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\centerline{{\it  Class 23} {\bf (Wednesday March 11)}}
 
\begin{description}
\item[SUMMARY] Applications of Cauchy-Goursat and simply-connected domains
\vspace{-12pt}
\item[CURRENT READING] Brown \& Curchill pages 104-111
\vspace{-12pt}
\item[NEXT READING] Brown \& Curchill pages 116-123
\end{description}

Previously we learned that the {\bf Cauchy-Goursat Theorem} implies that,\\
{\it given $f$ is continuous in a domain $D$ if any one of the following is true, then so are the others:}
\begin{description}
\item[(a)] $f$ has an antiderivative in $D$, called $F(z)$, such that $F'(z)=f(z)$
\item[(b)] the integrals of $f(z)$ along contours lying entirely in $D$ extending from $z_1$ to $z_2$ all have the same value, namely $F(z_2)-F(z_1)$
\item[(c)] the integrals of $f(z)$ around closed contours lying entirely in $D$ all have value zero
\end{description}
\underline{\bf Examples}\\
Evaluate the following integrals. In each case sketch the contour you used and
explain how you evaluated the integral, i.e. what idea are you using. Also answer the questions.\\
$1. {\displaystyle \int_{-i}^i \frac{dz}{z}} $
\vfill
Does the integral value change depending on the path from $-i$ to $i$? \\
Does $1/z$ have an antiderivative? In what domain?

2.  ${\displaystyle \oint_{|z|=1} \frac{dz}{z}} $
\vfill
Does the integral value change depending on what direction one takes along $|z|=1$? \\
What would the value of the integral be if the contour were traversed twice in the clockwise direction?

{\bf Loop}\\
A {\bf loop} or simple closed contour is said to be traversed in the {\bf positive sense} if an observer traversing the contour in that direction
would always see the {\it interior} of the contour on their left.

%\hspace{2in} {\displaystyle \oint_{C_r} \Log z dz} 


\newpage

{\bf Simply-Connected and Multiply-Connected Domains}\\
A {\bf simply-connected} domain $D$ is one such that {\it every} simple closed contour (i.e. loop) lying in  $D$ encloses only points of $D$\\

A domain which is {\it not} simply-connected, is called {\bf multiply-connected}.\\

\underline{\sc Groupwork}\\
Sketch and classify the following domains as {\bf simply-connected} or {\bf multiply-connected}
\begin{description}
\item[(a)] $| \mbox{Im } z | < 1$
\item[(b)] $1 < |z| < 2$
\item[(c)] {\bf C} $\setminus \lbrace $  Re$(z) >0 \cap $ Im $(z)=0 \rbrace$
\item[(d)]  $|z| < 4 \setminus \lbrace |z-i|<.5  \cup |z+i|<.5 \rbrace$
\end{description}
\vspace{2in}


FACT: Simply-connected domains have the property that every loop in $D$ can be continuously deformed in $D$ to a single point.\\

FACT: ${\displaystyle \int_{C} f(z) dz \equiv 0}$ if $C$ is a point.\\

{\bf Deformation Invariance Theorem}\\
Let $f$ be an analytic function in a domain $D$ containing loops $\Gamma_0$ and $\Gamma_1$. If these loops can be continuously deformed into each other by passing through points only in $D$, then
$${\displaystyle \oint_{\Gamma_0} f(z) \, dz = \oint_{\Gamma_1} f(z) \, dz}$$

\underline{\bf Examples}\\
Evaluate the integral below, where $\Gamma$ is some closed contour. \\[12pt]
${\displaystyle \oint_{\Gamma} \frac{dz}{z-z_0}}$
\vfill

Does it matter where $z_0 \in {\bf C}$ is?

%\underline{\bf QUESTION OF THE DAY (Due: 4/15)}\\
%Evaluate the following integral 
%$$
%{ \displaystyle \oint_{|z|=4} \frac{z}{(z-1)(z+2)} dz }
%$$
%where the contour is traversed {\it twice} in the clockwise direction.\\
%
% [HINT: Use partial fractions. What does the direction of traversal do to the sign of the value of the integral?]
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