For what simple piecewise smooth loops is this integral zero? I'm trying to solve the following problem:

For what simple piecewise smooth loops $\gamma$  does the following equation hold:
  $$\int\limits_\gamma \frac{1}{z^2 + z + 1}  \, \mathrm{d}z= 0$$

I'm allowed to appeal to Cauchy's integral theorem, and I have determined that the roots of the denominator are $\frac{-1}{2} \pm \sqrt{3}i$. But I am not exactly sure how to proceed, or even what the ideal solution would be. For example, I know that the integral is $0$ over any loop that is homotopic to a constant loop (at a point in the domain of the function), but I'm not sure of what else I should say.

The roots of $z^{3}-1$ are $e^{k(2\pi i/3)}$ for $k=0,1,2$ and
$$
            z^3-1 = (z-1)(z^2+z+1).
$$
So $(z-e^{2\pi i/3})(z-e^{-2\pi i/3})=z^2+z+1$, and
$$
     \frac{1}{(z-e^{2\pi i/3})(z-e^{-2\pi i/3})}=\frac{A}{z-e^{2\pi i/3}}+\frac{B}{z-e^{-2\pi i/3}},
$$
where $A$, $B$ are easily seen to satisfy $A=-B$.
Assuming $\gamma$ does not pass through the points $e^{\pm 2\pi i/3}$, you get zero iff the curve $\gamma$ has a winding number around $e^{-2\pi i/3}$ that is the same as the winding number around $e^{2\pi i/3}$.
