I am trying to get from

$$\frac{z^7 + 1}{z^2(z^4+1)}$$


$$\frac{1}{z^2} + z - \frac{z+z^2}{1+z^4}.$$

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?

  • $\begingroup$ First decompose it into normal partial fractions. Then add certain terms together. $\endgroup$ – Hui Yu Jul 17 '13 at 13:44
  • $\begingroup$ I don't think that's what's done. The context is integrating the initial term over a closed curve. Thus the purpose of doing this strange decomposition is to say "aha, the terms $z$ and $\frac{1}{z^2}$ both have anti-derivatives, so their integral is zero. Then the next step is to decompose the remaining term in order to integrate it. I know how a PFD is done, but I'm asking myself how the above is done directly. $\endgroup$ – Pascal Engeler Jul 17 '13 at 13:49
  • $\begingroup$ Yes. That is how I would do it. Do you know how to calculate the coefficients of a Taylor series expansion? Or calculate the residue? This is just a higher order residue. $\endgroup$ – Hui Yu Jul 17 '13 at 13:51
  • $\begingroup$ Yes and yes, but I don't see the context... $\endgroup$ – Pascal Engeler Jul 17 '13 at 13:59

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 $\frac{z^7+1}{z^2(z^4+1)}$.


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.

  • $\begingroup$ Is this guaranteed to "work", as in $ha$ and $hb$ having lower degree than $h$? $\endgroup$ – ShreevatsaR Jul 17 '13 at 13:53
  • $\begingroup$ This is how partial fractions should be taught in school. $\endgroup$ – OR. Jul 17 '13 at 13:53
  • 1
    $\begingroup$ @ShreevatsaR Read the last sentence in the answer. $\endgroup$ – OR. Jul 17 '13 at 13:54
  • $\begingroup$ @RGB: Read the last word in my comment. :-) I read the answer to the end, which is why I asked: note that I asked for lower degree than $h$, not lower degree than $f$ and $g$. To answer my own question: I guess the answer is no, it is not guaranteed (e.g. start with $h = 1$), but that doesn't matter. $\endgroup$ – ShreevatsaR Jul 17 '13 at 14:07
  • $\begingroup$ I assumed you were asking the wrong question since it is clear you can't expect $ha$ having lower degree than $h$ unless $a=0$. $\endgroup$ – OR. Jul 17 '13 at 14:10


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


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