Question

How do you write the partial fraction decomposition of the rational expression `1/(x^2+x+1)`?

Answer 1

There is no breakdown into simpler fractions with Real coefficients, but with Complex coefficients we find:

`1/(x^2+x+1)= (sqrt(3)i)/(3(x+1/2-sqrt(3)/2i))-(sqrt(3)i)/(3(x+1/2+sqrt(3)/2i))`

Explanation:

Note that `(x^2+x+1)(x-1) = x^3-1`

The zeros of `x^2+x+1` are the non-Real Complex cube roots of `1`, `omega` and `omega^2 = bar(omega)`

`omega = -1/2+sqrt(3)/2i` is the primitive Complex cube root of `1`.

`1/(x^2+x+1)`

`= A/(x-omega^2) + B/(x-omega)`

`=(A(x-omega)+B(x-omega^2))/(x^2+x+1)`

`=((A+B)x - (Aomega+Bomega^2))/(x^2+x+1)`

Equating coefficients:

`{ (A+B=0), (Aomega+Bomega^2 = -1) :}`

Substituting `B=-A` in the second of these equations we get:

`Aomega-Aomega^2=-1`

Rearranging, we find:

`A = 1/(omega^2-omega) = 1/(-sqrt(3)i) = sqrt(3)/3i`

So:

`1/(x^2+x+1)`

`= (sqrt(3)i)/(3(x-omega^2))-(sqrt(3)i)/(3(x-omega))`

`= (sqrt(3)i)/(3(x+1/2-sqrt(3)/2i))-(sqrt(3)i)/(3(x+1/2+sqrt(3)/2i))`