Calculate $\int_{|z|=2}\frac{1}{z^3+z^2+z+1}$

Question

I want to calculate $$\int_{|z|=2}\frac{1}{z^3+z^2+z+1}$$ where $|z|=2$ is run counterclockwise.

What I did: $$\int_{|z|=2}\frac{1}{z^3+z^2+z+1}=\int_{|z|=2}\frac{1}{(z+1)(z^2+1)} $$ Using Cauchy's integral formula

$$f(-1)=\frac{1}{2\pi i}\int_{|z|=2} \frac{f(z)}{z+1}dz$$

where $f(z)=\frac{1}{z^2+1}$

Therefore $$\int_{|z|=2}\frac{1}{z^3+z^2+z+1}=\pi i$$

However I realized that $f$ is not holomorphic in $1,-i,i$ (the roots of $z^3+z^2+z+1$) and this made me doubt my result. What is the correct answer?

Answer 1

Use partial fractions to write $$\frac{1}{(z+1)(z^2+1)}$$ as a sum of three functions each with a simple pole interior to the circle $|z| = 2$.

Answer 2

The poles $-1,\,\pm i$ all have modulus $1<2$, so we get contributions from all of them. The poles are also all first-order. The $z=-1$ residue is $\frac{1}{(-1)^2+1}=\frac12$. The $z=\pm i$ residue is $\frac{1}{(1\pm i)(\pm 2i)}=\frac{-1\mp i}{4}$. So the integral is $2\pi i(\frac12+\frac{-1}{2})=0$.