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\centerline{{\it  Class 26} {\bf (Monday March 30)}}
 
\begin{description}
\item[SUMMARY] Applications and Implications of Cauchy's Integral Formula
\vspace{-12pt}
\item[CURRENT READING] Brown \& Curchill pages 123-125
\vspace{-12pt}
\item[NEXT READING] Brown \& Curchill pages 125-129
\end{description}


\centerline{\bf Applications of Cauchy's Integral Formula}

Let $C$ be a simple closed (positively oriented) contour. If $f$ is 
analytic in some simply connected domain $D$ containing $C$ and $z_0$ is
{\bf any point inside} of $C$, then
$${\displaystyle \int_{C} \frac{f(z)}{z-z_0} dz} = 2 \pi i f(z_0)$$
and
$${\displaystyle \int_{C} \frac{f(z)}{(z-z_0)^{m}} dz = \frac{2 \pi i}{(m-1)!} f^{(m-1)}(z_0)}
$$
These two results lead to a number of other results. Actually, the two 
formulas are just restatement of one formula, known as the {\it generalized 
Cauchy Integral Formula}. Can you see how the first expression ({\bf CIF}) is 
just a special case ($m = ??$) of the second one?\\

\underline{\bf Examples}\\
We have rewritten the integral formulas in the way above so that we can use them to actually evaluate integrals. Let's to do the following two.\\ 
${\displaystyle \oint_{C} \frac{e^{5z}}{z^3} \, dz} =$
\vfill
(where $C$ is $|z|=1$ traversed once clockwise)\\
${\displaystyle \int_{C} \frac{2z+1}{z(z-1)^2} \, dz} =$
\vfill
(where $C$ is given in the sketch)
\newpage
There are numerous theorems which directly follow from Cauchy's Integral Formula. I have listed a {\it few} of the more famous ones below...\\

\centerline{\bf Implications of Cauchy's Integral Formula}

{\bf Morera's Theorem} \\
 If $f(z)$ is continuous in a simply-connected region $R$ and if $\oint_C f(z) dz = 0$ around {\it every} simple closed curve $C$ in $R$, then $f(z)$ is analytic in $R$.\\
(NOTE: Morera's Theorem is the converse of the Cauchy-Goursat theorem.)\\
\vfill
{\bf Cauchy's Inequality}\\
 If $f(z)$ is analytic inside and on a circle of radius $r$ and centered at $z=z_0$ then $$| f^{(n)}(z_0) | \le \frac{M \cdot n!}{r^n} \qquad \qquad n = 0, 1, 2, \ldots$$
where $M$ is an upper bound on $|f(z)|$ on $C$
%(i.e. a real number $M$ such that $|f(z)| < M$ on $C$)
\vfill

{\bf Liouville's Theorem}\\
Suppose that for all $z$ in the entire complex plane, if $f(z)$ is analytic and
bounded, (i.e. $|f(z)| < M$ for some real constant $M$) then $f(z)$ must be a constant.
\vfill

{\bf Fundamental Theorem of Algebra}\\
Every polynomial equation $P(z) = a_0 + a_1 z + a_2 z^2 + \cdots  + a_n z^n = 0$ with degree $n \ge 1$ and $a_n \neq 0$ has atleast one root.
\vfill

{\bf Gauss' mean value theorem}\\
If $f(z)$ is analytic inside and on a circle $C$ with center $z_0$ and radius $r$ then $f(z_0)$ is the mean of the values of $f(z)$ on $C$, namely
$${\displaystyle f(z_0) = \frac{1}{2 \pi} \int_0^{2 \pi} f(z_0 + r e^{i \theta}) d\theta }$$

{\bf Maximum modulus theorem}\\
If $f(z)$ is analytic inside and on a simple closed curve $C$ and is not identically equal to a constant, then the maximum value of $|f(z)|$ occurs on $C$.
\vfill

{\bf Minimum modulus theorem}\\
If $f(z)$ is analytic inside and on a simple closed curve $C$ and $f(z) \neq 0$ inside $C$, then the minimum value of $|f(z)|$ occurs on $C$.
\vfill

{\bf The Argument Theorem}\\
Let $f(z)$ be analytic inside and on a simple closed curve $C$ except for a finite number of poles inside $C$. Then $${\displaystyle \frac{1}{2 \pi i} \oint \frac{f'(z)}{f(z)} dz = N - P}$$
where $N$ and $P$ are the number of zeroes and poles of $f(z)$ inside $C$
\vfill

{\bf Rouche' Theorem}\\
If $f(z)$ and $g(z)$ are analytic inside and on a simple closed curve $C$ and if $|g(z)| < |f(z)|$ on $C$, then $f(z) + g(z)$ and $f(z)$ have the same number of zeros inside of $C$. 
\vfill
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\newpage
The results we will be considering are
\begin{itemize}
\item {\bf Cauchy's Inequality}
\item {\bf Liouville's Theorem}
\item {\bf Cauchy's Residue Theorem}
\end{itemize}

The first two can be derived from the generalised CIF almost immediately.\\

\underline{\bf Cauchy's Inequality}\\
Let $f$ be analytic inside and on a circle $|z-z_0|=R$, taken in the positive sense, and denoted by $C_R$, then the derivatives of $f$ as $z_0$ satisfy
$$|f^{n}(z_0)| \le \frac{n! M}{R^n} \qquad \qquad (n=1,2,3,\ldots)$$
where $|f(z)| \le M$ for all $z$ on $C_R$.\\

\underline{\bf Proof}:
(Use the generalized CIF to try and bound $f^{(n)}(z_0)$)
\vfill

\underline{\bf Liouville's Theorem}\\
If $f$ is {\it entire} and {\it bounded} in the complex plane, then $f(z)$ is constant throughout the plane.\\

%{\bf Proof:}\\
\underline{\bf Proof}:
(Use Cauchy's Inequality when $n=1$ and assume the radius has to be big enough to include the entire plane, e.g. ${R \rightarrow \infty}$)
\vfill
%{\bf Pregnant Pause:} Take 5 minutes and think about the proof. To illustrate that you really DO understand it, try and explain the proof to your nearest neighbor.

\underline{\bf Cauchy's Residue Theorem}\\
If $f$ is analytic on a simple (positively oriented) closed contour $C$ and everywhere inside $C$ except the finite number of points $z_1, z_2, \cdots z_n$ inside $C$, then
$${\displaystyle \oint_{C} f(z) \, dz = 2 \pi i \sum_{k=1}^n \mbox{\bf Res}(z_k)}$$

Before we talk about what a {\bf residue} of a function $f(z)$ at $z_k$, denoted by Res$(f,z_k)$ or Res$(z_k)$ we will need to have a discussion about {\bf poles},{\bf zeroes} and {\bf singularities}.

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