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+#LyX 2.3 created this file. For more info see http://www.lyx.org/
+\lyxformat 544
+\begin_document
+\begin_header
+\save_transient_properties true
+\origin unavailable
+\textclass book
+\use_default_options true
+\maintain_unincluded_children false
+\language spanish
+\language_package default
+\inputencoding auto
+\fontencoding global
+\font_roman "default" "default"
+\font_sans "default" "default"
+\font_typewriter "default" "default"
+\font_math "auto" "auto"
+\font_default_family default
+\use_non_tex_fonts false
+\font_sc false
+\font_osf false
+\font_sf_scale 100 100
+\font_tt_scale 100 100
+\use_microtype false
+\use_dash_ligatures true
+\graphics default
+\default_output_format default
+\output_sync 0
+\bibtex_command default
+\index_command default
+\paperfontsize default
+\spacing single
+\use_hyperref false
+\papersize default
+\use_geometry false
+\use_package amsmath 1
+\use_package amssymb 1
+\use_package cancel 1
+\use_package esint 1
+\use_package mathdots 1
+\use_package mathtools 1
+\use_package mhchem 1
+\use_package stackrel 1
+\use_package stmaryrd 1
+\use_package undertilde 1
+\cite_engine basic
+\cite_engine_type default
+\biblio_style plain
+\use_bibtopic false
+\use_indices false
+\paperorientation portrait
+\suppress_date false
+\justification true
+\use_refstyle 1
+\use_minted 0
+\index Index
+\shortcut idx
+\color #008000
+\end_index
+\secnumdepth 3
+\tocdepth 3
+\paragraph_separation indent
+\paragraph_indentation default
+\is_math_indent 0
+\math_numbering_side default
+\quotes_style french
+\dynamic_quotes 0
+\papercolumns 1
+\papersides 1
+\paperpagestyle default
+\tracking_changes false
+\output_changes false
+\html_math_output 0
+\html_css_as_file 0
+\html_be_strict false
+\end_header
+
+\begin_body
+
+\begin_layout Standard
+Sean 
+\begin_inset Formula $\Omega$
+\end_inset
+
+ un dominio, 
+\begin_inset Formula $f\in{\cal H}(\Omega)$
+\end_inset
+
+ y 
+\begin_inset Formula $Z(f):=\{z\in\Omega:f(z)=0\}$
+\end_inset
+
+, 
+\begin_inset Formula $Z(f)$
+\end_inset
+
+ tiene un punto de acumulación en 
+\begin_inset Formula $\Omega$
+\end_inset
+
+ si y sólo si 
+\begin_inset Formula $\exists a\in\Omega:\forall k\in\mathbb{N},f^{(k)}(a)=0$
+\end_inset
+
+, si y sólo si 
+\begin_inset Formula $f$
+\end_inset
+
+ es idénticamente nula.
+\end_layout
+
+\begin_layout Description
+\begin_inset Formula $1\implies2]$
+\end_inset
+
+ Sean 
+\begin_inset Formula $a\in Z(f)'\cap\Omega$
+\end_inset
+
+ y 
+\begin_inset Formula $D(a,\rho)\subseteq\Omega$
+\end_inset
+
+, existe 
+\begin_inset Formula $\{a_{n}\}_{n}\subseteq D(a,\rho)\setminus\{a\}$
+\end_inset
+
+ con 
+\begin_inset Formula $a_{n}\to a$
+\end_inset
+
+.
+ Por el teorema de Taylor,
+\begin_inset Formula 
+\[
+f(z)=\sum_{n=0}^{\infty}c_{n}(z-a)^{n}
+\]
+
+\end_inset
+
+para 
+\begin_inset Formula $c_{n}:=\frac{f^{(n)}(a)}{n!}$
+\end_inset
+
+, y queremos ver que todos los 
+\begin_inset Formula $c_{n}$
+\end_inset
+
+ son nulos por inducción.
+ Para 
+\begin_inset Formula $n=0$
+\end_inset
+
+, 
+\begin_inset Formula $c_{0}=f(a)=\lim_{n}f(a_{n})=0$
+\end_inset
+
+.
