Residue at infinity of a function which approaches a finite limit at infinity

Question

Show that if $f(z)$ tends to a finite limit as $z$ tends to infinity, then the residue of $f(z)$ at infinity is $\lim_{z \rightarrow \infty} z^2 f'(z)$.

I know that the residue at infinity is given by $$ R(\infty) = -{\rm Residue\ of\ } \frac{1}{z^2}f\left(\frac{1}{z}\right), $$ but I'm not sure how to use this result. Any hints on how to proceed would be much appreciated.

Answer 1

If $f(z)$ is analytic for $|z|>R$ for some $R$, and if $\lim_{z\to \infty}f(z)=c\ne 0$, then $f(z)$ can be written in terms of the Laurent Series

$$f(z)=\sum_{n=0}^\infty a_nz^{-n} \tag 1$$

for $|z|>R$, where $a_0=c$.

The residue at infinity of $f(z)$ is given by

$$\text{Res}\left(f(z),z=\infty\right)=-\text{Res}\left(\frac{f(1/z)}{z^2},z=0\right) \tag 2$$

Using $(1)$ and $(2)$ reveals

$$\begin{align} \text{Res}\left(f(z),z=\infty\right)&=-\text{Res}\left(\sum_{n=0}^\infty a_n z^{n-2},z=0\right)\\\\ &=-a_1 \tag 3 \end{align}$$

Finally, note that

$$\lim_{z\to \infty}(z^2f'(z))=-\lim_{z\to \infty}\sum_{n=0}^\infty na_nz^{1-n}=-a_1 \tag 4$$

whence equating $(3)$ and $(4)$ reveals

$$\bbox[5px,border:2px solid #C0A000]{\text{Res}\left(f(z),z=\infty\right)=\lim_{z\to \infty}(z^2f'(z))}$$

as was to be shown!

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NOTE:

If $a_0=c=0$, then $\lim_{z\to \infty}zf(z)=a_1$ and therefore

$$\text{Res}\left(f(z),z=\infty\right)=-\lim_{z\to \infty}(zf(z))$$