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Smith}% %{Inductive inference: theory and methods}% %{Computing Surveys 15}{1983}{pp.237--269.} \newcommand{\BibItem}[6]{% \bibitem{#1} #2 \newblock #3 \newblock #4 #5 #6 } \newenvironment{Proof}[1]{\vspace*{-1.0ez}\begin{proof}#1}{\end{proof}} %%%%%%%% ``Theorem-like'' environments (defs, lemmas numbered like Theorems) % \newtheorem{Theorem}{Theorem}[section] % numbe by section %%\newtheorem{Theorem}{Theorem} % don't number by section \newtheorem{Definition}[Theorem]{Definition} \newtheorem{Proposition}[Theorem]{Proposition} \newtheorem{Lemma}[Theorem]{Lemma} \newtheorem{Corollary}[Theorem]{Corollary} \newtheorem{Conjecture}[Theorem]{Conjecture} \newtheorem{Question}[Theorem]{Question} % % to label, reference them % \newcommand{\ThmRef}[1]{\ref{thm:#1}} \newcommand{\ThmLabel}[1]{\label{thm:#1}} \newcommand{\DefRef}[1]{\ref{def:#1}} \newcommand{\DefLabel}[1]{\label{def:#1}} \newcommand{\PropRef}[1]{\ref{prop:#1}} \newcommand{\PropLabel}[1]{\label{prop:#1}} \newcommand{\LemRef}[1]{\ref{lem:#1}} \newcommand{\LemLabel}[1]{\label{lem:#1}} \newcommand{\CorRef}[1]{\ref{cor:#1}} \newcommand{\CorLabel}[1]{\label{cor:#1}} \newcommand{\ConjLabel}[1]{\label{conj:#1}} \newcommand{\ConjRef}[1]{\ref{conj:#1}} % %%%%%%%% %%%%%%%% Commands for labelling & referencing figures & equations (1:label) % \newcommand{\FigLabel}[1]{\label{fig:#1}} \newcommand{\EqnLabel}[1]{\label{eqn:#1}} \newcommand{\FigRef}[1]{\ref{fig:#1}} \newcommand{\EqnRef}[1]{\ref{eqn:#1}} % %%%%%%%% Commands for correctly labelling equations and figures %%%%%%%% by Chapter number %%%%%%%% \newcounter{CHAPTEREQN} \setcounter{CHAPTEREQN}{0} \renewcommand{\theequation}{\mboz{\arabic{CHAPTEREQN}.\arabic{equation}}} \newcommand{\reseteqn}{ \stepcounter{CHAPTEREQN}% \setcounter{equation}{0}% } \newcounter{CHAPTERFIG} \setcounter{CHAPTERFIG}{0} \renewcommand{\thefigure}{\mboz{\arabic{CHAPTERFIG}.\arabic{figure}}} \newcommand{\resetfig}{ \stepcounter{CHAPTERFIG}% \setcounter{figure}{0}% } %% Body of text for Document \begin{document} \centerline{\bf Complex Analysis Notes}\medskip \centerline{\bf Princeton Lectures In Analysis II}\rm\medskip \centerline{\bf Dan Singer}\rm\medskip\medskip\medskip \noindent\bf \underline{The Field $\complex$}\rm\medskip \noindent\bf Definition: \rm $\complex=\{a+bi:a,b\in\reals\}$\medskip \noindent \bf Addition: \rm $$(a+bi)+(a'+b'i)=(a+a')+(b+b')i.$$ This is associative and commutative.\medskip \noindent $\complex$ is a group under addition, with identity element $0+0i$ and inverse operation $$-(a+bi)=(-a)+(-b)i.$$\medskip \noindent \bf Multiplication: \rm $$(a+bi)(a'+b'i)=(aa'-bb')+(ab'+a'b)i.$$ This is associative and commutative: \medskip \noindent $\complex^*$ is a group under multiplication, with identity element $1+0i$ and inverse operation $$(a+bi)^{-1}={a\over a^2+b^2}+{-b\over a^2+b^2}i.