Strange Partial Fractions Decomposition I am trying to get from
$$\frac{z^7 + 1}{z^2(z^4+1)}$$
to
$$\frac{1}{z^2} + z - \frac{z+z^2}{1+z^4}.$$
The author did this by doing a partial fractions decomposition. I don't see how, however.. If I compute the partial fractions decomposition, I first find the roots of the denominator, but that's not what's done here. What he does is something that he calls "partial" partial fractions decomposition.
Any help?
 A: Indeed, it is possible to do a version of partial fractions without using linear factors: whenever we have _relatively_prime_ polynomials $f,g$, we can use the Euclidean algorithm for polynomials (with coefficients in a field such as $\mathbb R$) to find two other polynomials $a,b$ such that $af+bg=1$. Then for any polynomial $h$
$$
{h\over fg} \;=\; {h\cdot 1\over fg} \;=\; {h(af+bg)\over fg} \;=\; {ha\over g}+{hb\over f}
$$
And of course one can divide-with-remainder to make the numerators of smaller degree than the denominators.
A: This is another way of doing what is in Paul Garret answer, avoiding to compute more than we are looking for. 
You get the polynomial part $z$ by doing long division of $z^7+1$ by $z^2(z^4+1)$. 
You get the principal part corresponding to the factor $z^2$ in the denominator by computing the first two steps of long division of $z^7+1$ by $(z^4+1)$ but organizing the terms in decreasing degrees before dividing. The first two coefficients you get are the coefficients of $\frac{A}{z^2}+\frac{B}{z}$ in that order. The other fraction you can get subtracting $\frac{1}{z^2}+z$ from $\frac{z^7+1}{z^2(z^4+1)}$.
A: Hint
First by Euclidean divsion we have
$$\frac{z^7 + 1}{z^2(z^4+1)}=z-\frac{z^3-1}{z^2(z^4+1)}$$
Now we know that the partial fractional decompostion takes the form
$$\frac{z^3-1}{z^2(z^4+1)}=\frac{a}{z^2}+\frac{b}{z}+F(z)$$
where $0$ isn't a pole for the fraction $F$ and with routine calculus we find $a=-1$ and $b=0$.
Finally to find $F$ just do
$$F(z)=\frac{z^3-1}{z^2(z^4+1)}+\frac{1}{z^2}=\frac{z+z^2}{1+z^4}$$
