If \(f\) is holomorphic on a domain and not identically zero, each zero \(a\) of \(f\) has a finite order \(m\ge1\), and \(f(z)=(z-a)^mg(z)\) with \(g\) holomorphic and \(g(a)\ne0\); hence the zeros of \(f\) are isolated. At an isolated singularity \(a\), \(f\) has a Laurent expansion \(f(z)=\sum_{n\in\ZZ}c_n(z-a)^n\) on a punctured disc around \(a\). The singularity is removable if \(c_n=0\) for all \(n<0\), a pole if \(c_n\ne0\) for at least one and only finitely many \(n<0\), and essential if \(c_n\ne0\) for infinitely many \(n<0\). The residue of \(f\) at \(a\) is \(\Res_af=c_{-1}\). Riemann’s theorem: the singularity is removable if and only if \(f\) is bounded on a punctured neighbourhood of \(a\). The Casorati–Weierstrass theorem: if \(a\) is an essential singularity, the image of every punctured disc around \(a\) is dense in \(\CC\). A function is meromorphic on \(\Omega\) if it is holomorphic on \(\Omega\) except at isolated poles.
Zeros and their orders (Tag PR-5A64G)
Let \(\Omega\subseteq\CC\) be a connected open set, let \(f\) be analytic on \(\Omega\) and not identically zero, and let \(z_0\in\Omega\) with \(f(z_0)=0\). Then there exist a unique integer \(n\ge1\), a neighborhood \(U\subseteq\Omega\) of \(z_0\), and a function \(g\) holomorphic and nonvanishing on \(U\) such that \[ f(z) = (z-z_0)^n g(z)\quad\text{for all } z\in U. \] This \(n\) is the order of the zero \(z_0\), and it is characterized by each of the following:
\(f^{(k)}(z_0)=0\) for \(0\le k\le n-1\) and \(f^{(n)}(z_0)\neq0\);
the Taylor expansion of \(f\) at \(z_0\) has the form \(f(z) = \sum_{k\geq n} c_k (z-z_0)^k\) with \(c_n\neq 0\).
Write \(f(z)=\sum_{k\ge0}c_k(z-z_0)^k\) on a disc \(U\) about \(z_0\), with \(c_k=f^{(k)}(z_0)/k!\). If every \(c_k\) vanished, \(f\) would vanish on \(U\), and by the identity theorem on the connected set \(\Omega\), \(f\) would be identically zero. So there is a least \(n\) with \(c_n\neq0\), and \(n\ge1\) because \(c_0=f(z_0)=0\); this \(n\) satisfies both characterizations. Put \(g(z)\coloneqq\sum_{k\ge n}c_k(z-z_0)^{k-n}\), which converges on \(U\), is holomorphic there, and satisfies \(f(z)=(z-z_0)^ng(z)\) and \(g(z_0)=c_n\neq0\); shrinking \(U\), continuity makes \(g\) nonvanishing on \(U\). If also \(f(z)=(z-z_0)^mh(z)\) near \(z_0\) with \(h(z_0)\neq0\) and, say, \(m>n\), then \(g(z)=(z-z_0)^{m-n}h(z)\) near \(z_0\), forcing \(g(z_0)=0\); hence \(n\) is unique.
Holomorphic functions have isolated zeros (Tag PR-EWOP5)
Let \(f\colon\CC\to \CC\) be holomorphic and not identically zero. Then the zeros of \(f\) are isolated: every \(z_0\) with \(f(z_0)=0\) has a punctured neighborhood on which \(f\) has no zeros.
By PR-5A64G, near a zero \(z_0\) we have \(f(z)=(z-z_0)^ng(z)\) with \(n\ge1\) and \(g\) holomorphic and nonvanishing on a neighborhood \(U\) of \(z_0\). For \(z\in U\setminus\theset{z_0}\) both factors are nonzero, so \(f(z)\neq0\).
Laurent expansion on an annulus (Tag T-4XPWL)
Let \(z_0\in\CC\), \(0\le r<R\le\infty\), and let \(f\) be holomorphic on the annulus \[ A\coloneqq\ts{z\in\CC\st r<\abs{z-z_0}<R}. \] Then \(f\) has a two-sided expansion \[ f(z)=\sum_{n\in\ZZ}c_n(z-z_0)^n \] converging locally uniformly on \(A\), with coefficients \[ c_n=\frac{1}{2\pi i}\int_{\abs{z-z_0}=\rho}\frac{f(z)}{(z-z_0)^{n+1}}\dz \] for any \(r<\rho<R\), independent of \(\rho\). The expansion is unique.
