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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.

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3 Answers 3

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.

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Thanks already solved it. – organgina Apr 10 '12 at 4:31
+1 If it doesn't feel like cheating, you can also find the values of $u$ and $v$ by using l'Hospital's rule: $$ u=\lim_{z\to\omega}\frac{z-\omega}{z^2-z+1}=\lim_{z\to\omega}\frac1{2z-1}\ldots$‌​$ – Jyrki Lahtonen Apr 10 '12 at 5:15
@Jyrki: Neat. I don’t think that I’ve seen anyone do that before. – Brian M. Scott Apr 10 '12 at 5:19

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


Hint 2

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


Hint 3

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

for some $A$ and $B$.

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An alternate starting point:

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

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