\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}