$$\medskip \noindent One check that multiplication is distributive. Hence $\complex$ is a field.\medskip \noindent \bf Complex Conjugation: \rm $$\overline{a+bi}=a-bi.$$ One can check that complex conjugation is a field isomorphism, i.e. that $\overline{z+w}=\overline{z}+\overline{w}$ and $\overline{zw}=\overline{z}\overline{w}$ for all $z,w\in\complex$.\medskip \noindent \bf Norm: \rm $$||a+bi||=\sqrt{a^2+b^2},$$ $$z\overline{z}=||z||^2.$$\medskip \noindent \bf Real and Imaginary Parts: \rm $$\re(z)={z+\overline{z}\over 2}, \qquad \im(z)={z-\overline{z}\over 2i}$$\medskip \noindent\bf Lemma: \rm $\re(z)\le ||z||$.\medskip \noindent\bf Proof: \rm This follows from $a\le \sqrt{a^2+b^2}$. \EndOfProof\medskip \noindent\bf Triangle Inequality: \rm For all $z,w\in\complex$, $||z+w||\le ||z||+||w||$.\medskip \noindent\bf Proof: \rm $$||z+w||^2=(z+w)(\overline{z}+\overline{w})=||z||^2+z\overline w+w\overline{z}+||w||^2=$$ $$||z||^2+2\re(z\overline{w})+||w||^2\le ||z||^2+2||z\overline{w}||+||w||^2=$$$$||z||^2+2||z||||w||+||w||^2=(||z||+||w||)^2.$$ \EndOfProof\medskip \noindent\bf Corollary: \rm For all $z,w\in\complex$, $|||z||-||w|||\le ||z-w||$.\medskip \noindent\bf Proof: \rm This follows from $||z||\le ||z-w||+||w||$ and $||w||\le ||w-z||+||z||$. \EndOfProof\medskip \noindent\bf Euler's Notation: \rm For $\theta\in\reals$, $$e^{i\theta}=\cos\theta+i\sin\theta.$$ Trigonometric identities yield $$e^{i\theta_1}e^{i\theta_2}=e^{i(\theta_1+\theta_2)}.$$\medskip \noindent\bf Polar Form: \rm Every $z\in\complex$ lies on a circle of radius $r\ge 0$ about the origin and can be expressed in the form $z=re^{i\theta}$ where $r>0$ and $\theta\in\reals$. In fact, $r=||z||$ and $\theta$ is any angle satisfying $r\cos\theta=\re z$ and $r\sin\theta=\im z$. Using Euler's notation we can see that complex multiplication can be interpreted in terms of rotation and dilation.\medskip \noindent\bf Solutions to $z^n=c$ where $c\not=0$: \rm Write $c=re^{i\theta}$ where $r>0$. We seek all $z=se^{i\psi}$ with $s>0$ satisfying $$s^ne^{in\psi}=re^{i\theta}.$$ We must have $s=r^{1\over n}$ and $e^{in\psi}=e^{i\theta}$. This forces $n\psi=\theta+2k\pi$ where $k\in\ints$, or $\psi={\theta\over n}+{2k\over n}\pi$, which yields $$z=r^{1\over n}e^{i\left({\theta\over n}+{2k\over n}\pi\right)}.$$ There are $n$ distinct values of $z$, corresponding to $0\le k0:\exists N:n\ge N\implies ||z_n-z||<\epsilon.$$\medskip \noindent\bf Example: \rm $\lim_{n\rightarrow\infty}{\left({1+n\over 2n}+{2n^5\over 3n^5-1000}i\right)}={1\over 2}+{2\over 3}i$.