If \(r=0\), so that \(z_0\) is an isolated singularity of \(f\), then the principal part \(\sum_{n<0}c_n(z-z_0)^n\) has no nonzero terms if and only if \(z_0\) is removable, finitely many if and only if \(z_0\) is a pole, and infinitely many if and only if \(z_0\) is essential.
See [@Ahl79, Chapter 5, Section 1.3, p. 184].
Isolated singularities and their classification (Tag D-IWIA5)
Let \(z_0\in\CC\) and \(r>0\), and let \(f\) be holomorphic on the punctured disc \(D_r(z_0)\setminus\{z_0\}\). Then \(z_0\) is an isolated singularity of \(f\). If \(f(z)=\sum_{k\in\ZZ}c_k(z-z_0)^k\) is the Laurent expansion of \(f\) on \(D_r(z_0)\setminus\{z_0\}\), the order of \(f\) at \(z_0\) is \[ v_{z_0}(f)\coloneqq\inf\{k\in\ZZ:c_k\neq0\}\in\ZZ\cup\{-\infty,+\infty\}, \] with \(v_{z_0}(f)=+\infty\) when every \(c_k\) is \(0\).
Let \(z_0\) be an isolated singularity of \(f\). Then:
\(z_0\) is a removable singularity if and only if \(v_{z_0}(f)\ge0\);
\(z_0\) is a pole if and only if \(v_{z_0}(f)\in\ZZ\) and \(v_{z_0}(f)<0\), and then the order of the pole is \(-v_{z_0}(f)\);
\(z_0\) is an essential singularity if and only if \(v_{z_0}(f)=-\infty\).
If \(f=g\) near \(z_0\) with \(g\) holomorphic on a disc about \(z_0\), the Taylor series of \(g\) is a Laurent expansion of \(f\), so by uniqueness of Laurent coefficients \(c_k=0\) for \(k<0\). Conversely, if \(c_k=0\) for \(k<0\), the power series \(\sum_{k\ge0}c_k(z-z_0)^k\) converges on \(D_r(z_0)\) and agrees with \(f\) off \(z_0\).
This is the equivalence of a pole of order \(n\) with a Laurent expansion \(\sum_{k\ge-n}c_k(z-z_0)^k\), \(c_{-n}\neq0\), proved in D-AUD6K.
An essential singularity is one that is neither removable nor a pole; by (a) and (b) this happens exactly when \(c_k\neq0\) for infinitely many \(k<0\), that is, when \(v_{z_0}(f)=-\infty\).
The function \(\sin(1/z)\) has an essential singularity at \(0\). Its Laurent expansion on \(\CC\setminus\{0\}\) is \[ \sin(1/z)=\sum_{m\ge0}\frac{(-1)^m}{(2m+1)!}\,z^{-(2m+1)}, \] which has nonzero coefficients \(c_k\) for infinitely many \(k<0\), so \(v_0(\sin(1/z))=-\infty\).
Removable singularities (Tag D-BQLJV)
Let \(z_0\in\CC\), let \(r>0\), and let \(f\) be holomorphic on the punctured disc \(D_r(z_0)\setminus\{z_0\}\). The point \(z_0\) is a removable singularity of \(f\) if there exist \(0<\rho\le r\) and a holomorphic function \(g\colon D_\rho(z_0)\to\CC\) such that \(f(z)=g(z)\) for \(0<\abs{z-z_0}<\rho\).
(Riemann's removable singularity theorem)
Let \(f\) be holomorphic on \(D_r(z_0)\setminus\{z_0\}\). The following are equivalent.
\(z_0\) is a removable singularity of \(f\).
\(\lim_{z\to z_0}(z-z_0)f(z)=0\).
\(f\) is bounded on \(D_\rho(z_0)\setminus\{z_0\}\) for some \(0<\rho\le r\).
(a)\(\Rightarrow\)(c). The function \(g\) is continuous on \(D_\rho(z_0)\), hence bounded on \(D_{\rho/2}(z_0)\), and \(f=g\) there away from \(z_0\).
(c)\(\Rightarrow\)(b). If \(\abs{f}\le M\) near \(z_0\), then \(\abs{(z-z_0)f(z)}\le M\abs{z-z_0}\to0\).
