# 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?

Many thanks.

• $\frac{1}{z-3} = -\frac13 \frac{1}{1-(z/3)}$ 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

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 $$0 < |z-z_0| for some $$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.

• 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
• @Muno The Riemann sphere, $\mathbb{C}\cup \{\infty\}$. – Daniel Fischer Apr 12 '17 at 14:36
• @Simon I'm not sure that your first two paragraphs are correct. I wrote a question in this post: math.stackexchange.com/questions/3321549/… – 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 $z = 0$.

• 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