Let $z_1,\dots,z_n\in\mathbb{C}$ be different points and let $f:\mathbb{C}\setminus\{z_1,\dots,z_n\}\to\mathbb{C}$ be a holomorphic function.

Suppose that $\lim\limits_{z\to\infty}f(z)=0$. Prove that

$$\lim\limits_{z\to\infty}zf(z)=\sum\limits_{k=1}^n \operatorname{Res}_{z_k}(f)$$

Clearly, we can look at $\lim\limits_{z\to 0}\frac{1}{z}f(\frac{1}{z})$. I have read something about residue at infinity, but I don't see how to use it.


Let $R>0$ be a radius such that all the singularities $z_j$ are inside the disk $D(0,R)$. (Center $0$ and radius $R$) Hence by the residue theorem, one has

$$2i\pi \sum\limits_{k=1}^n \operatorname{Res}_{z_k}(f) = \int_{C(0,R)^+} f(z) dz.$$ Now perform a change of variable $z=1/w$ to get $$\int_{C(0,R)^+} f(z) dz = -\int_{C(0,1/R)^-} \frac{1}{w^2}f\left(\frac{1}{w}\right) dw= \int_{C(0,1/R)^+} \frac{1}{w^2}f\left(\frac{1}{w}\right) dw.$$ Since the $z_j$ were inside $D(0,R)$, it is clear that the $1/z_j$ are now outside $D(0,1/R)$. Hence, the only singularity of $\frac{1}{w^2}f \left( \frac{1}{w}\right)$ inside $D(0,1/R)$ is $0$. Hence, by the residue theorem again, one has $$\int_{C(0,1/R)^+} \frac{1}{w^2}f\left(\frac{1}{w}\right) dw = 2i\pi\operatorname{Res}_{0}\left(w \mapsto \frac{1}{w^2}f\left(\frac{1}{w}\right)\right).$$ If we put everything together, we get $$\sum\limits_{k=1}^n \operatorname{Res}_{z_k}(f) =\operatorname{Res}_{0}\left(w \mapsto \frac{1}{w^2}f\left(\frac{1}{w}\right)\right).$$ Now the only thing you have to prove is that $0$ is a simple pole of $w \mapsto \frac{1}{w^2}f\left(\frac{1}{w}\right).$ Indeed if you do this, you will have $$\operatorname{Res}_{0}\left(w \mapsto \frac{1}{w^2}f\left(\frac{1}{w}\right)\right) = \lim_{w\to 0} w \frac{1}{w^2}f\left(\frac{1}{w}\right)=\lim_{z\to \infty}zf(z)$$ and you are done. Are you able to do that ?

  • $\begingroup$ (+1) Really nice! $\endgroup$ – José Carlos Santos Jun 20 '18 at 9:49
  • $\begingroup$ Thanks, I understand it in full, except for the part that I need to handle... I mean I tried to use Taylor expansion for $f$ around $0$ but when using $\frac{1}{w^2}f(\frac{1}{w})$ I got a Laurent expansion with infinitely many negative coefficients. $\endgroup$ – user554578 Jun 20 '18 at 10:29
  • $\begingroup$ Ok. On $\mathbb{C} \backslash D(0,R)$, you can write $f(z)$ as a Laurent series right ? Then, you can use your hypothesis $\lim_{z\to \infty} f(z) = 0.$ This will imply that your Laurent series has a specific form, etc ... $\endgroup$ – C. Dubussy Jun 20 '18 at 11:32

Those are initial thoughts... Unfortunately not a solution

Consider the domain $D$ limited by a circle $C(R)$ centered at the origin (with clockwise orientation) and with a radius $R$ large enough in order for the disk $D^\prime$ it limits to contain $\{z_1,\dots,z_n\}$ and small circles $C_1, \dots, C_n$ (oriented counterclockwise) centered respectively on $z_1, \dots, z_n$ all contained in $D^\prime$. As $f$ is holomorphic in $D$, you have $$\begin{aligned} 0=\frac{1}{2i \pi}\int_{C(R)} f + \sum\limits_{k=1}^n \int_{C_n} f &= \frac{1}{2i \pi}\int_{C(R)} f+\sum\limits_{k=1}^n Res_{z_k}(f)\\ &= -R \int_0^{2 \pi} f(Re^{2i \pi t}) \ dt+ \sum\limits_{k=1}^n Res_{z_k}(f) \end{aligned}$$

As $\lim\limits_{z\to\infty}f(z)=0$, for all $\epsilon >0$ you can find $r >0$ such that for $\vert z \vert \ge r$ you have $\vert f(z) \vert \le \epsilon$. Hence if $R \ge r$...

I don't know how to move further from there ...


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