(b)\(\Rightarrow\)(a). Define \(h(z)\coloneqq(z-z_0)^2f(z)\) for \(0<\abs{z-z_0}<r\) and \(h(z_0)\coloneqq0\). Then \(h\) is holomorphic away from \(z_0\), and at \(z_0\) \[ \frac{h(z)-h(z_0)}{z-z_0}=(z-z_0)f(z)\to0, \] so \(h\) is holomorphic on \(D_r(z_0)\) with \(h(z_0)=h'(z_0)=0\). Its Taylor expansion is therefore \(h(z)=\sum_{k\ge2}a_k(z-z_0)^k\) on \(D_r(z_0)\), and \(g(z)\coloneqq\sum_{k\ge2}a_k(z-z_0)^{k-2}\) is holomorphic on \(D_r(z_0)\) and equals \(f\) on the punctured disc.
In terms of the Laurent expansion \(f(z)=\sum_{k\in\ZZ}c_k(z-z_0)^k\) on \(D_r(z_0)\setminus\{z_0\}\), the point \(z_0\) is a removable singularity if and only if \(c_k=0\) for all \(k<0\), that is, if and only if the order \(v_{z_0}(f)\) satisfies \(v_{z_0}(f)\ge0\).
Riemann's removable singularity theorem (Tag T-ZZJDP)
Let \(\Omega\subseteq\CC\) be open, let \(z_0\in\Omega\), and let \(f\) be holomorphic on \(\Omega\setminus\{z_0\}\). The following are equivalent.
\(z_0\) is a removable singularity of \(f\).
\(f\) is bounded on \(\{z : 0<\abs{z-z_0}<\delta\}\) for some \(\delta>0\).
\((z-z_0)f(z)\to0\) as \(z\to z_0\).
\(f\) extends to a holomorphic function on \(\Omega\).
\(f\) extends to a continuous function on \(\Omega\).
The Laurent expansion of \(f\) on a punctured disc \(\{z : 0<\abs{z-z_0}<\delta\}\subseteq\Omega\) has vanishing principal part: \(f(z)=\sum_{k\geq0}c_k(z-z_0)^k\) there.
Poles and their order (Tag D-AUD6K)
Let \(z_0\in\CC\), let \(r>0\), and let \(f\) be holomorphic on the punctured disc \(D_r(z_0)\setminus\{z_0\}\). For an integer \(n\ge1\), \(f\) has a pole of order \(n\) at \(z_0\) if there exist \(0<\rho\le r\) and a holomorphic function \(h\colon D_\rho(z_0)\to\CC\) with \(h(z_0)\neq0\) such that \[ f(z)=(z-z_0)^{-n}h(z)\qquad\text{for } 0<\abs{z-z_0}<\rho. \] \(f\) has a pole at \(z_0\) if it has a pole of some order \(n\ge1\) there, and a pole of order \(1\) is a simple pole.
Let \(f\) be holomorphic on \(D_r(z_0)\setminus\{z_0\}\) and let \(n\ge1\). The following are equivalent.
\(f\) has a pole of order \(n\) at \(z_0\).
There is \(0<\rho\le r\) such that \(f\) has no zeros in \(D_\rho(z_0)\setminus\{z_0\}\) and \(g\coloneqq1/f\), extended by \(g(z_0)\coloneqq0\), is holomorphic on \(D_\rho(z_0)\) with a zero of order $n$ at \(z_0\).
The Laurent expansion of \(f\) on \(D_r(z_0)\setminus\{z_0\}\) has the form \(f(z)=\sum_{k\ge-n}c_k(z-z_0)^k\) with \(c_{-n}\neq0\).
Moreover, \(f\) has a pole at \(z_0\) if and only if \(\abs{f(z)}\to\infty\) as \(z\to z_0\).
(a)\(\Rightarrow\)(b). Since \(h(z_0)\neq0\) and \(h\) is continuous, shrinking \(\rho\) makes \(h\) nonvanishing on \(D_\rho(z_0)\). Then \(1/f(z)=(z-z_0)^n\cdot(1/h(z))\) for \(0<\abs{z-z_0}<\rho\), and the right side is holomorphic on \(D_\rho(z_0)\), vanishes at \(z_0\), and has \(1/h(z_0)\neq0\).
(b)\(\Rightarrow\)(a). Write \(g(z)=(z-z_0)^n k(z)\) with \(k\) holomorphic near \(z_0\) and \(k(z_0)\neq0\); shrinking \(\rho\), \(k\) is nonvanishing, and \(h\coloneqq1/k\) satisfies \(f(z)=(z-z_0)^{-n}h(z)\) with \(h(z_0)\neq0\).
