Calculate residue at essential singularity
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?
residue-calculus
edited Jul 2 '13 at 16:31
Andrea Orta
Willem BeekWillem Beek
$\begingroup$ $\frac{1}{z-3} = -\frac13 \frac{1}{1-(z/3)}$ expand the latter into a power series. Multiply. I'm not sure whether you get something nice or ugly. $\endgroup$ – Daniel Fischer♦ Jul 2 '13 at 15:59
$\begingroup$ 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 ! $\endgroup$ – Willem Beek Jul 2 '13 at 16:24
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 $-\sin \left ( \frac{1}{3} \right )$.
EDIT: As Daniel Li has pointed out, there is something wrong with my first two paragraphs. In fact, my choice of notation was quite poor ! I did not intend the $R$ of the second paragraph to be necessarily the same as the $R$ of the first paragraph. I only meant to convey on the one hand, a generic "punctured disk" type annulus centred at $0$, and on the other, a generic "proper" annulus (i.e. with strictly positive inner radius) centred at $0$. However, I certainly should have clarified this by not re-using the letter $R$, within the same argument ! In fact, if there is any overlap between the two annuli, then the two Laurent series must coincide, so in order to have two distinct Laurent series, we would actually need that the $r$ of the second paragraph be not less than the $R$ of the first paragraph.
The problem in the answer of Cocopuffs, I believe, is that they try to use the Laurent series in the annulus $|z|>3$, where they should instead use the Laurent series in the annulus $0<|z|<3$. The function has isolated singularities at $0$ and at $3$, and is otherwise analytic.
SimonSimon
$\begingroup$ 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. $\endgroup$ – Daniel Fischer♦ Jun 24 '14 at 10:05
$\begingroup$ That is a very nice elegant way to do it :) $\endgroup$ – Simon Jun 24 '14 at 10:48
$\begingroup$ 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; $\endgroup$ – Willem Beek Nov 16 '15 at 12:32
$\begingroup$ @Muno The Riemann sphere, $\mathbb{C}\cup \{\infty\}$. $\endgroup$ – Daniel Fischer♦ Apr 12 '17 at 14:36
$\begingroup$ @Simon I'm not sure that your first two paragraphs are correct. I wrote a question in this post: math.stackexchange.com/questions/3321549/… $\endgroup$ – Daniel Li Aug 13 '19 at 0:09
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 $z = 0$.
CocopuffsCocopuffs
$\begingroup$ 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? $\endgroup$ – Willem Beek Jul 2 '13 at 16:29
$\begingroup$ @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 $\endgroup$ – Cocopuffs Jul 2 '13 at 16:35
$\begingroup$ Ok thank you! I can't upvote our answer because I don't have 15 rep yet, just so that you know $\endgroup$ – Willem Beek Jul 2 '13 at 16:38
$\begingroup$ 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) $\endgroup$ – Daniel Fischer♦ Jun 24 '14 at 10:01
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