% !TeX root = ../../script.tex \documentclass[../../script.tex]{subfiles} \begin{document} \section{Laurent Series} \begin{defi}[Classification of isolated singularities] Let $f: U \rightarrow \cmpln$ and $U \subset \cmpln$ open. Then $z_0 \in \cmpln \setminus \set{U}$ is said to be an isolated singularity if there exists an $\epsilon > 0$ such that $\oball[\epsilon](z_0) \setminus \set{z_0} \subset U$. An isolated singularity $z_0$ is said to be \begin{enumerate}[(i)] \item \textit{removable} if $f$ can be analytically continued on $U \cup \set{z_0}$ \item a \textit{pole} if $\exists m \ge 1$ such that \[ (z- z_0)^m f(z) \] has a removable singularity in $z_0$. The smallest such $m$ is the order of the pole. \item \textit{essential} if it is neither removable nor a pole of finite degree. \end{enumerate} \end{defi} \begin{eg} \begin{enumerate}[(i)] \item The function $f(z) = \frac{\sin z}{z}$ is holomorphic on $\cmpln \setminus \set{0}$, and has a removable singularity in $z_0 = 0$. An analytic continuation of $f$ on all of $\cmpln$ is given by \[ z \longmapsto \sum_{n = 0}^{\infty} (-1)^n \frac{z^{2n}}{(2n + 1)!} \] \item Let $g: U \rightarrow \cmpln$ be holomorphic with $g(z_0) \ne 0$ for $z_0 \in U$. The function \[ f(z) = \frac{g(z)}{(z - z_0)^m} \] has a pole of $m$-th degree in $z_0$. \item Consider the function \begin{align*} f: \cmpln \setminus \set{0} &\longrightarrow \cmpln \\ z &\longmapsto e^{\rec{z}} \end{align*} $f$ has an essential singularity in $z_0 = 0$. The power series representation of $f$ is \[ f(z) = \sum_{n=0}^{\infty} \frac{1}{n!} \rec{z^n} \] This doesn't remove the sungularity in $z_0$, and the pole is of infinite order \[ z^k \abs{f(z)} \conv{z \rightarrow 0} \infty, \quad \forall k \in \natn \] \end{enumerate} \end{eg} \begin{thm}[Riemann's Theorem] An isolated singularity $z_0 \in U$ of a holomorphic function $f: U \setminus \set{z_0} \rightarrow \cmpln$ is removable if and only if $f$ is bounded in a punctured neighbourhood of $z_0$, i.e. \[ \exists \epsilon > 0, c \ge 0: \quad \abs{f(z)} \le c \quad \forall z \in \set[0 < \abs{\zeta - z_0} < \epsilon]{\zeta \in \cmpln} \] \end{thm} \begin{proof} If $f$ can be analytically continued on $U \cup \set{z_0}$, then this continuation is continuous in $z_0$ and thus bounded in a neighbourhood of $z_0$. Inversely, if there exists some $c \ge 0$ and $\epsilon > 0$ such that \begin{equation} \abs{f(z)} \le c \quad \forall z \in \set[0 < \abs{\zeta - z_0} < \epsilon]{\zeta \in \cmpln} \end{equation} Define the function \begin{equation} \begin{split} g: U &\longrightarrow \cmpln \\ z &\longmapsto \begin{cases} (z - z_0)^2 f(z), & z \ne z_0 \\ 0, & z = z_0 \end{cases} \end{split} \end{equation} Then \begin{align} \lim_{z \rightarrow z_0} \frac{\abs{g(z) - g(z_0)}}{\abs{z - z_0}} = \lim_{z \rightarrow z_0} \frac{\abs{z - z_0}^2 \abs{f(z)}}{\abs{z - z_0}} = \lim_{z \rightarrow z_0} \left( \abs{z - z_0} \abs{f(z)} \right) = 0 \end{align} Thus $g$ is holomorphic on $U$ with $g(z_0) = g'(z_0) = 0$, meaning that \begin{equation} g(z) = \sum_{n=2}^{\infty} c_n (z - z_0)^n \end{equation} with $c_n \in \cmpln$. So the function \begin{align} \tilde{f}: U &\longrightarrow \cmpln \\ z &\longmapsto \sum_{n=2}^{\infty} c_n (z - z_0)^{n-2} = \sum_{n=0}^{\infty} c_{n+2} (z - z_n)^n \end{align} is a holomorphic continuation of $f$ on $U \cup \set{z_0}$. \end{proof} \begin{defi}[Laurent Series] If we define the coefficients $c_n \in \cmpln$ for $n \in \intn$, and $z, z_0 \in \cmpln$, then the series \[ \sum_{n \in \intn} c_n (z - z_0)^n := \underbrace{\sum_{n=1}^{\infty} c_{-n} (z - z_0)^{-n}}_{\text{Analytic part}} + \underbrace{\sum_{n=0}^{\infty} c_n (z - z_0)^n}_{\text{Principal part}} \] is said to be a Laurent series. It converges absolutely if the parts do so. If $\frac{1}{r} \in [0, \infty]$ is the convergence radius of the principal part and $R \in [0, \infty]$ the convergence radius of the analytic branch, then the Laurent series converges on the annulus \[ K_{r,R}(z_0) := \set[r < \abs{z - z_0} < R]{z \in \cmpln} \] and is holomorphic. \end{defi} \begin{lem} If the series $f(z) := \sum_{n \in \intn} c_n (z - z_0)^n$ converges on $\cball[r,R](z_0)$, then for $\rho \in (r, R)$ \[ c_n = \rec{2\pi i} \oint_{|z - z_0| = \rho} \frac{f(z)}{(z - z_0)^{n+1}} \dd{z}, \quad n \in \intn \] \end{lem} \begin{proof} Due to the uniform convergence of the series on $\cball[r,R](z_0)$, we have \begin{equation} \begin{split} \oint_{|z - z_0| = \rho} \frac{f(z)}{(z - z_0)^{n+1}} \dd{z} = &\sum_{k \in \intn} c_k \oint_{|z - z_0| = \rho} (z - z_0)^{k-n-1} \dd{z} \\ = &\sum_{k \in \intn} c_k \cdot 2\pi i \delta_{k-n-1, -1} = 2\pi i \cdot c_n \end{split} \end{equation} with $\delta_{i,j}$ the Kronecker delta, defined as \begin{equation} \delta_{i, j} := \begin{cases} 1, & i = j \\ 0, & i \ne j \end{cases} \end{equation} \end{proof} \begin{thm} Let $f: \cball[r,R](z_0) \rightarrow \cmpln$ be holomorphic, then \[ f(z) = \sum_{n \in \intn} c_n (z - z_0)^n \] with \[ c_n = \rec{2\pi i} \oint_{|z - z_0| = \rho} \frac{f(z)}{(z - z_0)^{n+1}} \dd{z}, \quad n \in \intn, ~\rho \in (r, R) \] \end{thm} \begin{proof} W.l.o.g. we set $z_0 = 0$. Similar to the proof of Cauchy's theorem, we can prove Cauchy's theorem for annuli. To do that we define the following integration path \begin{center} \begin{tikzpicture}[scale=0.6] \draw[fill] (-7, 0) node[below] {$z_0$} circle [radius=0.05]; \draw (-7, 0) circle [radius=4]; \draw (-7, 0) circle [radius=1.5]; \draw (-7, 0) -- node[right] {$r$} ({-7 + 1.5*cos(70)}, {0 + 1.5*sin(70)}); \draw (-7, 0) -- node[above left=0.2] {$R$} ({-7 + 4*cos(110)}, {0 + 4*sin(110)}); \draw[fill] (-7, -2.75) node[left] {$z$} circle [radius=0.05]; \draw (-7, -2.75) -- node[below] {$\epsilon$} ({-7 + 1*cos(10)}, {-2.75 + 1*sin(10)}); \begin{scope}[decoration={ markings, mark=at position 0.5 with {\arrow{latex}}} ] \draw[thick, postaction={decorate}, domain=180:0] plot ({-7 + 1*cos(\x)}, {-2.75 + 1*sin(\x)}); \draw[thick, postaction={decorate}, domain=0:-180] plot ({-7 + 1*cos(\x)}, {-2.75 + 1*sin(\x)}); \end{scope} \draw[very thick, ->, >={latex}] (-2, 0) -- (2, 0); \draw[fill] (7, 0) node[below] {$z_0$} circle [radius=0.05]; \draw (7, 0) circle [radius=4]; \draw (7, 0) circle [radius=1.5]; \draw (7, 0) -- node[right] {$r$} ({7 + 1.5*cos(70)}, {0 + 1.5*sin(70)}); \draw (7, 0) -- node[above left=0.2] {$R$} ({7 + 4*cos(110)}, {0 + 4*sin(110)}); \draw[fill] (7, -2.75) node[left] {$z$} circle [radius=0.05]; \begin{scope}[decoration={ markings, mark= between positions 0.2 and 0.8 step 0.3 with {\arrow{latex}}} ] \draw[thick, postaction={decorate}, domain=5:355, samples=50] plot ({7 + 1.75*cos(\x)}, {0 + 1.75*sin(\x)}); \draw[thick, postaction={decorate}, domain=-2:-358, samples=50] plot ({7 + 3.75*cos(\x)}, {0 + 3.75*sin(\x)}); \end{scope} \begin{scope}[decoration={ markings, mark= at position 0.5 with {\arrow{latex}}} ] \draw[thick, postaction={decorate}] ({7 + 1.75*cos(355)}, {0 + 1.75*sin(355)}) -- ({7 + 3.75*cos(357.5)}, {0 + 3.75*sin(357.5)}); \draw[thick, postaction={decorate}] ({7 + 3.75*cos(2)}, {0 + 3.75*sin(2)}) -- ({7 + 1.75*cos(5)}, {0 + 1.75*sin(5)}); \end{scope} \draw ({7 + 1.5*cos(200)}, {0 + 1.5*sin(200)}) node[right] {$\delta$} -- ({7 + 1.75*cos(200)}, {0 + 1.75*sin(200)}); \draw ({7 + 3.75*cos(200)}, {0 + 3.75*sin(200)}) -- ({7 + 4*cos(200)}, {0 + 4*sin(200)}) node[left] {$\delta$}; \end{tikzpicture} \end{center} The two parallel path segments in the right figure are actually overlapping. They have been drawn next to each other for visual clarity. Now we can write \begin{equation} \begin{split} f(z) = &\rec{2\pi i} \oint_{|\zeta - z| = \epsilon} \frac{f(\zeta)}{\zeta - z} \dd{\zeta} \\ = &\rec{2\pi i} \oint_{|\zeta| = R - \delta} \frac{f(\zeta)}{\zeta - z} \dd{\zeta} - \rec{2\pi i} \oint_{|\zeta| = r + \delta} \frac{f(\zeta)}{\zeta - z} \dd{\zeta} \\ = &\rec{2\pi i} \oint_{|\zeta| = R - \delta} \frac{f(\zeta)}{\zeta} \rec{1 - \frac{z}{\zeta}} \dd{\zeta} + \rec{2\pi i} \rec{z} \oint_{|\zeta| = r + \delta} f(\zeta) \rec{1 - \frac{\zeta}{z}} \dd{\zeta} \end{split} \end{equation} We can now make use of the geometric series: \begin{subequations} \begin{equation} \rec{1 - \frac{z}{\zeta}} = \sum_{n=0}^{\infty} \left( \frac{z}{\zeta} \right)^n, \quad \abs{z} < \abs{\zeta} \end{equation} \begin{equation} \rec{1 - \frac{\zeta}{z}} = \sum_{n=0}^{\infty} \left( \frac{\zeta}{z} \right)^n, \quad \abs{\zeta} < \abs{z} \end{equation} \end{subequations} Thus we get \begin{equation} \begin{split} f(z) = &\rec{2\pi i} \oint_{|\zeta| = R - \delta} \frac{f(\zeta)}{\zeta} \sum_{n=0}^{\infty} \frac{z^n}{\zeta^n} \dd{\zeta} + \rec{2\pi i} \rec{z} \oint_{|\zeta| = r + \delta} f(\zeta) \sum_{n=0}^{\infty} \frac{\zeta^n}{z^n} \dd{\zeta} \\ = &\sum_{n=0}^{\infty} z^n \left(\rec{2\pi