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I know you can calculate a residue at an essential singularity by just writing down the Laurent series and look at the coefficient of the $z^{-1}$ term, but what can you do if this isn't so easy?

For instance (a friend came up with this function): what is the residue at $z = 0$ of the function $\dfrac{\sin\left(\dfrac{1}{z}\right)}{z-3}$?

The Laurent series of the sine is $\displaystyle \frac{1}{z} - \frac{1}{6z^{3}} + \frac{1}{120z^{5}} - \cdots + \cdots$

but if you divide by $(z-3)$, you get $\displaystyle \frac{1}{z(z-3)} - \frac{1}{6(z-3)z^{3}} + \frac{1}{120(z-3)z^{5}}+\cdots$

Now the series isn't a series solely "around" $z$! How to proceed further? Or shouldn't you try to write down the Laurent series?

Many thanks.

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$\frac{1}{z-3} = -\frac13 \frac{1}{1-(z/3)}$ expand the latter into a power series. Multiply. I'm not sure whether you get something nice or ugly. – Daniel Fischer Jul 2 '13 at 15:59
Thank you very much @daniel !But you'll get an infinite sum you can't compute I believe.. you will get -$\frac{1}{3}$ ($\frac{1}{z}$ - $\frac{1}{6z^{3}}$ + $\frac{1}{120z^{5}}$ - ..)(1 + $\frac{z}{3}$ + $\frac{z^{2}}{9}$ + $\frac{z^{3}}{27}$ ..) since we're looking for the coëfficients of $z^{-1}$, we have to compute $\frac{1}{z}$ * 1 + $\frac{1}{6z^{3}}$ * $\frac{z^{2}}{9}$ + $\frac{1}{120z^{5}}$ * $\frac{z^{4}}{81}$ and so on, I think! That would be $\sum\limits_{i=1}^n \frac{(-1)^{i}}{i!*3^{2i-1}}$ ? Is this calculable? I wouldn't know how to do this ! – Willem Beek Jul 2 '13 at 16:24
up vote 7 down vote accepted

I think there are some mistakes here.

In fact the residue of $f(z)$ at an isolated singularity $z_0$ of $f$ is defined as the coefficient of the $(z-z_0)^{-1}$ term in the Laurent Series expansion of $f(z)$ in an annulus of the form $0 < |z-z_0|<R$ for some $R > 0$ or $R = \infty$.

If you have another Laurent Series for $f(z)$ which is valid in an annulus $r < |z-z_0|< R$ where $r > 0$, then it might differ from the first Laurent Series, and in particular the coefficient of $(z-z_0)^{-1}$ might be different, and hence not equal to the residue of $f(z)$ at $z_0$.

In this example, $\sin \left ( \frac{1}{z} \right )$ has Laurent series $\sum_{k=0}^{\infty} (-1)^k \frac{z^{-2k-1}}{(2k+1)!} = \frac{1}{z} - \frac{1}{3! z^3} + \frac{1}{5! z^5} - \ldots$ which is valid in the annulus $0 < |z| < \infty$, and for $1/(z-3)$ we have $\frac{1}{z-3} = -\frac{1}{3} \frac{1}{1 - \frac{z}{3}} = -\frac{1}{3} \sum_{k=0}^{\infty} \left (\frac{z}{3}\right )^k$ which is valid in the annulus $0 < |\frac{z}{3}| < 1$, i.e. $0 < |z| < 3$.

The product of these two Laurent series gives the Laurent series of the product of $\sin \left( \frac{1}{z} \right )$ and $1/(z-3)$ which is valid in the intersection of these two annuli, i.e. in the annulus $0 < |z| < 3$.

The coefficient of $z^{-1}$ in that product is given by $-\frac{1}{3} \sum_{k=0}^{\infty} \frac{(-1)^k}{9^k (2k+1)!}$ which we recognise as $-\sin \left( \frac{1}{3} \right )$. Thus the residue of $\frac{\sin \left ( \frac{1}{z} \right )}{z - 3}$ at $0$ is $-\sin \left ( \frac{1}{3} \right )$.

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You're right. Another method is to note that the sum of the residues in $\widehat{\mathbb{C}}$ is $0$, the other answer shows the residue in $\infty$ is $0$, and the residue in $3$ is easily seen to be $\sin \frac{1}{3}$, from which $$\operatorname{Res}\left(\frac{\sin \frac{1}{z}}{z-3}; 0\right) = -\sin \frac{1}{3}$$ follows. – Daniel Fischer Jun 24 '14 at 10:05
That is a very nice elegant way to do it :) – Simon Jun 24 '14 at 10:48
It is a long time ago, but still: thank you very much! At the time I didn't have the rep. yet to upvote your answer I believe; – Willem Beek Nov 16 '15 at 12:32
You are most welcome ! :) And thank you very much for the upvote ! – Simon Nov 16 '15 at 17:51

We can look at the power series $$\frac{1}{\frac{1}{z} - 3} = \frac{z}{1 - 3z} = z + 3z^2 + ...$$ and $$\sin(z) = z - \frac{1}{6}z^3 + ...$$ so $$\frac{\sin(z)}{\frac{1}{z} - 3} = \Big(z + 3z^2 + ...\Big)\Big(z - \frac{1}{6}z^3 +...\Big) = z^2 + 3z^3 + ...$$ and $$\frac{\sin(\frac{1}{z})}{z - 3} = \frac{1}{z^2} + \frac{3}{z^3} +...$$ has residue $0$ at $z = 0$.

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Thank you very much, I didn't come up with looking at $\frac{1}{\frac{1}{z} - 3}$ ! Do you think Daniels approach works as well when you're able to calculate the sum I described in my comment to his answer above? – Willem Beek Jul 2 '13 at 16:29
@WillemBeek It will work, but it will be easier if you write $\frac{1}{z-3}$ as a power series in $\frac{1}{z}$ so everything is in negative powers of $z$. That is essentially what this answer is – Cocopuffs Jul 2 '13 at 16:35
Ok thank you! I can't upvote our answer because I don't have 15 rep yet, just so that you know – Willem Beek Jul 2 '13 at 16:38
This is incorrect, you have computed (part of) the Laurent series of $\frac{\sin (1/z)}{z-3}$ in the annulus $3 < \lvert z\rvert < \infty$, you can't read off the residue in $0$ from that (directly). (cc @WillemBeek) – Daniel Fischer Jun 24 '14 at 10:01

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