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Find the residue at $z=0$ for complex function $1/(z^2\sin z$)

I know $z=0$ is a pole of order 3 but can't seem to calculate the residue value for it.

Américo Tavares
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DJ_
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2 Answers2

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Try some power series expansion around zero:

$$\sin z=z-\frac{z^3}{6}+\ldots\implies\frac{1}{\sin z}=\frac{1}{z\left(1-\frac{z^2}{6}+\ldots\right)}=\frac{1}{z}\left(1+\frac{z^2}{6}+\frac{z^4}{120}+\ldots\right)\implies$$

$$\frac{1}{z^2\sin z}=\frac{1}{z^3}\left(1+\frac{z^2}{6}+\ldots\right)=\frac{1}{z^3}+\frac{1}{6z}+\ldots$$

DonAntonio
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    How did it become from 1-$z^2$/6 + ... to 1+$z^2$/6+... – DJ_ Apr 20 '13 at 00:19
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    @DJ_ : $$\frac{1}{1-z}=1+z+z^2+z^3+\ldots;,;;|z|<1$$ Note that we're only interested in the corresponding powers of $,z,$ that at the end will give the value of the residue (i.e., in the coefficient of $,\frac{1}{z},$ in the corresponding Laurent Series), and above we even got more: we actually showed $,z=0,$ indeed is a pole of order $;3$...:) – DonAntonio Apr 20 '13 at 00:24
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You can reference the result for a pole of order $3$ at $z=0$:

$$\text{Res}_{z=0} \frac{1}{z^2 \sin{z}} = \frac{1}{2!} \lim_{z \rightarrow 0} \left[\frac{d^2}{dz^2} \frac{z}{\sin{z}}\right]$$

$$\frac{d}{dz} \frac{z}{\sin{z}} = \frac{\sin{z} - z \cos{z}}{\sin^2{z}} $$

$$\begin{align}\frac{d^2}{dz^2} \frac{z}{\sin{z}} &= z \left(\csc ^3{z}+\cot ^2{z} \csc{z}\right)-2 \cot{z} \csc{z}\\ &=\left (\frac{1}{z^3} + \frac{1}{2} + \frac{1}{z^3} - \frac{1}{2}\right ) - \frac{2}{z^2} + \frac{1}{3} + O(z)\\ &= \frac{1}{3}\end{align}$$

Therefore the residue is $1/6$.

Ron Gordon
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    Unless there's some slick trick, calculating that second derivative's limit (or substitution $,z=0,$) looks awful, l'Hospital included... – DonAntonio Apr 20 '13 at 00:26
  • I wouldn't say it's awful. The second derivative is my limit of casual hand calculation. – Ron Gordon Apr 20 '13 at 00:36
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    Well, you had to resource to power series to make it nicer: for me that's, if not nasty, at least pretty ugly, in particular because you first had to differentiate and then use a little power series. I'd rather use the series from the beginning. Thanks for showing the work. – DonAntonio Apr 20 '13 at 00:46
  • @DonAntonio: you are very right about that. It is not always the case and sometimes it does work out to something simple. But not here. – Ron Gordon Apr 20 '13 at 00:57