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    Meromorphic functions

    A meromorphic function on an open set is holomorphic except on a discrete set, at each point of which it has a pole.

    Meromorphic function

    Let \(\Omega\subseteq\CC\) be open. A function \(f\) is meromorphic on \(\Omega\) if there exists a set \(P\subseteq\Omega\) with no limit point in \(\Omega\) such that \(f\) is holomorphic on \(\Omega\setminus P\) and has a pole at each point of \(P\).

    A set \(P\subseteq\Omega\) with no limit point in \(\Omega\) meets each compact subset of \(\Omega\) in finitely many points, so \(P\) is finite or countable and can be listed as a finite or infinite sequence \((z_n)\).

    Meromorphic functions on the Riemann sphere are rational

    Meromorphic on $\CC$ with no essential singularity at $\infty$ means rational.Suggest a slogan

    Let \(f\) be meromorphic on \(\CC\) such that \(\infty\) is either a removable singularity or a pole of \(f\), in the sense of singularities at infinity. Then \(f\) is a rational function: \(f=P/Q\) for polynomials \(P,Q\in\CC[z]\) with \(Q\neq0\).

    Since \(\infty\) is removable or a pole, \(f\) has no poles in \(\ts{\abs z > R}\) for some \(R\), and the poles in the compact disc \(\abs z\leq R\) are isolated, hence finitely many: \(z_1,\ldots,z_N\). Let \(P_j\) be the principal part of \(f\) at \(z_j\), a polynomial in \(\inverseof{(z-z_j)}\) without constant term, and let \(P_\infty\) be the polynomial part of the Laurent series of \(f\) on \(\abs z>R\) with its constant term omitted, so that \(f - P_\infty\) is bounded near \(\infty\). Then \(g \coloneqq f - P_\infty - \sum_j P_j\) has only removable singularities in \(\CC\), so it extends to an entire function, and it is bounded near \(\infty\) because each \(P_j\to 0\) as \(z\to\infty\). By Liouville’s theorem \(g\) is constant, so \(f = g + P_\infty + \sum_j P_j\) is rational.

    Taylor expansion with holomorphic remainder

    After truncating a Taylor series, the remainder still factors holomorphically by the next power of $z-z_0$.Suggest a slogan

    Let \(\Omega\subseteq\CC\) be open, let \(z_0\in\Omega\), let \(f\) be analytic on \(\Omega\), and let \(n\geq1\). Then there is a function \(R_n\) analytic on \(\Omega\) such that for all \(z\in\Omega\), \[ f(z)=\sum_{k=0}^{n-1}\frac{f^{(k)}(z_0)}{k!}(z-z_0)^k+R_n(z)(z-z_0)^n. \]

    Define \(R_n(z)\coloneqq\bigl(f(z)-\sum_{k=0}^{n-1}\frac{f^{(k)}(z_0)}{k!}(z-z_0)^k\bigr)(z-z_0)^{-n}\) for \(z\in\Omega\setminus\{z_0\}\). On a disc about \(z_0\) in \(\Omega\), \(f(z)=\sum_{k\ge0}\frac{f^{(k)}(z_0)}{k!}(z-z_0)^k\), so there \(R_n(z)=\sum_{k\ge n}\frac{f^{(k)}(z_0)}{k!}(z-z_0)^{k-n}\) for \(z\neq z_0\). This power series has the same radius of convergence, so setting \(R_n(z_0)\coloneqq f^{(n)}(z_0)/n!\) makes \(R_n\) analytic at \(z_0\); away from \(z_0\) it is a quotient of analytic functions with nonvanishing denominator.

    The argument principle counts zeros minus poles of a meromorphic function, and the residue theorem applies to meromorphic functions with finitely many poles in the region, where each residue is the coefficient \(c_{-1}\) of a Laurent series with finitely many negative terms.

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