\documentclass{amsart}
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\begin{document}

\title{On the Existence of the Roots of Polynomial Equations}
\maketitle


\textbf{Phil}  Today let's talk about polynomial equations.  Polynomial
equations have been one of the central subject in the history of
mathematics.  For a linear equation $x + a = 0$, obviously there
is no problem about the existence of a solution.  The existence
of a solution for a quadratic equation $x^2 + ax + b = 0$ is
harder than the existence of a solution for a cubic equation
$x^3 + ax^2 + bx + c$. Therefore, let's first consider the cubic equation.\\

\textbf{Stephan}  Did you say that cubic equations are simpler than
quadratic equations?\\

\textbf{Phil}  Yes, in fact any polynomial equation of an odd degree is
simpler than a quadratic equation for the existence of a solution.

\vspace{0.3cm}
\begin{figure}
\centering \includegraphics{quad.gif}
\caption{Graph of a quadratic polynomial}
\end{figure}
\vspace{0.3cm}

\begin{figure}
\centering \includegraphics{cube.gif} 
\caption{Graph of a cubic polynomial}
\end{figure}
\vspace{0.3cm}


\textbf{Phil} Consider, for example, the equation  $x^2 - 2x + 3 = 0$.
  Since   $x^2 - 2x + 3= (x - 1)^2 + 2 \ge 2$
for any real number $x$, there is no solution unless we count
complex numbers in.  The graph of $y = x^3 + ax^2 + bx + c$ 
varies with the values of the coefficients $a, b, c$.  However,
one common feature is that if the value of $x$ is sufficiently
small ( $x < 0$ with sufficiently great $|x|$), then the value of
$y$ is  negative; if the value of $x$ is sufficiently large, that is,
 $x > 0$ and $|x|$  is sufficient great, then the value of $y$ is positive. 
 On the other hand, $y = f(x)$ is a continuous function.
  Therefore, $x$ varies from the sufficiently small value to  the sufficiently
 great value, the curve of the graph should cross the $x$-axis. 
The $x$ coordinate of this crossing point is a solution of the
 equation $f(x) = 0$.\\

     Now, back to quadratic equations, let's try to prove the
existence of a solution for the equation $x^2 + x + 1 = 0$,
without actually solving it using square roots.  This attempt
will help us to prove the existence of a root for any polynomial
equation of degree $n$, whether it can be solved by radical roots
or not.\\
 
First, tell me about the graph of $y = x^2 + x + 1$. \\

\textbf{Jessica}  The graph is as shown here.  This graph never crosses
the x axis, so that there is no real number solution.\\

\textbf{Phil}  Right!  Therefore, as a next step we try to draw
its graph for complex numbers x and y.  As you know, the set of
all complex numbers are represented by a plane, so that we need
a four-dimensional space for the graph.  But we can draw sections of
the graph as follows.\\

First, notice that the image of the origin $0 + 0i$ in the complex plane
is $1 + 0i$ in the complex plane.  The complex
plane is the union of the origin and the concentric circles
$$
   C_r = \{ x \mid |x| = r\} = \{ u + iv \mid u^{2} + v^{2} = r^{2}\}
$$
for all positive real numbers $r$.  Therefore, we draw the
image of $C_r$ mapped by the function $y = x^{2} + x + 1$, that is
$$
     fC_r = \{ y \mid y = x^{2} + x + 1\; \mbox{and $|x| = r$} \} .
$$

\textbf{Phil} Stephan, can you express $C_r$ and $fC_r$ with only
one parameter, say $\theta$.  Try first $C_r$.\\

\textbf{Stephan}  All right.  first, the simplest circle $C_1$,
which is the unit circle around the $0+0i$ can be expressed by
 the parameter $\theta$ that represents the angle between the
vector $(0+0i, 1+0i)$ and the vector $(0+0i, u+vi)$.  That is,  
 $$
   C_1 = \{ u + vi \mid u = \mathrm{cos}\theta,  v = \mathrm{sin}\theta,
 0 \leq \theta \leq 2\pi \} .
$$
Therefore, 
 $$
   C_r = \{ u + vi \mid u = r\mathrm{cos}\theta,  v = r\mathrm{sin}\theta,
 0 \leq \theta \leq 2\pi\} .
$$
For $fC_r$, I have to calculate $f(u+vi) = (u+vi)^{2} + (u+vi) + 1$.\\

