# Calculate residue of pole with order 5

I am trying to evaluate the residue of $$\frac{(z^6-1)^2}{z^5(2z^4-5z^2+2)}$$ at z=0.

Is there a way to do this without having to do the long derivative calculation?

I know we have the formula $$Res(f,0)=\lim_{z\to0}\frac{1}{(5-1)!}\frac{\mathrm{d}^{5-1}f}{\mathrm{d}z^{5-1}}z^5f(z)$$. But finding the 5th derivative of this function would take a long time and be pretty messy. My first try was to find the residues at the other poles and at infinity since all the residues must sum to zero, but when I tried to do this the residue at infinity ended up being the same as the residue at zero.

I've also considered finding the Laurent expansion but the Partial Fraction Decomposition is also long.

• A possibility would be to find enough of the early terms of the Laurent expansion of $1/(z^5(2z^4-5z^2+2))$. It begins from the $z^{-5}$-term. Therefore taking the numerator $(z^6-1)^2=1-2z^6+z^{12}$ into account you see that you only need to include the constant term. The next term, $-2z^6$, has high enough degree to cancel the fifth degree poly already. Dec 26, 2021 at 19:47
• (Cont'd) You only need to figure out the terms in the Taylor expansion $$\frac1{2z^4-5z^2+2}=a_0+a_2z^2+a_4z^4+\cdots.$$ Unless I made a mistake actually only $a_4$ is important. Dec 26, 2021 at 19:50

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.

• More or less what I suggested in a comment. But written more precisly. +1 Dec 26, 2021 at 19:54

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.)

Here's a purely algebraic proof. The residue of the original expression is the coefficient of $$z^4$$ in the Taylor series of $$\frac{1}{2z^4-5z^2+2}$$ $$={1\over2}(1-{5\over2}z^2+z^4)^{-1}$$ $$={1\over2}(1+{5\over2}z^2-z^4+({5\over2}z^2-z^4)^2+\dots)$$ $$={1\over2}(1+{5\over2}z^2+(-1+{25\over4})z^4+\dots)$$ So the answer is $$21\over8$$.