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\centerline{{\it  Class 22} {\bf (Monday March 9)}}
 
\begin{description}
\item[SUMMARY] Path Independence and the Cauchy-Goursat Theorem
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\item[CURRENT READING] Brown \& Curchill pages 104-111
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\item[NEXT READING] Brown \& Curchill pages 116-123
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We have been practicing our complex integration skills. Today we will learn how
to use {\bf antiderivatives} and the {\bf Cauchy-Goursat Theorem} to evaluate
contour integrals more efficiently (simply!)


Recall that the value of  ${\displaystyle \int_C  2 \overline{z}^2 \ dz}$ {\bf did} depend on the contour when we evaluated it in the previous class.

However the value of ${\displaystyle \int_C  2 {z}^2 \ dz}$ {\bf did not} depend on the contour chosen and in each case the integral is equal to $\frac{-32}{3}$

How is $f(z)=2 z^2$ a different function than $g(z)= 2 \overline{z}^2$?\\
$2 z^2$ is a \rule{2in}{.01mm} while $2 \zc^2$ is a \rule{2in}{.01mm}.\\

{\bf Independence of Path}\\
{\bf THEOREM:} Suppose that the function $f(z)$ is continuous in a domain (open connected set) $D$ and has an antiderivative $F(z)$ throughout $D$, i.e. $dF/dz=f(z)$ at each point in $D$. Then for {\it every} contour $\Gamma$ lying in $D$ connecting $z_1$ to $z_2$ we have the result
$${\displaystyle \int_{\Gamma} f(z) \, dz = F(z_2) - F(z_1)}$$
In otherwords, the value of the integral is {\bf independent of the path} chosen to link $z_1$ and $z_2$


So, getting back to the difference between $f(z)=2 z^2$ and $g(z)=2 \overline{z}^2$ we know that $2 z^2$ has the property that it is continuous and is equal to the derivative of $2 z^3/3$ on the open set $z \in {\bf C}$. $g(z)$ has no 
similar antiderivative. (This is not surprising, since $g(z)$ doesn't have a derivative either.)\\

\underline{\bf Example}\\
${\displaystyle \int_C 2 z^2 \ dz} =$ where $C$ is a contour from $2$ to $-2$
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Note that according to the above theorem the function $F(z)$ will be {\bf analytic} and {\bf continuous} on the domain $D$ (since it has a derivative at every point of the open set $D$).

A corollary of the above theorem is the famous\\
{\bf Cauchy-Goursat Theorem:}
If $f(z)$ is analytic at all points interior to and on any simple closed contour ${\Gamma}$, then
$${\displaystyle \oint_{\Gamma}  f(z) dz  } = 0.$$
(Is this result surprising?)

\underline{\sc Groupwork}\\
1. ${ \displaystyle \int_C \sin (iz) + 2  e^{z} \, dz, \qquad \mbox{ where } C \mbox{is a contour joining } 0 \mbox{ to } \pi i }$ 
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2.${ \displaystyle \oint_C \frac{dz}{z^2-4} \, dz, \qquad \mbox{ where } C: |z|=1}$
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3. 
$
{ \displaystyle \int_{-i}^i \frac{dz}{z^2} \qquad \mbox{ and } \qquad  \oint_{C_r} \frac{dz}{z^2} \mbox{ where } C_r: |z|=1 } 
$
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