# Trouble calculating residues

• June 25th 2009, 04:08 AM
vsywod
Trouble calculating residues
Hello,

I'ld like to calculate the reside of $f(z):=\frac{z^{n-1}}{\sin^n(z)}$ at $z_0=0$.

Using the definition I get $\text{res}_{0}f = \frac{1}{2\pi i} \int_\kappa \frac{z^{n-1}}{\sin^n(z)} dz$, but I cant compute this integral.

My next idea was to develop the function into a Laurent-Series and use the coefficent $a_{-1}$ but thats basically the same calculation.

I would like to get some help on how to solve this and similar problems.

Thank you!
• June 25th 2009, 05:05 AM
shawsend
Quote:

Originally Posted by vsywod
Hello,

I'ld like to calculate the reside of $f(z):=\frac{z^{n-1}}{\sin^n(z)}$ at $z_0=0$.

Using the definition I get $\text{res}_{0}f = \frac{1}{2\pi i} \int_\kappa \frac{z^{n-1}}{\sin^n(z)} dz$, but I cant compute this integral.

My next idea was to develop the function into a Laurent-Series and use the coefficent $a_{-1}$ but thats basically the same calculation.

I would like to get some help on how to solve this and similar problems.

Thank you!

It's a simple pole right? $lim_{z\to 0} zf(z)=1$ and that's the residue although you can use the formula $(k+1)\frac{g^{(k)}}{h^{(k+1)}}$. The integral formula should only be used by those who love Complex Analysis and want dessert.
• June 25th 2009, 05:37 AM
chisigma
Remembering the 'infinite product'...

$\sin z = z \cdot \prod_{k=1}^{\infty} (1-\frac{z^{2}}{k^{2} \pi^{2}})$ (1)

... we obtain...

$\sin^{n} z = z^{n} \cdot \prod_{k=1}^{\infty} (1-\frac{z^{2}}{k^{2} \pi^{2}})^{n}$ (2)

... so that is...

$f(z)= \frac{1}{z\cdot \prod_{k=1}^{\infty} (1-\frac{z^{2}}{k^{2} \pi^{2}})^{n}}$ (3)

... and finally ...

$Res_{z=0} f(z) = \lim_{z \rightarrow 0} z\cdot f(z) = 1$ (4)

Kind regards

$\chi$ $\sigma$
• June 25th 2009, 05:40 AM
vsywod
Quote:

Originally Posted by shawsend
It's a simple pole right?

Well, actually I'm not sure it's a simple pole. I guess it is, but how can I show it?
I'ld have to give a holomorphic function $g$ thats doesn't vanish in 0 and which satisfies
$f(z) = z^{-1} \cdot g(z)$.
This would mean $g(z) = \frac{z^n}{\sin^n(z)}$ which is clearly holomorphic outside of 0. But why is $\lim_{z\to 0} g(z) = 1$?
I think this is a stupid question, but my head is really spinning at the moment...
Thank you!
• June 25th 2009, 06:33 AM
shawsend
Quote:

Originally Posted by vsywod
Well, actually I'm not sure it's a simple pole. I guess it is, but how can I show it?
I'ld have to give a holomorphic function $g$ thats doesn't vanish in 0 and which satisfies
$f(z) = z^{-1} \cdot g(z)$.
This would mean $g(z) = \frac{z^n}{\sin^n(z)}$ which is clearly holomorphic outside of 0. But why is $\lim_{z\to 0} g(z) = 1$?
I think this is a stupid question, but my head is really spinning at the moment...
Thank you!

. . . you makin' me come to Mars ain't ya. You know how I hate Mars . . . where's my Calculus text? Let's see, the limit of a composite is the composite at the limit so $w^n$ is continuous at 0 and thus $\lim_{z\to 0}\frac{z^n}{\sin(z)^n}=\left(\lim_{z\to 0} \frac{z}{sin(z)}\right)^n=1$. I don't know. Is that ok?
• June 25th 2009, 08:07 AM
vsywod
Yes, I got it myself now, thank you :)