# Principal Part of Laurent series' expansion of $f(z)=\frac{\sin(z^3)}{(1-\cos z)^3}$

I need to calculate principal part of the Laurent series expansion of $f$ at $z_0=0$ with $$f(z)=\frac{\sin(z^3)}{(1-\cos z)^3}$$

I can see that $f$ has a pole of order 3 at $z_0=0$ , and also that $f$ is an odd function, therefore the expansion will be $$a_{-3}z^{-3}+a_{-1}z^{-1}$$ but after this I am stuck. I tried using some trigonomteric identities, but they seem to make things more complicated.

I calculated $a_{-1}$ using the residue formula and it seems to be 2 , but I think there has to be a simpler way to deal with this.

Any help or advice towards the soluution would be welcome.

• Would the Taylor expansion help you ? – Claude Leibovici Aug 30 '14 at 8:36
• @ClaudeLeibovici The Taylor expansion of sinz comes in hand, but $(1- \cos z)^3$ results in its Taylor expansion raised in the third – helplessKirk Aug 30 '14 at 8:46
• Let me put it in an answer. – Claude Leibovici Aug 30 '14 at 8:48

Hoping that the Taylor series could help you to some extent, let us start with $$\sin(y)=y-\frac{y^3}{6}+O\left(y^4\right)$$ So $$\sin(z^3)=z^3-\frac{z^9}{6}+O\left(z^{10}\right)$$ Now $$\cos(z)=1-\frac{z^2}{2}+\frac{z^4}{24}+O\left(z^5\right)$$ $$1-\cos(z)=\frac{z^2}{2}-\frac{z^4}{24}+O\left(z^5\right)$$ So $$\frac{\sin(z^3)}{(1-\cos z)^3}=\frac{8}{z^3}+\frac{2}{z}+\frac{4 z}{15}+O\left(z^3\right)$$

• This is very simple! Thank you! I didn't know that I could do this though. I thought that I had to calculate $(1-cosz)^3$ by taking the cauchy product of the series representation of $(1-cosz)$ with itself times 3 formally. – helplessKirk Aug 30 '14 at 9:04
• You are very welcome ! Binomial theorem was the trick ! Cheers :-) – Claude Leibovici Aug 30 '14 at 9:20
• @ClaudeLeibovici How do you take the quotient between the two series? – hjhjhj57 Oct 3 '14 at 6:11
• @Javier. There are different ways : perform the long division (tedious) or search for the expansion of $x^n f(x)$, $n$ being such that there is a finite limit at $x=0$ or generalized Taylor. – Claude Leibovici Oct 3 '14 at 6:31
• For example, for $\frac{1}{ax^2+O(x^4)}$ what'd you suggest? Should I expand the Taylor series of $\frac{x^2}{ax^2+O(x^4)}$ and then divide by $x^2$? – hjhjhj57 Oct 3 '14 at 6:47

Notice that $$1-\cos(z)=2\sin^{2}(z/2)$$ so that

$$f(z)=\dfrac{\sin(z^3)}{8\sin^6(z/2)}.$$

Thus, by means of the fundamental L'Hopital involving the sine function, there holds $$\displaystyle \lim_{z\rightarrow a}f(z)(z-a)^6=\frac{\sin(a^3)}{\cos^2(a/2)}$$ whenever $$a$$ is a zero of $$\sin^{2}(z/2)$$.

Therefore, the poles of $$f$$ are of order $$6$$.