residue of this function at infinity

Question

How do I calculate the residue of $\dfrac{\sin(z)}{z}$ at infinity ? I tried to use Wikipedia definition for the case, $\lim_{|z|\rightarrow\infty}f(z)=0$ then $Res(f,\infty)=-\lim_{|z|\rightarrow\infty}z.f(z)$, but this gives $Res(\dfrac{\sin(z)}{z},\infty)=-\lim_{|z|\rightarrow\infty}z.\dfrac{\sin(z)}{z}=\sin(z)$ which is undefined, so is the residue undefined, what does that mean ?

Answer 1

Here, you don't have $\lim\limits_{\lvert z\rvert\to\infty} f(z) = 0$, so you need another way to compute the residue at $\infty$.

You can use the definition

$$\operatorname{Res}(f;\infty) = \operatorname{Res}(\tilde{f}; 0)$$

where $\tilde{f}(z) = -\frac{1}{z^2}f\left(\frac{1}{z}\right)$, or the fact that the sum of all residues of a function holomorphic on $\widehat{\mathbb{C}}\setminus \{z_1,\dotsc, z_k\}$ is always $0$.

The former yields

$$\operatorname{Res} \left(\frac{\sin z}{z}; \infty\right) = \operatorname{Res}\Biggl(-\frac{1}{z^2}\frac{\sin \frac{1}{z}}{1/z}; 0\Biggr) = \operatorname{Res}\Biggl( -\frac{1}{z}\sin \frac{1}{z}; 0\Biggr) = 0$$

using the Laurent expansion of $\sin \frac{1}{z}$ (or the fact that $\tilde{f}$ is an even function), and the latter yields $\operatorname{Res}\Bigl(\frac{\sin z}{z};\infty\Bigr) = 0$ since $\frac{\sin z}{z}$ is an entire function, and hence the residue at $\infty$ is the only residue of that function on the whole sphere, hence $0$.