Calculate residue at essential singularity

Question

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.

Answer 1

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.

Answer 2

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 $z = 0$.