+ Si 
+\begin_inset Formula $c_{0}=\dots=c_{k-1}=0$
+\end_inset
+
+, tenemos
+\begin_inset Formula 
+\[
+\frac{f(z)}{(z-a)^{k}}=c_{k}+\sum_{n=k+1}^{\infty}c_{n}(z-a)^{n-k}.
+\]
+
+\end_inset
+
+Sea 
+\begin_inset Formula $g_{k}(z):=\sum_{n=k+1}^{\infty}c_{n}(z-a)^{n-k}$
+\end_inset
+
+ una función holomorfa en 
+\begin_inset Formula $D(a,\rho)$
+\end_inset
+
+ con 
+\begin_inset Formula $g_{k}(a)=0$
+\end_inset
+
+, entonces 
+\begin_inset Formula $\frac{f(z)}{(z-a)^{k}}=c_{k}+g_{k}(z)$
+\end_inset
+
+, pero 
+\begin_inset Formula $c_{k}+g_{k}(a_{n})=0$
+\end_inset
+
+
+\begin_inset Note Note
+status open
+
+\begin_layout Plain Layout
+¿por qué?
+\end_layout
+
+\end_inset
+
+ y, tomando límites, 
+\begin_inset Formula $0=c_{k}+\lim_{n}g_{k}(a_{n})=c_{k}+g_{k}(a)=c_{k}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Description
+\begin_inset Formula $2\implies3]$
+\end_inset
+
+ Sea 
+\begin_inset Formula $A:=\{z\in\Omega:\forall k\in\mathbb{N},f^{(k)}(z)=0\}\neq\emptyset$
+\end_inset
+
+, pues 
+\begin_inset Formula $a\in A$
+\end_inset
+
+.
+ Como
+\begin_inset Formula 
+\[
+A=\bigcap_{k=0}^{\infty}\{z\in\Omega:f^{(k)}(z)=0\},
+\]
+
+\end_inset
+
+
+\begin_inset Formula $A$
+\end_inset
+
+ es intersección de cerrados y por tanto cerrado, ahora bien, sean 
+\begin_inset Formula $z\in A$
+\end_inset
+
+ y 
+\begin_inset Formula $D(z,\rho)\subseteq\Omega$
+\end_inset
+
+, por el teorema de Taylor, para 
+\begin_inset Formula $w\in D(z,\rho)$
+\end_inset
+
+,
+\begin_inset Formula 
+\[
+f(w)=\sum_{n=0}^{\infty}\frac{f^{(k)}(z)}{n!}(w-z)^{n}=0,
+\]
+
+\end_inset
+
+luego 
+\begin_inset Formula $f^{(k)}(w)=0$
+\end_inset
+
+ y 
+\begin_inset Formula $D(z,\rho)\subseteq A$
+\end_inset
+
+, con lo que 
+\begin_inset Formula $A$
+\end_inset
+
+ es abierto.
+ Por conexión, 
+\begin_inset Formula $A=\Omega$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Description
+\begin_inset Formula $3\implies1]$
+\end_inset
+
+ Trivial.
+\end_layout
+
+\begin_layout Standard
+El 
+\series bold
+principio de identidad para funciones holomorfas
+\series default
+ afirma que si dos funciones holomorfas en un dominio 
+\begin_inset Formula $\Omega$
+\end_inset
+
+ coinciden en un subconjunto de 
+\begin_inset Formula $\Omega$
+\end_inset
+
+ con algún punto de acumulación en 
+\begin_inset Formula $\Omega$
+\end_inset
+
+, entonces coinciden en todo 
+\begin_inset Formula $\Omega$
+\end_inset
+
+.
+ En efecto, sean 
+\begin_inset Formula $f,g\in{\cal H}(\Omega)$
+\end_inset
+
+ y 
+\begin_inset Formula $h=f-g$
+\end_inset
+
+, si 
+\begin_inset Formula $f$
+\end_inset
+
+ y 
+\begin_inset Formula $g$
+\end_inset
+
+ coinciden en un subconjunto de este tipo, entonces 
+\begin_inset Formula $Z(h)'\cap\Omega\neq\emptyset$
+\end_inset
+
+ y por tanto 
+\begin_inset Formula $h$
+\end_inset
+
+ es idénticamente nula, luego 
+\begin_inset Formula $f=g$
+\end_inset
+
+.
