%\documentstyle[12pt,palatino]{article}
\documentstyle[12pt]{article}
\topmargin=-1.25in
\oddsidemargin=0in
%\evensidemargin=0in
\textwidth=6.5in
\textheight=10.5in
\parindent=0in
%\parskip=10pt
\pagestyle{empty}

%\newfont{\cmss}{cmss12.tfm}
\tolerance=3000
%\setcounter{chapter}{1}

\begin{document}

\input 312top

\centerline{ Class 12 {\bf (Monday February 9)}}

\begin{description}
\item[SUMMARY] The Complex Exponential and other elementary functions
\vspace{-12pt}
\item[CURRENT READING] Brown \& Curchill, pages 65-75
\vspace{-12pt}
\item[NEXT READING] Brown \& Curchill pages 75-82
\end{description}

Now that we know something about analytic functions in general we need to 
expand our repertoire of complex functions.\\

{\bf The Complex Exponential $e^z$}\\
The complex version of the exponential function is defined like this:

$e^z = e^{x+iy} = e^x (\cos y + i \sin y)$, where $|e^z| = e^x$ and arg$(e^z) = y + 2 k \pi$, $k=0,\pm1, \pm2, \ldots$

$\mbox{arg}(e^z) = y + 2k\pi \qquad \qquad (k=0,\pm1, \pm2, \ldots)$

\underline{\bf Exercise }\\
Show that $f(z) = e^z$ is an {\it entire function} and that $f'(z) = e^z$\\
Take some time ( 3 minutes) to try and prove this. You will have to answer the
questions:
\begin{itemize}
\item[1:] What is an entire function?
\item[2:] How do you show that a function is analytic?
\item[3:] Do the real and complex parts of $e^z$ obey the CRE?
\end{itemize}
\vspace{1in}

{\bf More Properties of $e^z$}\\
\begin{itemize}
\vspace{-12pt}
\item $e^z$ is never zero
\item $e^z = 1 \iff z = 2 \pi k i$
\item $e^{z_1} = e^{z_2} \iff z_1 = z_2 + 2 k \pi i, \qquad \mbox{ where } k \in Z$
\item $e^z$ is a periodic function with period $2 \pi i$
\end{itemize}

%\vspace*{1in}
Sketch a {\it fundamental region} for $e^z$ below
\newpage

{\bf Other elementary functions} \\

Once we have a handle on $\boldmath \exp z$ we can use it to define other functions, most notably $\boldmath \sin z$ and $\boldmath \cos z$

$$\sin z = {\displaystyle \frac{e^{iz} - e^{-iz}}{2i}}, \qquad \qquad \cos z = 
{\displaystyle \frac{e^{iz} + e^{-iz}}{2}}$$

\underline{\bf\sc Groupwork }\\
Show that $\frac{d}{d z} \cos z = - \sin z$ by using the definition of $\cos z$
\vspace{1in}

There a whole bunch of typical trigonometric identities which are valid for
complex trig functions. Most of these can be proved using the definitions 
involving exponentials. For example, $\mbox{tan} z$, $\mbox{sec }z$ are analytic everywhere except at the zeroes of $\cos z$.

\underline{\bf\sc Groupwork}\\
Find the zeroes of $\cos z$ and $\sin z$
\vspace{2in}

The usual rules of derivatives of the trig functions remain valid for their complex counterparts.

%\centerline{\bf Complex Trigonometric Identities}
{\bf Complex Trigonometric Identities}\\
\vspace{-12pt}

\begin{eqnarray}
\sin (z + 2 \pi) = \sin z, &  & \cos(z+2 \pi) = \cos z \nonumber \\ \nonumber
\sin( -z) = - \sin z, & & \cos (-z) = \cos z\\ \nonumber
\sin^2 z + \cos ^2 z = 1, & & \tan^2 z + 1 = \sec^2 z\\ \nonumber
\sin 2z = 2 \sin z \cos z, & & \cos 2z = \cos^2 - \sin^2 z\\  \nonumber
\sec z = {1 \over \cos z}, & & \tan z = {\sin z \over \cos z}\\  \nonumber
\frac{d}{dz} \tan z = \mbox{sec}^2 z, \frac{d}{dz} \sec z = \mbox{sec} z \tan z & & \frac{d}{dz} \sin z = \cos z, \frac{d}{dz} \sin z = \cos z  \nonumber
\end{eqnarray}

Similarly the hyperbolic trigonometric functions can be defined using the complex exponential and the newly-defined complex trig functions 
$$\mbox{sinh }  z = {\displaystyle \frac{e^{z} - e^{-z}}{2}}, \qquad \qquad \mbox{cosh } z = 
{\displaystyle \frac{e^{z} + e^{-z}}{2}}$$ 

%\centerline{hyperbolic trigonometric identities}
{\bf Complex Hyperbolic Trigonometric Identities}\\
%\vspace{-12pt}
\begin{eqnarray}
\sinh z = -i \sin iz, & & \cosh z = \cos (iz)\nonumber \\  \nonumber
\frac{d}{dz} \mbox{sinh } z = \mbox{cosh }z, &  & \frac{d}{d z} \mbox{cosh }z = \mbox{sinh } z  \nonumber
\end{eqnarray}


\end{document}
