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.

  • 4
    $\begingroup$ $\frac{1}{z-3} = -\frac13 \frac{1}{1-(z/3)}$ expand the latter into a power series. Multiply. I'm not sure whether you get something nice or ugly. $\endgroup$ 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 ! $\endgroup$ 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.

  • 10
    $\begingroup$ 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. $\endgroup$ Jun 24 '14 at 10:05
  • 1
    $\begingroup$ That is a very nice elegant way to do it :) $\endgroup$
    – Simon
    Jun 24 '14 at 10:48
  • 1
    $\begingroup$ 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; $\endgroup$ Nov 16 '15 at 12:32
  • 1
    $\begingroup$ @Muno The Riemann sphere, $\mathbb{C}\cup \{\infty\}$. $\endgroup$ Apr 12 '17 at 14:36
  • 1
    $\begingroup$ @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 $z = 0$.

  • $\begingroup$ 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? $\endgroup$ 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 $\endgroup$ Jul 2 '13 at 16:38
  • 6
    $\begingroup$ 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) $\endgroup$ Jun 24 '14 at 10:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.