%% This document created by Scientific Notebook (R) Version 3.5 %% Starting shell: article \documentclass{article} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \usepackage{amssymb} %TCIDATA{OutputFilter=LATEX.DLL} %TCIDATA{Version=5.00.0.2570} %TCIDATA{} %TCIDATA{Created=Wednesday, February 10, 1999 13:29:48} %TCIDATA{LastRevised=Friday, November 16, 2007 12:56:16} %TCIDATA{} %TCIDATA{} %TCIDATA{CSTFile=On line bluem.cst} %TCIDATA{PageSetup=72,72,72,72,0} %TCIDATA{Counters=arabic,1} %TCIDATA{AllPages= %H=36 %F=36,\PARA{038
\QTR{small}{Matematick\U{e1} anal\U{fd}za III online - Integr\U{e1}lny po\U{10d}et funkci\U{ed} komplexnej premennej - Cauchyho integr\U{e1}lna veta\dotfill \thepage }} %} \newtheorem{theorem}{Theorem} \newtheorem{acknowledgement}[theorem]{Acknowledgement} \newtheorem{algorithm}[theorem]{Algorithm} \newtheorem{axiom}[theorem]{Axiom} \newtheorem{case}[theorem]{Case} \newtheorem{claim}[theorem]{Claim} \newtheorem{conclusion}[theorem]{Conclusion} \newtheorem{condition}[theorem]{Condition} \newtheorem{conjecture}[theorem]{Conjecture} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{criterion}[theorem]{Criterion} \newtheorem{definition}[theorem]{Definition} \newtheorem{example}[theorem]{Example} \newtheorem{exercise}[theorem]{Exercise} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{notation}[theorem]{Notation} \newtheorem{problem}[theorem]{Problem} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{remark}[theorem]{Remark} \newtheorem{solution}[theorem]{Solution} \newtheorem{summary}[theorem]{Summary} \newenvironment{proof}[1][Proof]{\textbf{#1.} }{\ \rule{0.5em}{0.5em}} \input{tcilatex} \begin{document} \author{A. U. Thor} \title{Lab Report} \date{The Date } \maketitle \begin{abstract} A Laboratory report created with Scientific Notebook \end{abstract} \section{Integr\'{a}lny po\v{c}et funkci\'{\i} komplexnej premennej} \begin{center} \begin{tabular}{|c|c|c|c|c|c|c|} \hline \textbf{% %TCIMACRO{\hyperref{Obsah}{}{}{mcindex.tex}}% %BeginExpansion \msihyperref{Obsah}{}{}{mcindex.tex}% %EndExpansion } & \textbf{% %TCIMACRO{\hyperref{Obsah kapitoly}{}{}{K3.tex}}% %BeginExpansion \msihyperref{Obsah kapitoly}{}{}{K3.tex}% %EndExpansion } & \textbf{% %TCIMACRO{\hyperref{Predch\'{a}dzaj\'{u}ca str\'{a}nka}{}{}{K31.tex}}% %BeginExpansion \msihyperref{Predch\'{a}dzaj\'{u}ca str\'{a}nka}{}{}{K31.tex}% %EndExpansion } & \textbf{% %TCIMACRO{\hyperref{Nasleduj\'{u}ca str\'{a}nka}{}{}{K33.tex}}% %BeginExpansion \msihyperref{Nasleduj\'{u}ca str\'{a}nka}{}{}{K33.tex}% %EndExpansion } & \textbf{% %TCIMACRO{\hyperref{Ot\'{a}zky}{}{}{Ot3.tex}}% %BeginExpansion \msihyperref{Ot\'{a}zky}{}{}{Ot3.tex}% %EndExpansion } & \textbf{% %TCIMACRO{\hyperref{Cvi\v{c}enia}{}{}{Cv3.tex}}% %BeginExpansion \msihyperref{Cvi\v{c}enia}{}{}{Cv3.tex}% %EndExpansion } & \textbf{% %TCIMACRO{\hyperref{Index}{}{}{Ind.tex}}% %BeginExpansion \msihyperref{Index}{}{}{Ind.tex}% %EndExpansion } \\ \hline \end{tabular} \end{center} \section{Cauchyho integr\'{a}lna veta} Hovor\'{\i}me, \v{z}e oblas\v{t} ohrani\v{c}en\'{a} jednou jednoduchou, spojitou, uzavretou krivkou sa naz\'{y}va \emph{jednoducho s\'{u}visl\'{a} oblas\v{t}.} Oblas\v{t} ohrani\v{c}en\'{a} dvomi jednoduch\'{y}mi, spojit% \'{y}mi, uzavret\'{y}mi a nepret\'{\i}naj\'{u}cimi sa krivkami naz\'{y}vame \emph{dvojn\'{a}sobne s\'{u}visl\'{a} oblas\v{t}, ....} Hovor\'{\i}me, \v{z}% e hranica oblasti je kladne orientovan\'{a}, ak pri pohybe po hranici v smere jej orient\'{a}cie vn\'{u}tro oblasti zost\'{a}va po na\v{s}ej \v{l}% avej ruke. Je mo\v{z}n\'{e} dok\'{a}za\v{t} nasleduj\'{u}cu vetu: \begin{theorem} \label{1}Nech $f$ je analytick\'{a} funkcia v jednoducho s\'{u}vislej oblasti $D.$ Ak $C$ je jednoduch\'{a}, po \v{c}astiach hladk\'{a} uzavret% \'{a} krivka v $D,$ potom \[ \int_{C}f\left( z\right) dz=0. \] \end{theorem} \begin{tabular}{|c|} \hline \textbf{% %TCIMACRO{\hyperref{D\^{o}kaz}{}{}{DO324.tex}}% %BeginExpansion \msihyperref{D\^{o}kaz}{}{}{DO324.tex}% %EndExpansion } \\ \hline \end{tabular}% \label{7} \ \begin{example} Vypo\v{c}\'{\i}tajte integr\'{a}l \[ \int_{C}\frac{2z^{2}-3z+4}{z+1}dz, \]% ak $\varphi :\left\langle 0,2\pi \right\rangle \longrightarrow \mathbf{C}% ,\,\,\varphi \left( t\right) =\frac{1}{2}\cos t+i\frac{1}{2}\sin t.$ \end{example} \begin{solution} Krivka $C$ je uzavret\'{a}, hladk\'{a} krivka, $C=\left\{ z\in \mathbf{C}% ;\,\left| z\right| =\frac{1}{2}\right\} .$ Nech $D=\left\{ z\in \mathbf{C}% ;\,\left| z\right| =\frac{3}{4}\right\} .$ $D$ je jednoducho s\'{u}visl\'{a} a $C\subset D.$ Funkcia $f\left( z\right) =\frac{2z^{2}-3z+4}{z+1}$ je analytick\'{a} v oblasti $D,$ teda $\int_{C}\frac{2z^{2}-3z+4}{z+1}dz=0.$ \thinspace $\square $ \end{solution} Cauchyho integr\'{a}lna veta implikuje skuto\v{c}nos\v{t}, \v{z}e integr\'{a}% l z analytickej funkcie v jednoducho s\'{u}vislej oblasti $D$ \emph{nez\'{a}% vis\'{\i} od integra\v{c}nej cesty.} Nech $C_{1},\,C_{2}$ s\'{u} dve jednoduch\'{e}, po \v{c}astiach hladk\'{e}, nepret\'{\i}naj\'{u}ce sa krivky s rovnak\'{y}m za\v{c}iato\v{c}n\'{y}m bodom $z_{1}$ aj rovnak\'{y}m koncov% \'{y}m bodom $z_{2}.$ Nech $C_{1}\subset D$ a $C_{2}\subset D.