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?
On
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.
On
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 $$F(z)=\frac{z^3-1}{z^2(z^4+1)}+\frac{1}{z^2}=\frac{z+z^2}{1+z^4}$$
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 $\frac{z^7+1}{z^2(z^4+1)}$.