(a)\(\Rightarrow\)(c). Expand \(h(z)=\sum_{j\ge0}b_j(z-z_0)^j\) on \(D_\rho(z_0)\) with \(b_0=h(z_0)\neq0\). Then \(f(z)=\sum_{k\ge-n}b_{k+n}(z-z_0)^k\) on \(D_\rho(z_0)\setminus\{z_0\}\), and by uniqueness of Laurent coefficients this is the Laurent expansion of \(f\), with \(c_{-n}=b_0\neq0\).
(c)\(\Rightarrow\)(a). The power series \(h(z)\coloneqq\sum_{k\ge-n}c_k(z-z_0)^{k+n}\) converges for \(0<\abs{z-z_0}<r\), hence on \(D_r(z_0)\), and defines a holomorphic function there with \(h(z_0)=c_{-n}\neq0\) and \(f(z)=(z-z_0)^{-n}h(z)\).
Finally, if \(f\) has a pole of order \(n\), then \(\abs{f(z)}=\abs{h(z)}/\abs{z-z_0}^n\to\infty\) because \(h(z_0)\neq0\). Conversely, if \(\abs{f(z)}\to\infty\), then \(f\) has no zeros on some \(D_\rho(z_0)\setminus\{z_0\}\), and \(1/f\) is holomorphic and bounded there with \(1/f(z)\to0\). By Riemann's removable singularity theorem, \(g\coloneqq1/f\) with \(g(z_0)\coloneqq0\) is holomorphic on \(D_\rho(z_0)\). It is not identically zero, so \(z_0\) is a zero of \(g\) of some order \(n\ge1\), and (b) holds.
Principal part and residue at an isolated singularity (Tag D-C3JIU)
Let \(f\) be holomorphic on a punctured disc \(D_r(z_0)\setminus\{z_0\}\), so that \(z_0\) is an isolated singularity of \(f\), and let \[ f(z)=\sum_{k=-\infty}^{\infty}a_k(z-z_0)^k \qquad (0<\abs{z-z_0}<r) \] be its Laurent expansion there. The principal part of \(f\) at \(z_0\) is \(\sum_{k\le -1}a_k(z-z_0)^k\), and the residue of \(f\) at \(z_0\) is \(\Res_{z=z_0}f\coloneqq a_{-1}\) [@Ahl79].
For every \(0<\rho<r\) one has \(\Res_{z=z_0}f=\frac{1}{2\pi i}\int_{\abs{z-z_0}=\rho}f(z)\dz\). If \(z_0\) is a pole of order $n\ge1$, then \(a_k = 0\) for \(k<-n\) and the principal part is \[ P(z)=\frac{a_{-n}}{(z-z_0)^n}+\cdots+\frac{a_{-1}}{z-z_0}; \] the function \(f-P\) equals \(\sum_{k\ge0}a_k(z-z_0)^k\) on \(D_r(z_0)\setminus\{z_0\}\), so it extends to a holomorphic function on \(D_r(z_0)\), and \(P\) is the unique polynomial in \((z-z_0)^{-1}\) without constant term with this property.
Essential singularity (Tag D-VKP6N)
Let \(f\) be holomorphic on a punctured disc \(D_r(z_0)\setminus\{z_0\}\). The isolated singularity \(z_0\) is an essential singularity of \(f\) if it is neither a removable singularity nor a pole of \(f\).
Let \(f(z)=\sum_{k\in\ZZ}c_k(z-z_0)^k\) be the Laurent expansion of \(f\) on \(D_r(z_0)\setminus\{z_0\}\). Then \(z_0\) is an essential singularity of \(f\) if and only if \(c_k\neq0\) for infinitely many \(k<0\), that is, if and only if the principal part \(\sum_{k<0}c_k(z-z_0)^k\) has infinitely many nonzero terms.
By the proposition on D-IWIA5, \(z_0\) is removable exactly when \(c_k=0\) for all \(k<0\), and a pole exactly when \(c_k\neq0\) for some but only finitely many \(k<0\). The remaining case is that \(c_k\neq0\) for infinitely many \(k<0\).
Casorati--Weierstrass theorem (Tag T-EO65T)
Let \(\Omega\subseteq\CC\) be open, let \(z_0\in\Omega\), and let \(f\) be holomorphic on \(\Omega\setminus\{z_0\}\) with an essential singularity at \(z_0\). Then for every \(\delta>0\) with \(\{z : 0<\abs{z-z_0}<\delta\}\subseteq\Omega\), the image \(f(\{z : 0<\abs{z-z_0}<\delta\})\) is dense in \(\CC\).
Meromorphic function (Tag D-7DFVJ)
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 \((z_n)\).