Is there an easy way to evaluate this complex integral without partial decomposition?

Question

$$\int_{C_2(0)}\frac{1}{z^2+z+1}\ dz$$

Where $C_2(0)$ is the open ball of radius 2, centred at 0, in the complex plane.

Using partial fractions and Cauchy's integral formula, I show'd the integral is equal to 0. However, the partial fractions part seemed unnecessarily long-winded to me.

Did I miss a trick by approaching the problem using partial fractions?

Answer 1

You might like this idea: That integral, by Cauchy's theorem, equals

$$\tag 1 \int_{\{|z|=R\}} \frac{1}{z^2+z+1}\, dz$$

for any $R>2.$ But the M-L estimate on this integral, as $R\to \infty,$ is bounded above by $2\pi R/(R^2-R-1) \to 0.$ Thus $(1)$ must equal $0.$

Answer 2

The poles of this function are $p_\pm = -1/2\pm \sqrt{3}/2i$. Since both are simple, the residues may be calculated by $\lim_{z \rightarrow p}(z-p)f(x)$ which give $\pm \sqrt{3}i$. The reside theorem then implies the integral is zero, since both poles have winding number 1.