Evaluate $\int_{\gamma} \frac{1}{z^2 + z +1} \,dz$, where $\gamma$ is the circle $|z|=2$.

Question

The question is given as an exercises in section 2.4 of Marsden’s Basic Complex Analysis. Based on the context, I am assuming there is a solution using Cauchy’s Integral formula, however I don’t see how to make it work. I tried factoring $z^2 + z + 1$, but because the roots are “inside” of our curve $\gamma$, it is not immediately obvious that Cauchy’s integral formula can be applied. We also cannot apply Cauchy’s Theorem, since $\gamma$ is not homotopic to a point on the region of analyticity of $\frac{1}{z^2 + z +1}$.

Any ideas?

Answer 1

Here are some remarks to get you started and almost finished. $$\frac{1}{z^2+z+1}=\frac{1}{(z-r_1)(z-r_2)}$$ $$=\frac{1}{(z-r_1)((z-r_1)+(r_1-r_2))}$$ $$=\frac{1}{(z-r_1)(r_1-r_2)(1+\frac{z-r_1}{r_1-r_2})}$$ $$=\frac{1-\frac{z-r_1}{r_1-r_2}+(\frac{z-r_1}{r_1-r_2})^2-...}{(z-r_1)(r_1-r_2)}.$$ In the Laurent series, the coefficient of $(z-r_1)^{-1}$ is $\frac{1}{r_1-r_2}$. I'll let you worry about convergence and what happens at the other singularity.

Answer 2

HINT: Using deformation principle for multiply-connected domain$$\oint_C \frac{1}{z^2+z+1}dz=\oint_{\gamma_1}\frac{1/(z-\omega^2)}{z-\omega}dz + \oint_{\gamma_2}\frac{1/(z-\omega)}{z-\omega^2}dz$$
where $\gamma_1: |z-\omega|=r_1$ and $\gamma_2: |z-\omega|=r_2$ are small circles, each enclosing just one singularity.