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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.

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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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