+ También, si 
+\begin_inset Formula $\Omega$
+\end_inset
+
+ es un dominio y 
+\begin_inset Formula $f\in{\cal H}(\Omega)$
+\end_inset
+
+ no es idénticamente nula, entonces todo punto de 
+\begin_inset Formula $Z(f):=\{z\in\Omega:f(z)=0\}$
+\end_inset
+
+ es aislado y 
+\begin_inset Formula $Z(f)$
+\end_inset
+
+ es numerable.
+\end_layout
+
+\begin_layout Standard
+Sean 
+\begin_inset Formula $f\in{\cal H}(\Omega)$
+\end_inset
+
+ y 
+\begin_inset Formula $a\in\Omega$
+\end_inset
+
+ con 
+\begin_inset Formula $f(a)=0$
+\end_inset
+
+, decimos que 
+\begin_inset Formula $f$
+\end_inset
+
+ tiene en 
+\begin_inset Formula $a$
+\end_inset
+
+ un 
+\series bold
+cero
+\series default
+ de 
+\series bold
+orden
+\series default
+ 
+\begin_inset Formula $\min\{n\in\mathbb{N}:f^{(n)}(a)\neq0\}$
+\end_inset
+
+.
+ Una función 
+\begin_inset Formula $f$
+\end_inset
+
+ holomorfa no idénticamente nula en en un dominio 
+\begin_inset Formula $\Omega$
+\end_inset
+
+ tiene un cero de orden 
+\begin_inset Formula $k\geq1$
+\end_inset
+
+ en 
+\begin_inset Formula $a\in\Omega$
+\end_inset
+
+ si y sólo si 
+\begin_inset Formula $\exists g\in{\cal H}(\Omega):(g(a)\neq0\land\forall z\in\Omega,f(z)=(z-a)^{k}g(z))$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Argument item:1
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $\implies]$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+Sea 
+\begin_inset Formula $D(a,\rho)\subseteq\Omega$
+\end_inset
+
+, existe 
+\begin_inset Formula $(c_{n})_{n\geq k}$
+\end_inset
+
+ con 
+\begin_inset Formula $c_{k}\neq0$
+\end_inset
+
+ y 
+\begin_inset Formula $f(z)=\sum_{n=k}^{\infty}c_{n}(z-a)^{n}$
+\end_inset
+
+ para cada 
+\begin_inset Formula $z\in D(a,\rho)$
+\end_inset
+
+.
+ Entonces
+\begin_inset Formula 
+\[
+g(z):=\begin{cases}
+\frac{f(z)}{(z-a)^{k}} & \text{si }z\neq a,\\
+c_{k} & \text{si }z=a
+\end{cases}
+\]
+
+\end_inset
+
+es holomorfa en 
+\begin_inset Formula $\Omega\setminus\{a\}$
+\end_inset
+
+ y cumple 
+\begin_inset Formula $g(z)=\sum_{n=k}^{\infty}c_{n}(z-a)^{n-k}$
+\end_inset
+
+ para todo 
+\begin_inset Formula $z\in D(a,\rho)$
+\end_inset
+
+, luego es holomorfa en 
+\begin_inset Formula $D(a,\rho)$
+\end_inset
+
+ y, por tanto en 
+\begin_inset Formula $a$
+\end_inset
+
+.