i} \oint_{|\zeta| = R - \delta} \frac{f(\zeta)}{\zeta^{n+1}} \dd{\zeta}\right) + \sum_{n=0}^{\infty} \frac{1}{z^{n+1}} \left(\rec{2\pi i} \oint_{|\zeta| = r + \delta} f(\zeta)\zeta^n \dd{\zeta}\right) \\ = &\sum_{n=0}^{\infty} c_n z^n + \sum_{n=1}^{\infty} z^{-n} \underbrace{\left(\rec{2\pi i} \oint_{|\zeta| = r + \delta} \frac{f(\zeta)}{\zeta^{-n + 1}} \dd{\zeta} \right)}_{= c_{-n}} \end{split} \end{equation} \end{proof} \begin{eg} Consider \[ f(z) = \frac{1}{z(z-1)} = \frac{1}{z-1} - \frac{1}{z} \] Using the geometric series we can then find for $\cball[0,1](0)$ \[ f(z) = -\rec{z} - \sum_{n=0}^{\infty} z^n \] the Laurent series of $f$ around $z_0 = 0$. For $\cball[0,1](1)$ we get \begin{align*} f(z) &= \rec{z-1} - \rec{z-1+1} = \rec{z-1} - \rec{1 - (1 - z)} \\ &= \underbrace{\rec{z-1}}_{\text{Principal part}} - \underbrace{\sum_{n=0}^{\infty} (1-z)^n}_{\text{Analytic part}} \end{align*} \end{eg} \begin{eg} \[ f(z) = e^{\rec{z}} = \sum_{n=0}^{\infty} \rec{n!}\left(\rec{z}\right)^n = 1 + \underbrace{\sum_{n=1}^{\infty} \rec{n!} \rec{z^n}}_{\text{Principal part}} \] converges on $\cball[0, \infty](0)$. \end{eg} \begin{thm} If $f: U \setminus\set{z_0} \rightarrow \cmpln$ has an essential singularity in $z_0 \in U$, then for every $\epsilon > 0$ the image $f(\oball[\epsilon](z_0) \setminus \set{z_0})$ is dense in $\cmpln$, i.e. \[ \forall \alpha \in \cmpln ~\exists \anyseqdef[z]{U \setminus \set{z_0}}: \quad z_n \longrightarrow z_0 \implies f(z_n) \longrightarrow \alpha \] \end{thm} \begin{proof} \reader \end{proof} \begin{rem} We have essentially noticed three things: \begin{enumerate}[(i)] \item \begin{align*} &f \text{ has a removable singularity in } z_0 \\ \iff &f \text{ is bounded in a neighbourhood of } z_0 \\ \iff &\lim_{\abs{z - z_0} \rightarrow 0} f(z) \text{ exists and is bounded} \end{align*} \item \begin{align*} &f \text{ has a pole of order } m \ge 1 \text{ in } z_0 \\ \iff &\lim_{\abs{z - z_0} \rightarrow 0} \abs{f(z)} = \infty \text{ and } \lim_{\abs{z - z_0} \rightarrow 0} (z- z_0)^m f(z) < \infty \end{align*} \item \begin{align*} &f \text{ has an essential singularity in } z_0 \\ \iff &\text{the set of accumulation points of } f(z) \text{ for } z \longrightarrow z_0 \text{ is all of } \cmpln \end{align*} \end{enumerate} \end{rem} \begin{defi} Let $U \subset \cmpln$ be a domain. For holomorphic $g, h: U \rightarrow \cmpln$ with $h \ne 0$ the function \begin{align*} f: U \setminus \set[h(z) = 0]{z \in U} &\longrightarrow \cmpln \\ z &\longmapsto \frac{g(z)}{h(z)} \end{align*} \sloppy is said to be a meromorphic function. Meromorphic functions are holomorphic on ${U \setminus \set{h(z) = 0}}$. If $z_0 \in U$ a root of order $m \in \natn$ of $h$ and a root of order $k \in \natn_0$ of $g$, then the isolated singularity in $z_0$ of $f$ is \begin{itemize} \item removable for $k \ge m$ \item a pole of order $m - k$ for $k < m$ \end{itemize} \end{defi} \end{document}