How to evaluate $\int_{|z|=2} \frac{1}{z^2+z+1} dz$

Question

I found the poles of $\displaystyle f(z) = \frac{1}{z^2+z+1}$ given by the solutions for $z^2 + z + 1$, $ \displaystyle z_1 = -\frac{1}{2}+i\frac{\sqrt{3}}{2}$ and $\displaystyle z_2 = -\frac{1}{2}-i\frac{\sqrt{3}}{2}$.

But when I use the Residue's Theorem, I get an indetermination. For example, for $z_1$ $$\displaystyle Res\left(\frac{1}{z^2+z+1}, -\frac{1}{2}+i\frac{\sqrt{3}}{2}\right)= \lim_{z \to -\frac{1}{2}+i\frac{\sqrt{3}}{2}} \left((z - \left(-\frac{1}{2}+i\frac{\sqrt{3}}{2}\right)\right)\left(\frac{1}{z^2+z+1}\right) $$$$= (0)(\frac{1}{0})$$

Is this the way I should treat this problem or should I try with something else?

Answer 1

Hint:

Just observe that \begin{align*} (z - (-\frac{1}{2}+i\frac{\sqrt{3}}{2}))(\frac{1}{z^2+z+1})&=(z-z_1)\times\frac 1{(z-z_1)(z-z_2)}\\ &=\frac 1{z-z_2}\qquad\text{for all }z\neq z_1 \end{align*} So, the limit you are looking for is $$\lim_{z\to z_1}\left[(z - z_1)\times\frac{1}{(z-z_1)(z-z_2)}\right]=\lim_{z\to z_1}\frac{1}{z-z_2}=\frac{1}{z_1-z_2}=\frac{1}{i\sqrt 3}=-\frac 1{\sqrt 3}i$$

Answer 2

Hint:

$$z^2+z+1=\left (z- \left(-\frac{1}{2}+i\frac{\sqrt{3}}{2}\right)\right )\left(z-\left( -\frac{1}{2}-i\frac{\sqrt{3}}{2}\right)\right)$$

Answer 3

Factorize: $z^2+z+1=\left(z-\frac{-1+i\sqrt3}{2}\right)\left(z-\frac{-1-i\sqrt3}{2}\right)$ $$\text{Re}\left(\frac{1}{z^2+z+1}, \frac{-1+i\sqrt3}{2}\right)$$$$=\left[\left(z-\frac{-1+i\sqrt3}{2}\right)\frac{1}{\left(z-\frac{-1+i\sqrt3}{2}\right)\left(z-\frac{-1-i\sqrt3}{2}\right)}\right]_{z=\dfrac{-1+i\sqrt3}{2}}$$ $$=\left[\frac{1}{\left(z-\frac{-1-i\sqrt3}{2}\right)}\right]_{z=\dfrac{-1+i\sqrt3}{2}}$$ Similarly, $$\text{Re}\left(\frac{1}{z^2+z+1}, \frac{-1-i\sqrt3}{2}\right)=\left[\frac{1}{\left(z-\frac{-1+i\sqrt3}{2}\right)}\right]_{z=\dfrac{-1-i\sqrt3}{2}}$$