Calculate residue of pole with order 5

Solution 1:

For practical calculations, I have never used that derivative formula to calculate any residue. What is useful is learning how to perform a limited Laurent expansion. One of the most useful "tricks" is the simple geometric series formula $\frac{1}{1-\zeta}=\sum_{n=0}^{\infty}\zeta^n$ provided $|\zeta|<1$. You should also be slightly comfortable with big Oh notation to keep track of the order of the various terms (I mean this is just an efficient tracker of how many terms you need to actually care about).

So: \begin{align} \frac{(z^6-1)^2}{z^5(2z^4-5z^2+2)}&=\frac{1}{2z^5}\cdot(z^6-1)^2\cdot\frac{1}{1-\left(\frac{5}{2}z^2-z^4\right)} \end{align} Notice how I have the $\frac{1}{z^5}$ term. Since I want the residue (i.e coefficient of $\frac{1}{z}$), I should expand terms atleast up to order $4$, meaning I should keep explicitly terms involving $z^4$; I can ignore $z^5$ or higher terms. The third term is precisely of the form $\frac{1}{1-\zeta}$ I mentioned previously, so we shall use this now: \begin{align} \frac{(z^6-1)^2}{z^5(2z^4-5z^2+2)}&=\frac{1}{2z^5}\cdot \left[1+\mathcal{O}(z^6)\right]\cdot \left[1+\left(\frac{5}{2}z^2-z^4\right)+ \left(\frac{5}{2}z^2-z^4\right)^2+ \mathcal{O}\left(\left(\frac{5}{2}z^2-z^4\right)^3\right)\right]\\ &=\frac{1}{2z^5}\cdot \left[1+\mathcal{O}(z^6)\right]\cdot \left[1+ \left(\frac{5}{2}z^2-z^4\right) + \frac{25z^4}{4}+\mathcal{O}(z^6)+\mathcal{O}(z^6)\right]\\ &=\frac{1}{2z^5}\cdot \left[1+\mathcal{O}(z^6)\right]\cdot \left[1+\frac{5}{2}z^2+\frac{21}{4}z^4+\mathcal{O}(z^6)\right]\\ &=\frac{1}{2z^5}\cdot \left[1+\frac{5}{2}z^2+\frac{21}{4}z^4+\mathcal{O}(z^6)\right] \end{align} I leave the final simplification to you to determine the coefficient of $\frac{1}{z}$, and hence the residue.

Note for example that in the first line, $(z^6-1)^2=z^{12}-2z^6+1$, but I simply shortened this to $1+\mathcal{O}(z^6)$, because that's more than enough (remember we only need to keep track of expansions up to $z^4$ term; anything higher will not contribute to the residue). Similarly, in the second step rather than fully expanding out $\left(\frac{5}{2}z^2-z^4\right)^2$, I write this briefly as $\frac{25}{4}z^4+\mathcal{O}(z^6)$, since that's more than enough. And so on.

Solution 2:

We will work in the power series ring $\Bbb Q[[z]]$ in the variable $z$ over $\Bbb Q$. Let us denote by $\operatorname{Coeff}_{z^k}\;f$ the coefficient in $z^k$ of some series $f$ in this ring. It may be useful to substitute $w=z^2$, then work in $\Bbb Q[[w]]$, but this is important only to have an easier typing. We may then work modulo $O(z^5)$, respectively modulo $O(w^3)$. Then the needed residue is: $$ \begin{aligned} &\operatorname{Res}_{z=0} \frac{(z^6-1)^2}{z^5(2z^4-5z^2+2)} \\ &\qquad= \operatorname{Coeff}_{z^4} \ \frac 12\cdot \frac{(z^6-1)^2}{z^4-\frac 52z^2+1}\\ &\qquad= \frac 12 \cdot \operatorname{Coeff}_{w^2} \ \frac{(w^3-1)^2}{w^2-\frac 52w +1}\\ &\qquad= \frac 12 \cdot \operatorname{Coeff}_{w^2} \ \frac{(w^3-1)^2}{w^2-\frac 52w +1} +O(w^3)\\ &\qquad= \frac 12 \cdot \operatorname{Coeff}_{w^2} \ \frac{(0-1)^2}{w^2-\frac 52w +1} +O(w^3)\\ &\qquad= \frac 16 \cdot \operatorname{Coeff}_{w^2} \ \left[\frac 4{1-2w} - \frac 1{1-\frac w2}\right] +O(w^3)\\ &\qquad= \frac 16 \cdot \operatorname{Coeff}_{w^2} \ \left[\ 4(1 + 2w + 4w^2+\dots) - \left(1+\frac w2+\frac {w^2}4+\dots\right)\ \right] +O(w^3)\\ &\qquad= \frac 16 \cdot \operatorname{Coeff}_{w^2} \left[\ (4-1) + \left(8-\frac 12\right)w+ \left(16-\frac 14\right)w^2+\dots \ \right]+O(w^3)\\ &\qquad= \frac 16 \cdot \left(16-\frac 14\right) = \color{blue}{\frac{21}8}\ . \end{aligned} $$ (The computation was done so that we could finally have the Laurent exansion in zero, but then the numerator $(w^3-1)^2$ has to be considered more carefully.)

Computer check, here sage:

sage: R.<z> = PowerSeriesRing(QQ)
sage: (z^6-1)^2 / z^5 / (2*z^4 - 5*z^2 + 2) + O(z^6)
1/2*z^-5 + 5/4*z^-3 + 21/8*z^-1 + 69/16*z + 261/32*z^3 + 1029/64*z^5 + O(z^6)

(This may be not wanted, but it is just a check, and it may be important to know this can be easily done with computational aid.)