Download - Complex at infinity
\documentclass[11pt,reqno]{amsart}\usepackage{amscd,amssymb,amsthm,epsfig}\setlength{\oddsidemargin}{0.0in}\setlength{\evensidemargin}{0.0in}\setlength{\topmargin}{0.25in}\setlength{\textheight}{8.0in}\setlength{\textwidth}{6.5in}\renewcommand{\baselinestretch}{1.2}\newcounter{ictr}\newenvironment{ilist}{\begin{list}{\textup{(\roman{ictr})}}{\usecounter{ictr}\setlength{\leftmargin}{0.6truein}\setlength{\itemsep}{0.0truein}\setlength{\labelwidth}{0.3truein}}}{\end{list}}\begin{document}\title{Limits Involving the Point at Infinity.}\author{Jerry DeVor Math 480}\email{[email protected]}\date{\today}\begin{abstract}This talk will examine the concept of limit with respect to the extended complex plane. This includes the so-called point at infinity. It will begin with a discussion of the sterographic projection of the extended complex plane onto a unit sphere. Then for complex values of z and w = f(z), the set of all z in the neighborhood of infinity will be defined. This will enable an analysis of limits of complex functions as specific variables approach infinity.\end{abstract}\maketitle\section{Traditional Concept of Limit in the Complex Plane}Let z be a complex value and w = f(z). Then \begin{equation*}\lim_{z\to z_{o}} f(z) = w_{o}. \end{equation*}means that for each positive real $\varepsilon$ and $\delta$\begin{equation*}|f(z) - w_{o}| < \varepsilon \end{equation*}whenever\begin{equation*}0 < |z - z_{o}| < \delta \end{equation*}The definition of limit provides a deleted $\delta$ neighborhood of $z_{o}$ in the complex plane for every $\epsilon$ neighborhood of $w_{o}$ in the w (or u-v) plane. Here every point z in the delta neighborhood of $z_{o}$ is mapped to the region encompassed by the $\varepsilon$ neighborhood of $w_{o}$. \section{The Riemann Sphere and Extended Complex Plane}Consider the complex plane as incident with the equator of a unit sphere centered at z = 0. Each point on the plane may be represented by the intersection of a line incident with the point z and the north pole of the sphere. This line is, of course, also incident with the surface of the sphere. Thus there is a one-to-one correspondence between points of the extended complex plane (which includes the point at inifinity) and points on the surface of the sphere.\subsection{Neighborhood of Infinity}Note that all points on the complex plane exterior to the unit circle correspond to points on the upper hemisphere of the Riemann sphere excluding the point at infinity. Consequently for arbitrarily small values of $\varepsilon$ all points on the plane exterior to the circle \begin{equation*}|z| = 1/\varepsilon \end{equation*}correspond to points near the north pole on the Riemann sphere. Considering all such points z as inhabiting a "neighborhood of infinity" on the complex plane provides for a mechanism to evaluate limits of complex functions hitherto unavailable for analyis.\section{Limits at Infinity}The expression of limit in the complex plane\begin{equation*}\lim_{z\to z_{o}} f(z) = w_{o} \end{equation*}when the variables $w_{o}$, $z_{o}$ or both are replaced by the point at infinity may now be evaluated. Indeed\begin{ilist}\item \begin{equation*}\lim_{z\to z_{o}} f(z) = \infty \Longleftrightarrow \lim_{z\to z_{o}} 1/f(z) = 0. \end{equation*}\item \begin{equation*}\lim_{z\to\infty} f(z) = w_{o} \Longleftrightarrow \lim_{z\to 0} f(1/z) = w_{o}.\end{equation*}\item \begin{equation*}\lim_{z\to\infty} f(z) = \infty \Longleftrightarrow \lim_{z\to 0} 1/f(1/z) = 0.\end{equation*}\end{ilist}Proof of (i):Note that the first of the two limits means that for any positive $\varepsilon$ there exists a positive $\delta$ such that \begin{equation*}|f(z)| > 1/\varepsilon \end{equation*}when\begin{equation*}0 1/\delta\end{equation*}. Then z lies within a neighborhood of infinity and \begin{equation*}|f(z)- w_{o}| < \varepsilon\end{equation*}represents a mapping of z to within an $\varepsilon$ neighborhood of $w_{o}$. Substituting 1/z for z yields: \begin{equation*}0 < |z-0| < \delta \Rightarrow |f(1/z) - w_{o}| < \varepsilon \end{equation*}Proof of (iii). In this case the statement says that \begin{equation*}|f(z)| > 1/\varepsilon \end{equation*}when\begin{equation*}|z| > 1/\delta\end{equation*}Substituting 1/z for z yields \begin{equation*}|1/z| > 1/\delta \Rightarrow |z| < \delta \Rightarrow 0 < |z-0| < \delta \end{equation*}and\begin{equation*}|f(1/z)| > 1/\varepsilon \Rightarrow 1/|f(1/z)| < \varepsilon\end{equation*}Thus\begin{equation*}0 < |z - 0| < \delta \Rightarrow |1/f(1/z)-0| < \varepsilon \end{equation*}and the required result is obtained.\section{2 Applications}Now a simple application using theorem (i).\begin{equation*}\lim_{z\to\ 1} 1/(z-1)^3 = \infty \end{equation*}since\begin{equation*}\lim_{z\to\ 1} 1/f(z) = lim_{z\to\ 1} 1/[1/(z-1)^3 = \lim_{z\to 1} (z-1)^3 = 0.\end{equation*}And theorem (ii) shows that\begin{equation*}\lim_{z\to\infty} 4z^2/(z-1)^2 = 4.\end{equation*}since \begin{equation*}\lim_{z\to\ 0} f(1/z) = \lim_{z\to\ 0} 4(1/z)^2/(1/z-1)^2 = \lim_{z\to\ 0} 4z^2/z^2(z-1)^2 = 4.\end{equation*}\end{document}