Complex-Residue Theorem

**WannaBe** · May 3rd 2010, 02:51 AM

How can I calculate the residue at 0 of the next function:
$f(z)=\frac{1}{z^2 \sin z}$ ??
Hope you'll be able to help me

Thanks a lot!

**HallsofIvy** · May 3rd 2010, 02:57 AM

By calculating the residue "at infinity" of $\frac{1}{f(z)}= z^2 sin(z)$ .

That is, write sin(z) as a Taylor's series, then multiply by $z^2$ . That, of course, will be a power series with only positive powers. Now invert the powers getting a Laurent series with only negative powers. The residue is the coefficient of the $z^{-1}$ term.

**piglet** · May 3rd 2010, 03:07 AM

what hallsofivy said. I thought i had another approach but on reflection don't think it would be correct

**WannaBe** · May 3rd 2010, 03:25 AM

Originally Posted by HallsofIvy

By calculating the residue "at infinity" of $\frac{1}{f(z)}= z^2 sin(z)$ .

That is, write sin(z) as a Taylor's series, then multiply by $z^2$ . That, of course, will be a power series with only positive powers. Now invert the powers getting a Laurent series with only negative powers. The residue is the coefficient of the $z^{-1}$ term.

Here is my try:
$z^2 sin(z)=z^2 (z-\frac{z^3}{3!} + \frac{z^5}{5!} -...)=$
$z^3 - \frac{z^5}{3!} +\frac{z^7}{5!} -...$
Hence:
$\frac{1}{z^2 sin(z) } = \frac{1}{z^3 - \frac{z^5}{3!} +\frac{z^7}{5!} -...} =$
$= \frac{1}{\sum_{n=3}^{\infty}\frac{z^n}{(n-2)!} }$
and this is where I got stuck in my calculation... We can't just right it with negative powers because all the sum is in negative power...

Hope you'll be able to correct my misunderstanding and help me continue...

Thanks a lot !

**chisigma** · May 3rd 2010, 05:55 AM

Let's start with the series expansion of the function...

$f(z)= \frac{z}{\sin z} = a_{0} + a_{1}\cdot z + a_{2}\cdot z^{2} + \dots$ (1)

... from the well know result...

$\frac{\sin z}{z} = \sum_{n=0}^{\infty} (-1)^{n}\cdot \frac{z^{2n}}{(2n+1)!} = 1 - \frac{z^{2}}{6} + \frac{z^{4}}{120} - \dots$ (2)

The most 'spontaneous' approach is to impose...

$(a_{0} + a_{1}\cdot z + a_{2}\cdot z^{2} + \dots) \cdot (1 - \frac{z^{2}}{6} + \frac{z^{4}}{120} - \dots)= 1$ (3)

... and that permits us to compute the $a_{n}$ in recursive way...

$a_{0}=1$

$a_{1} =0$

$a_{2} = \frac{a_{0}}{6} = \frac{1}{6}$

$a_{3} =0$

$a_{4} = \frac{1}{36} - \frac{1}{120} = \frac{7}{360}$

... and so one...

Now we can write...

$\frac{z}{\sin z} = 1 + \frac{z^{2}}{6} + \frac{7}{360}\cdot z^{4} + \dots$ (4)

... from which it follows...

$\frac{1}{z^{2}\cdot \sin z} = z^{-3} + \frac{z^{-1}}{6} + \frac{7}{360}\cdot z + \dots$ (5)

... so that the residue of $\frac{1}{z^{2}\cdot \sin z}$ in $z=0$ is $\frac{1}{6}$ ...

Kind regards

$\chi$ $\sigma$

**WannaBe** · May 3rd 2010, 10:32 AM

Actually I didn't understand your last result:
$ \frac{z}{\sin z} = 1 + \frac{z^{2}}{6} + \frac{7}{360}\cdot z^{4} + \dots $
It means that:
$ \frac{\frac{1}{z^2}}{\sin z} = 1 + \frac{(\frac{1}{z^2})^{2}}{6} + \frac{(\frac{1}{z^2})^4}{360}\cdot + \dots $

Which means the Residue is excatly 0, isn't it?

Thanks!

**chisigma** · May 3rd 2010, 09:23 PM

Originally Posted by WannaBe

Actually I didn't understand your last result:
$ \frac{z}{\sin z} = 1 + \frac{z^{2}}{6} + \frac{7}{360}\cdot z^{4} + \dots $
It means that:
$ \frac{\frac{1}{z^2}}{\sin z} = 1 + \frac{(\frac{1}{z^2})^{2}}{6} + \frac{(\frac{1}{z^2})^4}{360}\cdot + \dots $

Which means the Residue is excatly 0, isn't it?

Thanks!

If You start from $f(z) = \frac{z}{\sin z}$ , then is...

$\frac{1}{z^{2}\cdot \sin z} = \frac{f(z)}{z^{3}}$ (1)

Your Laurent expression is very interesting however, but the correct expression is...

$\frac{1}{z^{2}\cdot \sin \frac{1}{z^{2}}} = 1 + \frac{1}{6}\cdot z^{-4} + \frac{7}{360}\cdot z^{-8} + \dots$ (2)

The function represented in (2) has in $z=0$ an essential singularity, that is not the case for $\frac{1}{z^{2}\cdot \sin z}$ ...

Kind regards

$\chi$ $\sigma$

**WannaBe** · May 3rd 2010, 11:33 PM

Thanks...Sry for my mistake...Completely understandable!

Thread: Complex-Residue Theorem

LinkBack

Thread Tools

Display

Complex-Residue Theorem

Similar Math Help Forum Discussions

2 more questions in Residue Theorem-Complex

Complex Analysis - Residue Theorem well annoying

Residue Theorem Integrals (Complex Analysis)

Residue Theorem/Complex Methods Question

complex integrals--residue theorem

Search Tags