|z|=\frac{2}{3}}\frac{\sin z}{z e^{\frac{1}{z+2}}}$

Question

Compute the integral using residues $$\int_{|z|=\frac{2}{3}}\frac{\sin z}{z e^{\frac{1}{z+2}}}$$

I could not find any singular points of the function $\frac{\sin z}{z e^{\frac{1}{z+2}}} $so I decided to compute the integral using the residue at infinity.

Therefore I tried unsuccessfully to compute the Laurent series:

$\frac{\sin}{z}=1+\frac{z^2}{3!}+\frac{z^4}{5!}...$ and $e^{-\frac{1}{z+2}}=1-\frac{1}{(z+2)}+\frac{1}{(z+2)^2 2!}-\frac{1}{(z+2)^3 3!}...$

Then $\frac{\sin z}{z e^{\frac{1}{z+2}}}=(\frac{\sin}{z}=1+\frac{z^2}{3!}+\frac{z^4}{5!}...)(1-\frac{1}{z+2}+\frac{1}{(z+2)^2 2!}-\frac{1}{(z+2)^3 3!}...)$

However in order to compute $c_{-1}$ the coefficient of order $-1$ of the Laurent series would take me into infinite number of arithmetic operations.

Question:

How should I compute the integral above?

Thanks in advance!

Answer 1

Since your function has no non-removable singularities in the region $\left\{z\in\mathbb C\,\middle|\,\lvert z\rvert\leqslant\frac23\right\}$ (it has a singularity at $0$, but that's a removable one), your integral is equal to $0$.