I am stuck on the following problem :

$\,\,\,\,$*Problem*$\quad$The residue of an entire function at $\infty$ is $0$.

Solution: True. This follows from the definition of the residue at $\infty$ together with the Cauchy-Goursat Theorem. Another way to see this is to take the Taylor expansion for $f$ at $0$ $\displaystyle f(z)=\sum_{n=0}^\infty a_nz^n$ and replace $z$ with $1/z$ to get the Laurent expansion for $f(1/z)$ at $0$: $\displaystyle f(z)=\sum_{n=-\infty}^0a_{-n}z^n.$ Then we have that the residue at $\infty$ is given by the negative coefficient of the $1/z$ term of $$ \dfrac1{z^2}f\left(\dfrac1z\right)=\sum_{n=-\infty}^{-2}a_{-n-2}z^n, $$ which is clearly $0$.

I am having trouble to understand the last few lines (in italic) in the given solution. Can someone give lucid explanation? Thanks and regards to all.


The thing is that functions do not have residues, but rather differentials have residues. This is something which can be quite confusing in a first complex analysis class. The "residue of a function" is not invariant under a change of local parameter, but the residue of a differential is. For this reason, what is usually called the "residue at $0$ of $f(z)$" is actually the residue at $0$ of $f(z)dz$.

When you change the coordinate from $z$ to $w=1/z$, the differential $dz$ is transformed into $-dw/w^2$, which explains the change of sign and the extra factor. Thus,

$$f(z)dz = \frac{-1}{w^2} f(1/w) dw.$$

The "residue of $f$ at $\infty$" is the residue at $0$ of $\frac{-1}{w^2} f(1/w) dw$.

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