What is the complex partial fraction decomposition of $\frac{1}{z^2 - z + 1}$

Question

I need this result to compute a residue. I haven't been successful so far.

What I have tried: I have tried decomposing $\frac{1}{z^2 - z + 1} = \frac{A + Bi}{z - \omega} + \frac{C + Di}{z + \omega}$ where $\omega$ is the cube root of unity. I didn't get anything from this method.

Answer 1

Hint 1 Given that the degree of the polynomial is $2$, it is fundamental that

$$z^2-z+1=(z-\alpha)(z-\beta)$$

Hint 2

$$\left\{\begin{align*} &\alpha\beta=1\\ &\alpha+\beta=1 \end{align*}\right.$$

(Why?)

Hint 3

$$\frac{1}{(z-\alpha)(z-\beta)}= \frac{A}{(z-\alpha)}+\frac{B}{(z-\beta)}$$

for some $A$ and $B$.

Answer 2

By the quadratic formula the zeroes of $z^2-z+1$ are $\frac12(1\pm i\sqrt3)$, the two non-real cube roots of $1$. Let $\omega=\frac12(1+ i\sqrt3)$; the other root is $\overline\omega$, the complex conjugate of $\omega$. Then $z^2-z+1=$ $(z-\omega)(z-\overline\omega)$.

Now just set up the usual partial fractions computation:

$$\frac1{z^2-z+1}=\frac{u}{z-\omega}+\frac{v}{z-\overline\omega}\;,$$ so $u(z-\overline\omega)+v(z-\omega)=1$, and you have the system $$\left\{\begin{align*}&u+v=0\\&u\overline\omega+v\omega=-1\;.\end{align*}\right.$$ That ought to be pretty straightforward to solve.

Answer 3

An alternate starting point:

$$ 1-z+z^2 = \frac{z^3+1}{z+1}. $$