# Calculating the residue of the following function

1. May 5, 2012

### Septim

Greetings everyone,

I have a question about calculating the residue of the following function at z=0 point. I am new to complex analysis so I would like to hear all the possible methods. I tried to expand the function I was dealing with in Laurent series and was unable to do so. How can I expand the following function in Laurent series about z=0 so that I can read of the residue at z=0?

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

Many thanks

Last edited by a moderator: May 5, 2012
2. May 5, 2012

### DonAntonio

Re: Septim

I corrected your LaTeX:
$$\sin\pi z=\pi z-\frac{\pi^3z^3}{6}+...\Longrightarrow \frac{1}{\sin\pi z}=\frac{1}{\pi z\left(1-\left(\frac{\pi z}{\sqrt{6}}\right)^2+...\right)}=\frac{1}{\pi z}\left(1+\frac{\pi^2 z^2}{6}+\frac{\pi^4 z^4}{36}+...\right)\Longrightarrow \frac{1}{z^2\sin\pi z}=\frac{1}{\pi z^3}+\frac{\pi}{6z}+...\Longrightarrow\,\,res.= \frac{\pi}{6}$$

DonAntonio

3. May 6, 2012

### Septim

Re: Septim

Thanks you but how can I calculate the terms in the series in a systematic way ?

4. May 6, 2012

### DonAntonio

Re: Septim

Well, just as I showed you...! Of course, in this case we're contented with very few summands of $\,\,\sin\pi z\,\,$ , as we're interested

only in the coefficient of $\,\,\frac{1}{z}\,\,$ , but in general you can go as much as you want.

Of course, another way would be to develop the MacClaurin (in fact, Laurent) series directly for $\,\,\frac{1}{\sin\pi z}\,\,$ , but who's time for all those calculations...?

DonAntonio

5. May 6, 2012

### Septim

Re: Septim

But does neglecting some terms in the denominator not introduce a big error ? It may also affect the coefficient of 1/z, how can you be sure that this is not the case ? I am unfamiliar with handling the series btw do not get me wrong.

6. May 6, 2012

### DonAntonio

Re: Septim

Check it carefully and you'll get convinced that there's no error at all in that (perhaps in some calculations is, though), since not writing

more summands of the Taylor series just "cuts" the summands with higher powers of z in the final outcome, and those don't interest us

to calculate the function's residue (in this case, at zero).

DonAntonio

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