\medskip \noindent\bf Proof: \rm Let $\epsilon>0$ be given. We wish to find $N$ so that $n>N$ implies $||\left({1+n\over 2n}+{2n^5\over 3n^5-1000}i\right)-\left({1\over 2}+{2\over 3}i\right)||<\epsilon$, or equivalently $||{1\over 2n}+{2000\over 9n^5-3000}i||<\epsilon$. Given that $$\left|\left|{1\over 2n}+{2000\over 9n^5-3000}i\right|\right|\le \left|\left|{1\over 2n}\right|\right|+\left|\left|{2000\over 9n^5-3000}i\right|\right|=\left|{1\over 2n}\right|+\left|{2000\over 9n^5-3000}\right|,$$ it suffices to require ${1\over 2n}<{\epsilon\over 2}$ and ${2000\over 9n^5-3000}<{\epsilon\over 2}$. The first inequality occurs when $n>{2\over 2\epsilon}$. Given that $9n^5-3000>8n^5$ when, for examle, $n>10$, we have $${2000\over 9n^5-3000}<{2000\over 8n^5}\le {2000\over 8n}<{\epsilon\over 2}$$ when $n>{4000\over 8\epsilon}$. So we can choose any $N$ greater than all three of the numbers ${2\over 2\epsilon},10,{4000\over 8\epsilon}$. \EndOfProof\medskip \noindent\bf Example: \rm Let $z\in\reals$ satisfying $0<||z||<1$. Then $\lim_{n\rightarrow\infty}{z^n}=0$.\medskip \noindent\bf Proof: \rm Let $\epsilon>0$ be given. We wish to find $N$ so that $n> N$ implies $||z^n||<\epsilon$, or equivalently $\left({1\over ||z||}\right)^n>{1\over\epsilon}$. Write ${1\over ||z||}=1+\theta$ where $\theta>0$. By the Binomial Theorem, $\left({1\over ||z||}\right)^n=(1+\theta)^n\ge1+n\theta$. We wish to require $1+n\theta>{1\over\epsilon}$. We just need any natural $N$ satisfying $N> {{1\over\epsilon}-1\over \theta}={{1\over\epsilon}-1\over {1\over ||z||}-1}$. \EndOfProof\medskip \noindent\bf Theorem: \rm A convergent sequence cannot have two distinct limits.\medskip \noindent\bf Proof: \rm Suppose $z_n\rightarrow w$ and $z_n\rightarrow w'$ where $w\not=w'$. Then for each $n$ we have $||w-w'||\le ||w-z_n||+||z_n-w'||$, and for sufficiently large $n$, $||z_n-w||<{||w-w'||\over 2}$ and $||z_n-w'||<{||w-w'||\over 2}$, which implies $||w-w'||<||w-w'||$, a contradiction. \EndOfProof\medskip \noindent\bf Theorem: \rm Assume $(z_n)$ converges to $z$. Then every subsequence $(z_{n_k})$ converges to $z$.\medskip \noindent\bf Proof: \rm Let $\epsilon>0$ be given. Then there exists $N$ such that $k\ge N$ implies $||z_k-z||<\epsilon$, hence $k\ge N$ implies $n_k\ge k\ge N$ implies $||z_{n_k}-z||<\epsilon$. \EndOfProof\medskip \noindent\bf Theorem: \rm Assume that $(z_n)$ is convergent and that a subsequence $(z_{n_k})$ converges to $z$. Then $(z_n)$ converges to $z$.\medskip \noindent\bf Proof: \rm If $(z_n)$ converges to $w$ then $(z_{n_k})$ converges to $w$. By uniqueness of limits, $w=z$. Hence $(z_n)$ converges to $z$. \EndOfProof\medskip \noindent\bf Theorem: \rm If $z_n\rightarrow z$ then $||z_n||\rightarrow ||z||$.\medskip \noindent\bf Proof: \rm This follows from $|||z_n||-||z|||\le ||z_n-z||\rightarrow 0$.\qed\medskip \noindent\bf \underline{The Sum, Product, and Quotient Rules}\rm\medskip \noindent\bf Theorem: \rm Assume $\lim_{n\rightarrow\infty}{z_n}=z$ and $\lim_{n\rightarrow\infty}{w_n}=w$. Then:\medskip \noindent (1) $\lim_{n\rightarrow\infty}{z_n+w_n}=z+w$\medskip \noindent (2) $\lim_{n\rightarrow\infty}{z_nw_n}=zw$\medskip \noindent (3) When $w_n\not=0$ for all $n$ and $w\not=0$, $\lim_{n\rightarrow\infty}{z_n\over w_n}={z\over w}$.