Partial fraction expansion of a simple quotient.

Question

Consider partial fraction expansion of $$ \frac{1}{z^n-1}=\sum_{k=0}^{n-1}\frac{c_{nk}}{z-z_{nk}}, $$ where $z_{nk}$ are primitive roots of 1: $z_{nk}=e^{\frac{2\pi i}{n}k}$.

What is the general expression for the coefficients $c_{nk}$?

Answer 1

The point here is that both sides have a pole of multiplicity $1$ at $z_{n,k}$, so after multiplying both sides by $z-z_{n,k}$ you can remove the singularity. At $z_{n,k}$ only $c_{n,k}$ will contribute to the result on the right hand side. Thus

$$c_{n,k}=\lim_{z \to z_{n,k}} \frac{z-z_{n,k}}{z^n-1}=\lim_{z \to z_{n,k}} \frac{1}{nz^{n-1}}=\frac{1}{n} z_{n,k}.$$

To make sure I didn't screw it up, consider $n=2$ where $z_{2,0}=1$, $z_{2,1}=-1$, $c_{2,0}=1/2$, $c_{2,1}=-1/2$, which is consistent with the above.

Incidentally, this setup can be used as is to compute the partial fraction expansion of $\frac{p(z)}{q(z)}$ whenever $q$ has distinct complex roots. When there are repeated roots, something similar works (you can use this to compute the coefficient of $1/(z-z_0)^m$ where $m$ is the multiplicity, then subtract off that term, compute the coefficient of $1/(z-z_0)^{m-1}$, etc.)