% !TeX root = ../../script.tex \documentclass[../../script.tex]{subfiles} \begin{document} \section{Identity Theorem \& Analytic Continuation} \begin{defi} Let $f: U \rightarrow \cmpln$ be a function on $U \subset \cmpln$ and $n \in \natn$. $f$ has a root with multiplicity $n$ at $z_0$, if \begin{align*} f^{(k)} (z_0) &= 0, \quad \forall k = 0, \cdots, n - 1 \\ f^{(n)} (z_0) &= 0 \end{align*} If $f$ is holomorphic it can be written as \[ f(z) = \sum_{k=n}^{\infty} c_n (z - z_0)^k \] \end{defi} \begin{thm}[Identity Theorem] Let $U \subset \cmpln$ be a domain and $f: U \rightarrow \cmpln$ analytic. If \[ \set[f(z) = 0]{z \in \cmpln} \] has an accumulation point, i.e. \[ f(z_n) = 0, \quad (z_n)_{n \in \natn} \subset U, ~(z_n) \conv{n \rightarrow \infty} z_{\infty} \in U \] then $f = 0$ on $U$. \end{thm} \begin{proof} Since $f$ is analytic in $z_0 \in U$, $\exists \delta > 0$ such that \begin{equation} f(z) = \sum_{k=0}^{\infty} a_k (z - z_0)^k, \quad \forall z \in \oball[\delta](z_0) \end{equation} Because $z_0 \in U$ is a root of $f$ we can find that $a_0 = 0$. If $a_k \ne 0$ for some $k \ge 1$ then we can consider \begin{equation} m = \min\set[a_k \ne 0]{k \ge 1} \end{equation} Define \begin{equation} g(z) = \sum_{n=0}^{\infty} a_{n+m} (z - z_0)^n \end{equation} Then $g(z_0) \ne 0$ and \begin{equation} f(z) = (z - z_0)^m g(z) \end{equation} This function $g$ is analytic in $\oball[\delta](z_0)$, and thus continuous. This means $\exists \delta' < \delta$ such that $g$ doesn't vanish on $\oball[\delta'](z_0)$. We can conclude that $f$ doesn't vanish on $\oball[\delta'](z_0) \setminus \set{z_0}$ either. If $a_k = 0 ~~\forall k \in \natn$, then $f = 0$ on $\oball[\delta(z_0)]$. Now define the set \begin{equation} A = \set[f^{(k)}(z) = 0, \quad \forall k \in \natn_0]{z \in U} \end{equation} Since $f^{(n)}$ is continuous for all $n \in \natn_0$, we find \begin{equation} \begin{split} A &= \bigcap_{n \in \natn_0} \set[f^{(n)}(z) = 0]{z \in U} \\ &= \underbrace{\bigcap_{n \in \natn_0} \underbrace{\underbrace{\inv{\left(f^{(n)}\right)}}_{\text{continuous}} (\underbrace{\set{0}}_{\text{closed}})}_{\text{closed}}}_{\text{closed}} \end{split} \end{equation} But $A$ is also open. To prove this we consider a point $z_1 \in A$. Then the Taylor series of $f$ in $z_1$ is identical to the zero-function. But then $f = 0$ on a neighbourhood $V$ of $z_1$. However, since $f^{(n)}(z) = 0 ~~\forall n \in \natn_0$ and $z \in V$, we can use our previous results to conclude that $V \subset A$, making $A$ a closed set. $U$ can now be represented in terms of $A$: \begin{equation} U = A \cup (U \setminus A) \end{equation} This is the disjoint union of two open sets. Since $U$ is a domain (and thus connected) this can only be the case if \begin{equation} A = \set{U, \varnothing} \end{equation} Since $z_0 = 0$ we can conclude $A = U$. \end{proof} \begin{defi} If $V \subset U \subset \cmpln$, and there exist two holomorphic functions \begin{align*} f: V &\longrightarrow \cmpln \\ \tilde{f}: U &\longrightarrow \cmpln \end{align*} with the property \begin{align*} f(z) = \tilde{f}(z), \quad \forall z \in V \end{align*} then $\tilde{f}$ is said to be the analytic continuation of $f$ on $U$. \end{defi} \begin{rem} If the set $V$ has an accumulation point and if $U$ is a domain, then the analytic continuation $\tilde{f}$ of $f$ on $U$ is unique (This follows from the identity theorem). \end{rem} \begin{eg} \begin{enumerate}[(i)] \item $f(z) = \sum_{n=0}^{\infty} z^n$ is holomorphic on $\set[\abs{z} < 1]{z \in \cmpln}$. The function \[ \tilde{f}(z) = \frac{1}{1 - z} \] is an analytic continuation of $f$ on $\cmpln \setminus \set{1}$. \item We can also find the analytic continuation along a chain of circular disks: for $j \in \natn$ define the power series \[ f_j(z) := \sum_{n=0}^{\infty} a_n(j)(z - z_j)^n \] around $z_j \in \cmpln$ with convergence radius $\rho_j \in (0, \infty]$. If the disks overlap and the functions are compatible, i.e. \[ f_j(z) = f_k(z), \quad \forall z \in \oball[\rho_j](z_j) \cap \oball[\rho_k](z_k) \] then there is a unique holomorphic continuation on \[ \bigcup_{j \in \natn} \oball[\rho_j](z_j) \] \end{enumerate} \end{eg} \begin{defi}[Analytic continuation along curves] Let $\gamma: [t_0, t_1] \rightarrow \cmpln$ be a continuous curve and \[ f(z) = \sum_{n = 0}^{\infty} a_n (z - z_0)^n \] a converging power series around $z_0 = \gamma(t_0)$. Then the family of functions \[ f_t(z) := \sum_{n=0}^{\infty} a_n(t) (z - \gamma(t))^n, \quad t \in [t_0, t_1] \] is an analytic continuation of $f$ along $\gamma$ if \begin{itemize} \item $f_{t_0} = f$ \item $\forall t \in [t_0, t_1]$ exists a $\epsilon > 0$ such that for all $\abs{\tau} < \epsilon$ the functions $f_t$ and $f_{\min\set{t, \tau, t_1}}$ are compatible. \end{itemize} \end{defi} \begin{eg}[Complex Logarithm] The family \[ L_t(z) := it + \sum_{n=1}^{\infty} \frac{(-1)^n}{n} \left(e^{it}z - 1\right)^n, \quad t \in [0, \infty) \] is an analytic continuation of the main branch of the complex logarithm $L_0(z) = \Log(z)$ along the unit circle. This yields the secondary branches of the complex logarithm: \[ L_{2\pi n}(z) = 2\pi in + \Log(z) \] \end{eg} \end{document}