+ Así, 
+\begin_inset Formula $g\in{\cal H}(\Omega)$
+\end_inset
+
+, 
+\begin_inset Formula $g(a)\neq0$
+\end_inset
+
+ y 
+\begin_inset Formula $f(z)=(z-a)^{k}g(z)$
+\end_inset
+
+ para 
+\begin_inset Formula $z\in\Omega$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Argument item:1
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $\impliedby]$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+Sea 
+\begin_inset Formula $D(a,\rho)\subseteq\Omega$
+\end_inset
+
+, por el teorema de Taylor existe 
+\begin_inset Formula $(\alpha_{n})_{n}$
+\end_inset
+
+ tal que 
+\begin_inset Formula $g(z)=\sum_{n}\alpha_{n}(z-a)^{n}$
+\end_inset
+
+ para 
+\begin_inset Formula $z\in D(a,\rho)$
+\end_inset
+
+, con 
+\begin_inset Formula $\alpha_{0}=g(a)\neq0$
+\end_inset
+
+ y entonces 
+\begin_inset Formula $f(z)=\sum_{n}\alpha_{n}(z-a)^{n+k}$
+\end_inset
+
+ y por tanto 
+\begin_inset Formula $f^{(q)}(a)=0$
+\end_inset
+
+ para 
+\begin_inset Formula $q\in\{0,\dots,k-1\}$
+\end_inset
+
+ y 
+\begin_inset Formula $f^{(k)}(a)=k!\alpha_{0}\neq0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+
+\series bold
+Regla de L'Hôpital:
+\series default
+ Sean 
+\begin_inset Formula $\Omega$
+\end_inset
+
+ un dominio, 
+\begin_inset Formula $f,g\in{\cal H}(\Omega)$
+\end_inset
+
+ no idénticamente nulas y 
+\begin_inset Formula $a\in\Omega$
+\end_inset
+
+ con 
+\begin_inset Formula $f(a)=g(a)=0$
+\end_inset
+
+, entonces
+\begin_inset Formula 
+\[
+\lim_{z\to a}\frac{f(z)}{g(z)}=\lim_{z\to a}\frac{f'(z)}{g'(z)}.
+\]
+
+\end_inset
+
+
+\series bold
+Demostración:
+\series default
+ Por lo anterior, existen 
+\begin_inset Formula $m,n\in\mathbb{N}$
+\end_inset
+
+ y 
+\begin_inset Formula $F,G\in{\cal H}(\Omega)$
+\end_inset
+
+ con 
+\begin_inset Formula $f(z)=(z-a)^{m}F(z)$
+\end_inset
+
+, 
+\begin_inset Formula $g(z)=(z-a)^{n}G(z)$
+\end_inset
+
+ y 
+\begin_inset Formula $F(a),G(a)\neq0$
+\end_inset
+
+.
+ Como el conjunto de puntos donde 
+\begin_inset Formula $g$
+\end_inset
+
+ se anula está formado por puntos aislados y 
+\begin_inset Formula $g(a)=0$
+\end_inset
+
+, debe haber un disco perforado alrededor de 
+\begin_inset Formula $a$
+\end_inset
+
+ donde 
+\begin_inset Formula $g$
+\end_inset
+
+ no se anula.
+ También hay un disco perforado alrededor de 
+\begin_inset Formula $a$
+\end_inset
+
+ donde 
+\begin_inset Formula $g'$
+\end_inset
+
+ no se anula, por el mismo motivo si 
+\begin_inset Formula $g'(a)=0$
+\end_inset
+
+ o por continuidad si 
+\begin_inset Formula $g'(a)\neq0$
+\end_inset
+
+.
+ Sea entonces 
+\begin_inset Formula $D(a,\rho)\subseteq\Omega$
+\end_inset
+
+ con 
+\begin_inset Formula $g(z)\neq0$
+\end_inset
+
+ y 
+\begin_inset Formula $g'(z)\neq0$
+\end_inset
+
+ para 
+\begin_inset Formula $z\in D(a,\rho)\setminus\{a\}$
+\end_inset
+
+, para estos puntos,
+\begin_inset Formula 
+\begin{align*}
+\frac{f(z)}{g(z)} & =\frac{(z-a)^{m}F(z)}{(z-a)^{n}G(z)}, & \frac{f'(z)}{g'(z)} & =(z-a)^{m-n}\frac{mF(z)+(z-a)F'(z)}{nG(z)+(z-a)G'(z)}.
+\end{align*}
+
+\end_inset
+
+Tomando límites cuando 
+\begin_inset Formula $z\to a$
+\end_inset
+
+, si 
+\begin_inset Formula $m=n$
+\end_inset
+
+, ambos límites valen 
+\begin_inset Formula $\frac{F(a)}{G(a)}$
+\end_inset
+
+; si 
+\begin_inset Formula $m>n$
+\end_inset
+
+, ambos son nulos, y si 
+\begin_inset Formula $m<n$
+\end_inset
+
+, ambos son 
+\begin_inset Formula $\infty$
+\end_inset
+
+.
+\end_layout
+
+\end_body
+\end_document