# Understanding the Laurent expansion of a meromorphic function about $\infty$.

Suppose $$f:\hat{\mathbb{C}}\to\hat{\mathbb{C}}$$ were meromorphic, and suppose $$f$$ has a pole at $$\infty$$. I'm trying to understand the Laurent series of $$f$$ about $$\infty$$. By definition, $$f$$ has a pole at $$\infty$$ if $$f(1/z)$$ has a pole at $$0$$. If we let $$w=1/z$$, on an annulus around $$0$$, we have $$f(w)=\sum_{j=-n}^\infty a_jw^j=\sum_{j=-n}^\infty a_j(1/z)^j.$$ Thus, $$f(z)=\sum_{j=-n}^\infty a_jz^j$$ I'm wondering if this is the correct construction of the Laurent expansion of $$f$$ about infinity. My suspicion is raised because this looks like the expansion of $$f$$ about $$0$$, not $$\infty$$.

$$g(w) = f(1/w)$$ has a pole at $$w=0$$, so $$g(w) = \sum_{j=-n}^\infty a_jw^j$$ for $$0 < |w| < R$$ with some $$R > 0$$. Then $$f(z) = g(1/z) = \sum_{j=-n}^\infty a_j z^{-j} = \sum_{j=-\infty}^n a_{-j} z^j$$ for $$1/R < |z| < \infty$$, and that is the Laurent expansion of $$f$$ for the pole at infinity: A power series in $$1/z$$ plus the “principal part” which is a polynomial without constant term.
The residue at infinity is defined as $$\operatorname {Res}(f,\infty )=-\operatorname {Res}\left({1 \over z^{2}}f\left({1 \over z}\right),0\right) \, .$$ For $$f(z) = \sum_{j=-\infty}^n a_{-j} z^j$$ this is $$\operatorname {Res}(f,\infty ) = -a_{-1}$$.
You got the wrong result because $$f(w)=\sum_{j=-n}^\infty a_j w^j$$ holds for “small” $$w$$, so you cannot simply substitute $$w=1/z$$ by $$z$$ in that identity.
• The Laurent expansion of $f$ has an infinite number of negative indexed terms? I thought that a function has a pole at a point if and only if its Laurent expansion had a finite number of negative indexed terms? Commented Apr 26 at 7:02
• @TyPerkins: That is only true for poles at some $z \in \Bbb C$, not for poles at $\infty$. Commented Apr 26 at 7:04
• In the resulting Laurent expansion, do we consider the "principal" part of the expansion to then be the finite terms for which the exponent of $z$ is positive (and so the residue is still $a_{-1}$? Or is the principal part of the expansion still considered the negative indexed terms, and our residue becomes $a_1$? Commented Apr 26 at 7:15