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