$ Potom $% C=C_{1}+C_{2}^{-},$ kde $C_{2}^{-}$ m\'{a} opa\v{c}n\'{u} orient\'{a}ciu ako $C_{2},\,$je jednoduch\'{a} uzavret\'{a}, po \v{c}astiach hladk\'{a} krivka, ktor\'{a} sp\'{l}\v{n}a predpoklady %TCIMACRO{\hyperref{vety}{}{}{K32.tex#1}}% %BeginExpansion \msihyperref{vety}{}{}{K32.tex#1}% %EndExpansion , teda \[ 0=\int_{C}f\left( z\right) dz=\int_{C_{1}}f\left( z\right) dz+\int_{C_{2}^{-}}f\left( z\right) dz=\int_{C_{1}}f\left( z\right) dz-\int_{C_{2}}f\left( z\right) dz, \]% teda \[ \int_{C_{1}}f\left( z\right) dz=\int_{C_{2}}f\left( z\right) dz. \]% Tak sme dok\'{a}zali tvrdenie. \begin{lemma} \label{8}Ak je $f:D\left( \subset \mathbf{C}\right) \longrightarrow \mathbf{C% }$ analytick\'{a} funkcia definovan\'{a} v jednoducho s\'{u}vislej oblasti $% D,$ potom $\int_{C}f\left( z\right) dz$ v oblasti $D$ nez\'{a}vis\'{\i} od cesty. \end{lemma} Ak $z_{1}$ je za\v{c}iato\v{c}n\'{y} bod a $z_{2}$ koncov\'{y} bod krivky $% C, $ potom namiesto z\'{a}pisu $\int_{C}f\left( z\right) dz$ m\^{o}\v{z}me pou\v{z}i\v{t} z\'{a}pis $\int_{z_{1}}^{z_{2}}f\left( z\right) dz.$ \begin{description} \item[Pozn\'{a}mka] Nech $f:D\left( \subset \mathbf{C}\right) \longrightarrow \mathbf{C}$ je analytick\'{a} funkcia v jednoducho s\'{u}% vislej oblasti $D$ a nech $z_{1},\,z_{2}\in D.$ Nech pre ka\v{z}d\'{e} $z\in D$ plat\'{\i} $F\,^{\prime }\left( z\right) =f\left( z\right) ,$ potom funkciu $F$ naz\'{y}vame \emph{primit\'{\i}vnou funkciou} k funkcii $f$ na $% D.$ Ak m\'{a} analytick\'{a} funkcia $f$ primit\'{\i}vnu funkciu $F,$ potom \[ \int_{z_{1}}^{z_{2}}f\left( z\right) dz.=F\left( z_{2}\right) -F\left( z_{1}\right) . \] \end{description} \begin{example} Vypo\v{c}\'{\i}tajte integr\'{a}l \[ \int_{C}z^{2}dz \]% kde a) $C$ je \'{u}se\v{c}ka sp\'{a}jaj\'{u}ca bod $z_{1}=0$ s bodom $z_{2}=1+i,$ b) $C$ je krivka zlo\v{z}en\'{a} z dvoch \'{u}se\v{c}iek sp\'{a}jaj\'{u}cich body $z_{1}=0,\,z_{2}=1,\,z_{3}=1+i.$ \end{example} \begin{solution} a) Krivka $C$ sa d\'{a} parametrizova\v{t} takto $\varphi :\left\langle 0,1\right\rangle \longrightarrow \mathbf{C},\,\,\varphi \left( t\right) =t\left( 1+i\right) .$ Potom $\varphi ^{\prime }\left( t\right) =1+i,$ odkia% \v{l} potom m\'{a}me \[ \int_{0}^{1+i}z^{2}dz=\int_{0}^{1}\left[ t\left( 1+i\right) \right] ^{2}\left( 1+i\right) dt=\left( 1+i\right) ^{3}\left[ \frac{t^{3}}{3}\right] _{0}^{1}=\frac{1}{3}\left( -2+2i\right) . \]% Pou\v{z}it\'{\i}m pozn\'{a}mky tento v\'{y}sledok dostaneme priamo: preto% \v{z}e je $\frac{z^{3}}{3}$ primit\'{\i}vna funkcia ku funkcii $z^{2},$ m% \'{a}me \[ \int_{0}^{1+i}z^{2}dz=\left[ \frac{t^{3}}{3}\right] _{0}^{1+i}=\frac{1}{3}% \left( -2+2i\right) . \]% b) Teraz m\'{a}me $C=C_{1}+C_{2},$ kde% \[ C_{1}:\varphi _{1}:\left\langle 0,1\right\rangle \longrightarrow \mathbf{C}% ,\,\varphi _{1}\left( t\right) =t\text{ \ a \ }\varphi ^{\prime }\left( t\right) =1, \]% \[ C_{2}:\varphi _{2}:\left\langle 0,1\right\rangle \longrightarrow \mathbf{C}% ,\,\varphi _{2}\left( t\right) =1+it\text{\ a\ }\varphi ^{\prime }\left( t\right) =i, \] potom% \[ \int_{C}z^{2}dz=\int_{C_{1}}z^{2}dz+\int_{C_{2}}z^{2}dz=\int_{0}^{1}t^{2}dt+% \int_{0}^{1}\left( 1+it\right) ^{2}idt= \]% \[ =\frac{1}{3}+\left[ \frac{\left( 1+it\right) ^{3}}{3i}\right] _{0}^{1}=\frac{% 1}{3}\left( -2+2i\right) . \]% Porovnan\'{\i}m a) a b) vid\'{\i}me, \v{z}e n\'{a}\v{s} integr\'{a}l nez\'{a}% vis\'{\i} od cesty. Ale tento v\'{y}sledok sme o\v{c}ak\'{a}vali, preto\v{z}% e funkcia $f\left( z\right) =z^{2}$ je analytick\'{a} funkcia v celej komplexnej rovine $C$. \thinspace $\square $ \end{solution} Treba si uvedomi\v{t}, \v{z}e funkcia $f\left( z\right) =\left( \overline{z}% \right) ^{2}$ z %TCIMACRO{\hyperref{pr\'{\i}kladu}{}{}{K31.tex#2} }% %BeginExpansion \msihyperref{pr\'{\i}kladu}{}{}{K31.tex#2} %EndExpansion nie je analytick\'{a} funkcia a preto $\int_{C}\left( \overline{z}\right) ^{2}dz$ z\'{a}vis\'{\i} od integra\v{c}nej cesty. \subsection{Cauchyho integr\'{a}lna veta vo viacn\'{a}sobne s\'{u}visl\'{y}% ch oblastiach.} \begin{theorem} \label{4}Nech je $f:A\left( \subset \mathbf{C}\right) \longrightarrow \mathbf{C}$ je analytick\'{a} funkcia a $C_{0},C_{1},C_{2},\dots ,C_{n}$ s% \'{u} jednoduch\'{e}, po \v{c}astiach hladk\'{e}, uzavret\'{e} krivky orientovan\'{e} v jednom smere, ktor\'{e} sp\'{l}\v{n}aj\'{u} podmienky: a) $\overline{IntC_{i}}\subset IntC_{0},$ pre $i=1,2,\dots ,n$ b) $\overline{IntC_{i}}\cap \overline{IntC_{j}}=\emptyset ,$ pre ka\v{z}d% \'{e} $i\neq j,\,i,j=1,2,\dots ,n$ c) $\overline{IntC_{0}}\,\setminus \bigcup_{i=1}^{n}IntC_{i}\subset A.$ Potom \[ \int_{C_{0}}f\left( z\right) dz=\sum_{i=1}^{n}\int_{C_{i}}f\left( z\right) dz. \] \end{theorem} \begin{tabular}{|c|} \hline \textbf{% %TCIMACRO{\hyperref{D\^{o}kaz}{}{}{DO321.tex}}% %BeginExpansion \msihyperref{D\^{o}kaz}{}{}{DO321.tex}% %EndExpansion } \\ \hline \end{tabular}% \label{2} \begin{theorem} \label{5}(Veta o deform\'{a}cii integra\v{c}nej krivky.) Nech $C_{1},\,C_{2}$ s\'{u} dve uzavret\'{e}, po \v{c}astiach hladk\'{e} rovnako orientovan\'{e} krivky, ktor\'{e} sa nepret\'{\i}naj\'{u} a $C_{2}\subset IntC_{1}.$ Nech tieto krivky a mno\v{z}ina bodov le\v{z}iacich medzi nimi le\v{z}ia v oblasti $D.$ Ak $f:D\left( \subset \mathbf{C}\right) \longrightarrow \mathbf{% C}$ je analytick\'{a} funkcia, tak \[ \int_{C_{1}}f\left( z\right) dz=\int_{C_{2}}f\left( z\right) dz. \] \end{theorem} \begin{tabular}{|c|} \hline \textbf{% %TCIMACRO{\hyperref{D\^{o}kaz}{}{}{DO322.tex}}% %BeginExpansion \msihyperref{D\^{o}kaz}{}{}{DO322.tex}% %EndExpansion } \\ \hline \end{tabular}% \label{3} \begin{example} Vypo\v{c}\'{\i}tajte \[ \int_{C}\frac{1}{z-a}dz, \]% kde $C$ je jednoduch\'{a}, uzavret\'{a}, po \v{c}astiach hladk\'{a}, kladne orientovan\'{a} krivka komplikovan\'{e}ho tvaru, ktor\'{a} vo svojom vn\'{u}% tri obsahuje singul\'{a}rny bod $z=a$ funkcie $f\left( z\right) =\frac{1}{z-a% }.