We proved the following theorem in my Complex Variables class the other day.
Let $R(z)=\frac{P(z)}{Q(z)}$ be a rational function with $deg(P)<deg(Q)$ and let the roots of $Q(z)$ be $z_{i}$. Then $$R(z)=\sum_{i=1}^{n}P_{z_{i}}(z)$$ where $P_{z_{i}}(z)$ is the principal part of $R(z)$ at $z_{i}$.
This is a nice theorem, making partial fraction expansion very intuitive in many cases, including for real numbers. Recall that $f(z)=\frac{h(z)}{z-z_{0}}$ and $h(z)$ is analytic at $z_{0}$ then the principal part of $f(z)$ at $z_{0}$ is $\frac{h(z_{0})}{z-z_{0}}$. So for example, $$\frac{1}{z^{4}-1}=\frac{1}{z-1}\frac{1}{z+1}\frac{1}{z-i}\frac{1}{z+i}$$ is now easily expandable into: $$\frac{1}{4}\left(\frac{1}{z-1}-\frac{1}{z+1}+\frac{1}{z-i}-\frac{1}{z+i}\right)$$
Which does save time when compared to the traditional way of using undetermined coefficients. However for homework I had to find the partial fraction decomposition of $$\frac{1}{(z-1)^{2}(z^{4}-1)}$$ which has a third order pole at $z=1$. I am finding it very hard to find the principal part here in a nice way. I can figure out the contributions from the other poles like above, as they are all simple. I can find the other principal part with traditional methods, or by using a computer. But I can't figure out a nice way to compute it using the theorem above. Help?
Best Answer
It's convenient to write $z = 1 + t$, so the pole you're interested in is at $t=0$. Then $$\eqalign{\dfrac{1}{(z-1)^2(z^4-1)} &= \dfrac{1}{t^2(4 t + 6 t^2 + 4 t^3 + t^4)}\cr &= \dfrac{1}{4t^3(1 + 3t/2 + t^2+t^3/4)}\cr &= \dfrac{1}{4t^3}\left(1 - \left(\frac{3t}{2} + t^2 + \frac{t^3}{4}\right) + \left(\frac{3t}{2} + t^2 + \frac{t^3}{4}\right)^2 - \ldots\right)}$$