\textbf{Stephan}  Here we are.  
$$
fC_r = \{ s+ ti \mid \left\{ 
 \begin{array}{lll}
 s &=& r^{2}
(\mathrm{cos}^{2}\theta \mathrm{sin}^{2}\theta) + r\mathrm{cos}\theta + 1,\\
 t &=& 2r^{2}\mathrm{cos}\theta\mathrm{sin}\theta + r\mathrm{sin}\theta
\end{array} \right.
0 \leq \theta \leq 2\pi \} .
$$ 

\textbf{Phil}  Now, Jessica, you are good at computer programming, aren't you?
Can you write for us a programming to describe these $fC_r$ and
 give us their graphs at the next meeting.\\

\textbf{Jessica}  Yes, but by QBASIC I can view graphs on the screen,
but cannot get a hard copy of graphs.  So, I will use free graphics software
GNUPLOT to get hard copies and bring them at the next meeting.\\  
     
\textbf{Phil}  O.K. then,  Let r = 0.4, 0.8, 1.2, and a large value 12.   

 [At the next meeting}\\

\textbf{Jessica}  Here are the graphs of $fC_r$ for $r = 0.4, 0.8, 1.2$,
 and $12$.

\begin{figure}
\centering \includegraphics{ccub4.gif}
\caption{$fC_{0.4}$}
\end{figure}
\vspace{0.3cm}

{\begin{figure}
{\centering \includegraphics{ccub8.gif} \par}
\caption{$fC_{0.8}$}
\end{figure}\par}
\vspace{0.3cm}


\begin{figure}
\centering \includegraphics{ccub12.gif}
\caption{$fC_{1.2}$}
\end{figure}\par}
\vspace{0.3cm}

\begin{figure}
\centering \includegraphics{ccub120.gif}
\caption{$fC_{12}$}
\end{figure}
\vspace{0.3cm}


\textbf{Phil} Thank you very much.  Let us examine these graphs.
If $r$ is sufficiently small, like $r = 0.4$ in this case, then
$fC_r$ is a closed curve near the point $1 +0i$, as shown in Fig. 3. \\

Fig. 4 shows the image of $C_{0.8}$. In these graphs, the origin $0+0i$
is outside the close curves.\\

Now see Fig 5, which shows the image of $C_{1.2}$.  The origin is
inside the closed curve that is the image of $C_{1.2}$.  Therefore,
we conclude that between $r = 0.8$ and $r = 1.2$, there should be some
$r$ such that the origin is on the closed curve that is the image of $C_r$.\\

In order to apply the above method to the proof of general
polynomial equation, $f(x) =  x^n + a_{n-1}x^{n-1} + . . . + a_1x + a_0 = 0$,
where each $a_i$ is a complex number, we only  need to consider the images of
sufficiently small circle $C_r$ and large $C_r$.  If $a_0 = 0$, then obviously
 $x = 0$ is a solution of the equation.  Therefore, we
assume $a_0 \ne 0$.  If $r$ is sufficiently small, then as seen from
Fig. 3, the image of $C_r$ is a closed curve around $a_0$, so that the
origin is outside this closed curve.  In order to see the shape of the
image $fC_r$ for sufficiently large $r$.  we contract an image by
dividing each point of the image by $r^n$.  That is, we consider the
point set $\{ f(x)/r^n  \mid  |x|  = r \}$.   This set is clearly
 geometrically similar to $fC_r$, and contracted in the direction
 to the origin.  This means, in the above Fig. 6, the coordinate value
$12^2 = 144$ becomes 1.  The shape is close to a circular closed curve
that is winding twice around the origin, so that the origin is inside this
curve.

     In the general case of degree $n$, this curve is a closed curve
winding $n$ times around the origin.   This fact can be proved by 
$$
     \lim_{r\rightarrow\infty}(f(x)/r^n)
 = \lim_{r\rightarrow\infty}x^n/r^n = x'^n,
$$
where $x' = x/r$, so that $|x'| = 1$.\\

In conclusion, between the two extreme cases, the case where $r$ is
 sufficiently small and the case where $r$ is sufficiently large,
 there should be some $C_r$ such that the origin is on
the image $fC_r$.

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