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

Question

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 $\infty$.

Answer 1

$g(w) = f(1/w)$ 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 $\operatorname {Res}(f,\infty ) = -a_{-1}$.

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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.