$ \end{example} \begin{solution} Aplikujeme %TCIMACRO{\hyperref{vetu}{}{}{K32.tex#5} }% %BeginExpansion \msihyperref{vetu}{}{}{K32.tex#5} %EndExpansion a dostaneme, \v{z}e ak $C_{1}$ je krivka jednoduch\'{e}ho tvaru, jednoduch% \'{a}, uzavret\'{a}, po \v{c}astiach hladk\'{a}, kladne orientovan\'{a} napr% \'{\i}klad kru\v{z}nica so stredom v bode $z=a$ a s polomerom $R,$ kror\'{a} le\v{z}\'{\i} v $IntC,$ tak dostaneme \[ \int_{C}\frac{1}{z-a}dz=\int_{C_{1}}\frac{1}{z-a}dz. \]% Preto\v{z}e \[ C_{1}:\,\varphi _{1}:\left\langle 0,2\pi \right\rangle \longrightarrow \mathbf{C},\,\varphi _{1}\left( t\right) =a+Re^{it},\,\varphi _{1}^{\prime }\left( t\right) =iRe^{it}, \]% potom \[ \int_{C}\frac{1}{z-a}dz=\int_{C_{1}}\frac{1}{z-a}dz=\int_{0}^{2\pi }\frac{% iRe^{it}}{a+Re^{it}-a}dt=i[t]_{0}^{2\pi }=2\pi i.\,\square \] \end{solution} \begin{example} Vypo\v{c}\'{\i}tajte \[ \int_{C}\frac{1}{\left( z-a\right) ^{n}}dz,\,n\neq 1,\,n\in \mathbf{Z,} \]% kde $C$ je jednoduch\'{a}, uzavret\'{a}, po \v{c}astiach hladk\'{a}, kladne orientovan\'{a} krivka, ktor\'{a} vo svojom vn\'{u}tri obsahuje bod $z=a.$ \end{example} \begin{solution} Aplikujeme t\'{u} ist\'{u} met\'{o}du ako v predch\'{a}dzaj\'{u}com pr\'{\i}% klade a dostaneme \[ \int_{C}\frac{1}{\left( z-a\right) ^{n}}dz=\int_{C_{1}}\frac{1}{\left( z-a\right) ^{n}}dz=\int_{0}^{2\pi }\frac{iRe^{it}}{\left( a+Re^{it}-a\right) ^{n}}dt= \]% \[ =\frac{i}{R^{n-1}}\int_{0}^{2\pi }e^{i\left( 1-n\right) t}dt=\frac{i}{R^{n-1}% }\left[ \frac{e^{i\left( 1-n\right) t}}{i\left( 1-n\right) }\right] _{0}^{2\pi }=0, \]% pri\v{c}om sme pou\v{z}ili \[ e^{\left( 1-n\right) 2\pi i}=\cos \left( \left( 1-n\right) 2\pi \right) +i\sin \left( \left( 1-n\right) 2\pi \right) =1, \]% tak dostaneme \[ \int_{C}\frac{1}{\left( z-a\right) ^{n}}dz=\left\{ \begin{tabular}{ccc} $2\pi i$ & pre & $n=1$ \\ $0$ & pre & $n=0,-1,\pm 2,\dots $% \end{tabular}% \right. .\,\square \] \end{solution} \subsection{Cauchyho integr\'{a}lna formula.} \begin{theorem} \label{9}Nech $C$ je jednoduch\'{a}, uzavret\'{a}, po \v{c}astiach hladk\'{a}% , kladne orientovan\'{a} krivka a $f:IntC\cup C\longrightarrow \mathbf{C}$ je analytick\'{a} funkcia. Potom pre ka\v{z}d\'{e} $a\in IntC$ \[ f\left( a\right) =\frac{1}{2\pi i}\int_{C}\frac{f\left( z\right) }{z-a}dz. \] \end{theorem} \begin{tabular}{|c|} \hline \textbf{% %TCIMACRO{\hyperref{D\^{o}kaz}{}{}{DO323.tex}}% %BeginExpansion \msihyperref{D\^{o}kaz}{}{}{DO323.tex}% %EndExpansion } \\ \hline \end{tabular}% \label{6} Nasleduj\'{u}cu vetu uvedieme bez d\^{o}kazu. Dok\'{a}\v{z}eme ju %TCIMACRO{\hyperref{nesk\^{o}r}{}{}{K33.tex#7}}% %BeginExpansion \msihyperref{nesk\^{o}r}{}{}{K33.tex#7}% %EndExpansion . \begin{theorem} \label{10}Ka\v{z}d\'{a} funkcia $f\left( z\right) $ analytick\'{a} v uzavretej oblasti $\overline{D}$ m\'{a} v tejto oblasti deriv\'{a}cie v\v{s}% etk\'{y}ch r\'{a}dov a plat\'{\i} \[ f^{\left( n\right) }\left( a\right) =\frac{n!