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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?

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It's convenient to write $z = 1 + t$, so the pole you're interested in is at $t=0$. Then $$\eqalign{\dfrac{1}{(z-1)^2(z^4-1)} &= \dfrac{1}{t^2(4 t + 6 t^2 + 4 t^3 + t^4)}\cr &= \dfrac{1}{4t^3(1 + 3t/2 + t^2+t^3/4)}\cr &= \dfrac{1}{4t^3}\left(1 - \left(\frac{3t}{2} + t^2 + \frac{t^3}{4}\right) + \left(\frac{3t}{2} + t^2 + \frac{t^3}{4}\right)^2 - \ldots\right)}$$

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  • $\begingroup$ I don't see how to use this Robert. The pole at $t=0$ is not essential so the principal part should terminate unlike your expression? $\endgroup$ Commented Oct 18, 2013 at 0:46
  • $\begingroup$ The principal part is just the terms in negative powers of $t$. Anything coming from $\dfrac{1}{4t^3} \left(\dfrac{3t}{2}+t^2 + \dfrac{t^3}{4}\right)^p$ for $p \ge 3$ can't give you negative powers of $t$. $\endgroup$ Commented Oct 18, 2013 at 2:08
  • $\begingroup$ And so by our theorem above all of these should cancel each other? $\endgroup$ Commented Oct 18, 2013 at 2:14
  • $\begingroup$ ... or rather they should give the terms from the other poles. $\endgroup$ Commented Oct 18, 2013 at 5:58
  • $\begingroup$ I apologize, but I feel like we are talking on two different wavelengths here. The principal part of a function that is analytic except at a given point is the part of the Laurent series at this given point with terms of the form $1/z^{n}, n>0$. So as the pole in question is of order 3, there should only be three terms in the principal part. Or do you disagree? $\endgroup$ Commented Oct 18, 2013 at 6:45

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