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\centerline{{\it  Class 25} {\bf (Friday March 27)}}
 
\begin{description}
\item[SUMMARY] Cauchy's Integral Formula and Integral Examples
\vspace{-12pt}
\item[CURRENT READING] Brown \& Curchill pages 123-129
\vspace{-12pt}
\item[NEXT READING] Brown \& Curchill pages 125-129
\end{description}

{\bf 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
$$f(z_0) = {\displaystyle \frac{1}{2 \pi i} \int_{C} \frac{f(z)}{z-z_0} dz}
$$
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The CIF leads to some of the most astonishing results in complex analysis. 
It is a truly amazing idea; that the value of an analytic function at a point $z_0$
 in a simply-connected domain depends on values it takes on some closed 
contour $C$ encircling the point. \\

An alternative  proof of the result is reasonably straightforward and involves the continuity of $f(z)$ at every point in $D$ and the formula for bounding a contour integral. You might try reading it on page 123-124 of the text.\\
 

{\bf Higher Derivatives of Analytic Functions}\\
Here is the first of many amazing ideas derived from the {\bf CIF}. \\
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
$$
f'(z_0) = {\displaystyle \frac{1}{2 \pi i} \int_{C} \frac{f(z)}{(z-z_0)^{2}} dz}
$$
and in fact you should be able to write down a general formula for the $n^{th}$ derivative of $f(z)$ evaluated at $z_0$ in terms of a contour integral:
$$f^{(n)}(z_0) = \hspace{2in}$$

\bigskip
This is an amazing result, because it means that when a function is analytic then all of its higher derivatives exist and are also each analytic!
 
\underline{\bf Exercise}\\
${\displaystyle \int_{|2z|=3} \frac{e^{z^2}}{(z+i)(z+2)} \, dz} = $ \hspace{2in}
${\displaystyle \int_{|z|=3} \frac{e^{z^2}}{(z+i)(z+2)} \, dz} = $ 
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\underline{\sc Groupwork}\\
$
{\bf 1.} {\displaystyle \oint_C \frac{x^2-y^2}{2} + xy i \, dz \qquad C: |z - i|=2 \mbox{ counter-clockwise}}\\
$
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$
{\bf 2.} {\displaystyle \oint_C \overline{z} \, dz, \qquad C: |z|=2 \mbox{  clockwise}}\\
$
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$
{\bf 3.} {\displaystyle \oint_C \frac{dz}{z^2+\pi^2} \qquad C: |z|=3 \mbox{ counter-clockwise}}\\
$
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$
{\bf 4.} {\displaystyle \oint_C \frac{\sinh(2z)}{z^2+\pi^2} \, dz \qquad C: |z+i|=3 \mbox{ counter-clockwise} }\\
$
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$
{\bf 5.} {\displaystyle \oint_C \frac{dz}{(z-3)^4} \qquad C: | z-2|=2 \mbox{ twice counter-clockwise}}\\
$
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\underline{\sc Exercise}\\
Evaluate the following integral $ { \displaystyle \oint_{C} \frac{z+i}{z^3+2z^2} dz }
$
where the contour $C$ is
\begin{description}
\item[(a)] the circle $|z|=1$ traversed once counter clockwise
\item[(b)] the circle $|z+2-i|=2$ traversed once counter clockwise
\item[(c)] the circle $|z-2i|=1$ traversed once counter clockwise
\end{description}
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