}{2\pi i}\int_{C}\frac{f\left( a\right) }{\left( z-a\right) ^{n+1}}dz,\,n=1,2,\dots \]% kde $C$ je uzavret\'{a}, kladne orientovan\'{a}, po \v{c}astiach hladk\'{a} krivka, ktor\'{a} le\v{z}\'{\i} so svoj\'{\i}m vn\'{u}trom v $\overline{D}.$ \end{theorem} \begin{example} Vypo\v{c}\'{\i}tajte integr\'{a}l $\int_{C}\frac{e^{z}}{z\left( z-i\right) }% dz,$ kde $C$ je kru\v{z}nica so stredom v bode $z=i$ a s polomerom $\frac{1}{% 2}.$ \end{example} \begin{solution} Preto\v{z}e funkcia $f\left( z\right) =\frac{e^{z}}{z}$ je analytick\'{a} vo vn\'{u}tri kruhu $C$ a bod $z=i\in IntC,$ potom \[ \int_{C}\frac{e^{z}}{z\left( z-i\right) }dz=\int_{C}\frac{\frac{e^{z}}{z}}{% z-i}dz=2\pi i\left[ \frac{e^{z}}{z}\right] _{z=i}=2\pi e^{i}.\,\square \] \end{solution} \begin{example} Vypo\v{c}\'{\i}tajte integr\'{a}l $\int_{C}\frac{\sin z}{\left( z-i\right) ^{4}}dz,$ kde $C$ je jednoduch\'{a}, uzavret\'{a}, po \v{c}astiach hladk\'{a}% , kladne orientovan\'{a} krivka obsahuj\'{u}ca bod $i.$ \end{example} \begin{solution} Funkcia $\sin z$ je analytick\'{a} funkcia v $\mathbf{C}$. Bod $z=i\in IntC,$ potom \[ \int_{C}\frac{\sin z}{\left( z-i\right) ^{4}}dz=\frac{2\pi i}{3!}\left[ \frac{d^{3}\left( \sin z\right) }{dz^{3}}\right] _{z=i}=-\frac{\pi i}{3}\cos i=-\frac{\pi i}{3}\cosh 1.\,\square \] \end{solution} \begin{center} \begin{tabular}{|c|c|c|c|c|c|c|} \hline \textbf{% %TCIMACRO{\hyperref{Obsah}{}{}{mcindex.tex}}% %BeginExpansion \msihyperref{Obsah}{}{}{mcindex.tex}% %EndExpansion } & \textbf{% %TCIMACRO{\hyperref{Obsah kapitoly}{}{}{K3.tex}}% %BeginExpansion \msihyperref{Obsah kapitoly}{}{}{K3.tex}% %EndExpansion } & \textbf{% %TCIMACRO{\hyperref{Predch\'{a}dzaj\'{u}ca str\'{a}nka}{}{}{K31.tex}}% %BeginExpansion \msihyperref{Predch\'{a}dzaj\'{u}ca str\'{a}nka}{}{}{K31.tex}% %EndExpansion } & \textbf{% %TCIMACRO{\hyperref{Nasleduj\'{u}ca str\'{a}nka}{}{}{K33.tex}}% %BeginExpansion \msihyperref{Nasleduj\'{u}ca str\'{a}nka}{}{}{K33.tex}% %EndExpansion } & \textbf{% %TCIMACRO{\hyperref{Ot\'{a}zky}{}{}{Ot3.tex}}% %BeginExpansion \msihyperref{Ot\'{a}zky}{}{}{Ot3.tex}% %EndExpansion } & \textbf{% %TCIMACRO{\hyperref{Cvi\v{c}enia}{}{}{Cv3.tex}}% %BeginExpansion \msihyperref{Cvi\v{c}enia}{}{}{Cv3.tex}% %EndExpansion } & \textbf{% %TCIMACRO{\hyperref{Index}{}{}{Ind.tex}}% %BeginExpansion \msihyperref{Index}{}{}{Ind.tex}% %EndExpansion } \\ \hline \end{tabular} \end{center} \rule{6.5in}{0.04in} \textsl{Matematick\'{a} anal\'{y}za III} \section{Integr\'{a}lny po\v{c}et funkci\'{\i} komplexnej premennej} \end{document}