Inverse Z-transform with a complex root The z-transform of a signal is 
$$ X(z)=\frac{1}{z^2+z+1}$$
I attempted to solve for the the inverse z-transform by decomposing the denominator into complex roots, $\alpha$ and $\alpha^\ast$, to get 
$$\frac{1}{z^2+z+1} = \frac{A}{z-\alpha}+\frac{B}{z-\alpha^\ast}=\frac{\frac{-i\sqrt{3}}{3}}{z-\alpha}+\frac{\frac{i\sqrt{3}}{3}}{z-\alpha^\ast}$$
for $$\alpha = e^{2\pi i /3}  \ \ \text{and}  \ \ \alpha^\ast = e^{-2\pi i /3}$$
And ultimately this leads to a long process of manipulating a series to relate it to the definition of the z-tranform. I was wondering if there was a faster or easier way to solve this problem, perhaps with a more direct series expansion
 A: What you may do next is write $$\frac{A}{z-\alpha} + \frac{B}{z-\alpha^*} = \frac{A}{-\alpha}\frac{1}{1-\frac{z}{\alpha}} + \frac{B}{-\alpha^*}\frac{1}{1-\frac{z}{\alpha^*}} = \frac{A}{-\alpha}\sum_{n=0}^\infty \left( \frac{1}{\alpha^n} z^n \right) + \frac{B}{-\alpha^*}\sum_{n=0}^\infty \left( \frac{1}{\alpha^{*n}} z^n \right)$$
$$= \sum_{n=0}^\infty\left[ \frac{A}{-\alpha}\left( \frac{1}{\alpha^n} \right) +\frac{B}{-\alpha^*} \left( \frac{1}{\alpha^{*n}}\right)\right]z^n = \sum_{n=0}^\infty x_n z^n$$
Thus $$x_n = \left[ \frac{A}{-\alpha}\left( \frac{1}{\alpha^n} \right) +\frac{B}{-\alpha^*} \left( \frac{1}{\alpha^{*n}}\right)\right]$$
A: Notice that $(1-z)(1+z+z^2)=1-z^3$. (That's a trick worth noting whenever you're working with $1+z+z^2+\ldots z^n$.)
So $1/(z^2+z+1)$ is actually $(1-z)/(1-z^3)$.
$1/(1-z^3)$ is the sum of a geometric progression ie. $1+z^3+z^6+\ldots$.
So the final result is $1+z^3+z^6+\ldots-z(1+z^3+z^6+\ldots)$
So if we write $X(z)=\sum_{n=0}^\infty a_n z^n$ we have
$a_n=\cases{1&$n=0$ mod $3$\cr -1&$n=1$ mod $3$\cr0&$n=2$ mod 3}$
It's a surprisingly simple result!
