Residue of $f(z)=\frac{1}{z^2\,\sin(\pi z)}$ in $z=0$

I want to calculate the residue of the following function without using to the development in series of Laurent.

$$f(z)=\frac{1}{z^2\,\sin(\pi z)}$$

$z=0$ is a triple pole

$$\mathrm{Res}[f,0]=\lim_{z->0}\frac{1}{3!}\frac{\mathrm{d^2} }{\mathrm{d} x^2}\left [ \frac{1}{z^2\,\sin(\pi z)} z^3\right ]$$

Usually in these cases I developed before making the limit the trigonometric function. Until now I had only considered simple poles in these cases and I always took the first term of the development. in $z=0$ $$\sin(\pi z)=\pi t-\frac{\pi^3t^3}{6}+o(t^4)$$

If I replace only the first term the result is $0$, and if I replace the second the result is right $\frac{\pi}{6}$. $$\mathrm{Res}[f,0]=\lim_{z->0}\frac{1}{3!}\frac{\mathrm{d^2} }{\mathrm{d} x^2}\left [ \frac{1}{z^2\,\pi z-\frac{\pi^3z^3}{6}} z^3\right ]= \frac{\pi}{6}$$

$$\mathrm{Res}[f,0]=\lim_{z->0}\frac{1}{3!}\frac{\mathrm{d^2} }{\mathrm{d} x^2}\left [ \frac{1}{z^2\,\pi z} z^3\right ]= 0$$

why the two limits are different?

What is the rule? I need to replace up to the order of the pole or higher. Considering simple poles was fine always replace the first, this is why I make this assumption.

Someone can help me.

Thank you so much.

• Are you willing to grant that $\sin z/z$ is analytic at $0$ if it takes the value $1$ there? – Lubin Apr 1 '17 at 23:54
• Your question is unclear to me. – Zaid Alyafeai Apr 2 '17 at 1:32
• I want to calculate the residue in $z=0$ using the formula with the limit of the multiple poles. – Stefano Barone Apr 2 '17 at 5:49

Your formula is not true : $$\mathrm{Res}[f,0]=\lim_{z->0}\frac{1}{2!}\frac{\mathrm{d^2} }{\mathrm{d} z^2}\left [ \frac{1}{z^2\,\sin(\pi z)} z^3\right ]$$

So : $$\frac{\mathrm{d^2} }{\mathrm{d} z^2}\left [ \frac{1}{z^2\,\sin(\pi z)} z^3\right ]=\frac{\mathrm{d^2} }{\mathrm{d} z^2}\frac{z}{\sin(\pi z)}=\frac{\mathrm{d} }{\mathrm{d} z}\frac{\sin(\pi z)-z^2\pi\cos(\pi z)}{\sin^2(\pi z)}=z (\pi^2 \csc^3(\pi z) + π^2 \cot^2(\pi z) \csc(\pi z)) - 2 \pi \cot(\pi z) \csc(\pi z)$$

So : $$\mathrm{Res}[f,0]=\lim_{z->0}\frac{1}{2!}[z (\pi^2 \csc^3(\pi z) + π^2 \cot^2(\pi z) \csc(\pi z)) - 2 \pi \cot(\pi z) \csc(\pi z)]$$

By using $\cot(\pi z)=\frac{1}{\pi z}-\frac{\pi z}{3}+o(z)$ and $\csc(\pi z)=\frac{1}{\pi z}+\frac{\pi z}{6}+o(z)$, we get : $$\lim_{z->0}\frac{1}{2!}[z (\pi^2 \csc^3(\pi z) + π^2 \cot^2(\pi z) \csc(\pi z)) - 2 \pi \cot(\pi z) \csc(\pi z)]=\frac{\pi}{6}$$

$$\mathrm{Res}[f,0]=\frac{\pi}{6}$$

You could also write : $$\frac{z}{\sin(\pi z)}=\frac{1}{\pi-\frac{\pi^3z^2}{6}+o(z^3)}$$

$$\frac{\mathrm{d^2} }{\mathrm{d} z^2}\frac{z}{\sin(\pi z)}=\frac{\mathrm{d^2} }{\mathrm{d} z^2}\frac{1}{\pi-\frac{\pi^3z^2}{6}+o(z^3)}=\frac{\pi}{3}+o(1)$$

So :

$$\mathrm{Res}[f,0]=\lim_{z->0}\frac{1}{2!}\frac{\mathrm{d^2} }{\mathrm{d} z^2}\left [ \frac{1}{z^2\,\sin(\pi z)} z^3\right ]=\lim_{z->0}\frac{1}{2!}(\frac{\pi}{3}+o(1))=\frac{\pi}{6}$$

• Thank you so much for the very accurate answer. I wanted to avoid use derivatives and I prefer to write directly a series expansion. I just want to figure out how many terms of development I must put. Is right to reach the degree equal to the pole number? – Stefano Barone Apr 2 '17 at 19:51
• You have to reach at least this degree but I can't ensure you that it will be enough sometimes you need to take a higher degree because you still could have a limit of the form $\frac{0}{0}$ , but I don't have an example in mind. – Bérénice Apr 2 '17 at 20:20

A pure Laurent series approach. Consider \begin{aligned} \frac{1}{z^2\sin(\pi z)}&=\frac{1}{z^2}\frac{1}{\pi z-\frac{\pi^3z^3}{6}+O(z^5)}\\ &=\frac{1}{\pi z^3}\frac{1}{1-\frac{\pi^2z^2}{6}+O(z^4)}\\ &=\frac{1}{\pi z^3}\left[\frac{\pi^2z^2}{6}+O(z^4)\right]\\ &=\color\red{\frac{\pi}{6}}\frac{1}{z}+\boxed{O(z)}. \end{aligned}

All you need is to make sure that the remaining term has nothing to do with $$1/z$$, and that's why approximation $$\sin(\pi z)\sim \pi z+O(z^3)$$ fails.