1. ## Complex-Residue Theorem

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!

2. 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.

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

4. 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 !

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

$\chi$ $\sigma$

6. 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!

7. 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

$\chi$ $\sigma$

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