\medskip \noindent\bf Proof: \rm \medskip \noindent (1) We have $$||(z_n+w_n)-(z+w)||\le ||z_n-z||+||w_n-w||.$$ In order to make this quantity $<\epsilon$, it suffices to make $||z_n-z||<{\epsilon\over 2}$ and $||w_n-w||<{\epsilon\over 2}$. Given $\epsilon>0$, we will choose $N$ so that $n\ge N$ forces both inequalities.\medskip \noindent (2) We have $$||z_nw_n-zw||\le ||z_nw_n-zw||+||zw-zw_n||=||z_n-z||||w||+||w_n-w||||z||.$$ In order to make this quantity $<\epsilon$, it suffices to make $||z-z_n||<{\epsilon\over 2(1+||w||)}$ and $||w-w_n||<{\epsilon\over 2(1+||z||)}$. Given $\epsilon>0$, we will choose $N$ so that $n\ge N$ forces both inequalities.\medskip \noindent (3) We have $$\left|\left|{z_n\over w_n}-{z\over w}\right|\right| =\norm{{z_nw-zw_n\over ww_n}}\le {\norm{z_n-z}\norm{w}\over\norm{w}\norm{w_n}}+{\norm{w-w_n}\norm{z}\over\norm{w}\norm{w_n}}.$$ We will first show that the denominator contribution can be bounded above. Since $w_n\rightarrow w$ and $||w||>0$, there exists $N_1$ such that $n\ge N_1$ implies $|||w_n||-||w|||\le ||w_n-w||<{||w||\over 2}$, which implies $||w_n||>{||w||\over 2}$. Hence $n\ge N$ implies $${1\over\norm{w}\norm{w_n}}\le {2\over ||w||^2}.$$ Now let $\epsilon>0$ be given. Then there exists $N_2$ such that $n\ge N_2$ implies $||z-z_n||||w||<{\epsilon||w||^2\over 4}$ and $||w_n-w||||z||<{\epsilon||w||^2\over 4}$. Hence for any $n$ larger than both $N_1$ and $N_2$, $$\left|\left|{z_n\over w_n}-{z\over w}\right|\right| <\epsilon.$$ \EndOfProof \medskip \noindent\bf\underline{A Brief Review of the Topology of $\reals$}\rm\medskip \noindent \bf Least Upper Bound Axiom: \rm Every $S\subseteq\reals$ that has an upper bound has a least upper bound. \medskip \noindent\bf Example: \rm The set $(-\infty,1)$ has many upper bounds, including the number 1. None of the numbers in $(-\infty,1)$ is an upper bound, because if $t\in(-\infty,1)$ then ${t+1\over 2}\in(-\infty,1)$ as well, and since $t<{t+1\over 2}$, $t$ cannot be an upper bound. Therefore 1 is the least upper bound of $(-\infty,1)$.\medskip \noindent\bf Example: \rm Fix $\sigma>1$. Let $S=\{s_n:n\in\nats\}$ where $$s_n=\sum_{k=1}^n{{1\over k^\sigma}}={1\over 1^\sigma}+{1\over 2^\sigma}+\cdots+{1\over n^\sigma}.$$ The set $S$ is bounded above: for any $p\in\nats$ we have $$s_{2^p-1}=\sum_{i=1}^{p}{\left(\sum_{k=2^{i-1}}^{2^i-1}{1\over k^\sigma}\right)}\le \sum_{i=1}^{p}{\left(\sum_{k=2^{i-1}}^{2^i-1}{1\over (2^{i-1})^\sigma}\right)}=\sum_{i=1}^p{\left(1\over 2^{\sigma-1}\right)^{i-1}}=$$$${1-\left(1\over 2^{\sigma-1}\right)^{p}\over 1-{1\over 2^{\sigma}}}\le {1\over 1-{1\over 2^{\sigma}}}.$$ For any $n\in\nats$, $n\ge 2^p-1$ for some $p\in\nats$, hence $s_n\le s_{2^p-1}\le {1\over 1-{1\over 2^{\sigma}}}$ for all $n\in\nats$. So $S$ has a least upper bound.\medskip \end{document}