\documentclass[11pt]{article}

\usepackage{amsmath}
\usepackage{amsthm}
\usepackage{amssymb}
\input epsf
\oddsidemargin=-0.0cm
\evensidemargin=-0.0cm
\topmargin=-0.0cm
\textwidth=7in
\textheight=9in
\newtheorem{theorem}{Theorem}
\newtheorem{corollary}{Corollary}
\newtheorem{lemma}{Lemma}
\newtheorem{example}{Example}
\newtheorem{definition}{Definition}


\def\ds{\displaystyle}
\def\ul{\underline}


\title{Cauchy-Riemann equations}
%pagestyle{}
\author{March 31, 2011}
\date{}

\begin{document}
\maketitle
%\markboth{Thomae}{Thomae}

\begin{definition}
Let $f(z)$ be a complex valued function defined in a neighborhood of a point $z_1=x_1+iy_1$.  Then $f$ is \textbf{complex differentiable} at $z_1$ if 
\begin{equation}
f(z)=f(z_1)+p(z)(z-z_1),
\end{equation}
where $p$ is continuous at $z_1$.  We define the complex derivative of $f$ at $z_1$ to be the value of $p$ at $z_1$.
\begin{equation}
f'(z_1)=p(z_1).
\end{equation}
\end{definition}


\begin{theorem} Let $f(z)=u(x,y) + iv(x,y)$.  Then $f$ is complex differentiable at  $z_1$ if and only if $u$ and $v$ are real differentiable at $(x_1,y_1)$ and 
\begin{equation}
u_x(x_1,y_1)= v_y(x_1,y_1), \  u_y(x_1,y_1)=-v_x(x_1,y_1).
\end{equation}
These are called the Cauchy-Riemann equations.
\end{theorem}
\begin{proof}
Let us write equation (1) in matrix form:
\begin{equation}
\begin{bmatrix}
u(x,y)\\
v(x,y
\end{bmatrix}
= 
\begin{bmatrix}
u(x_1,y_1)\\
v(x_1,y_1)
\end{bmatrix}
+ 
\begin{bmatrix}
a(x,y) & -b(x,y)\\
b(x,y) & a(x,y)
\end{bmatrix}
\begin{bmatrix}
x-x_1\\
y-y_1
\end{bmatrix}
\end{equation}
We can write this line by line
\begin{align}
u(x,y)= & u(x_1,y_1) + a(x,y)(x-x_1) - b(x,y)(y-y_1)\\
v(x,y) = & v(x_1,y_1) + b(x,y)(x-x_1) + a(x,y)(y-y_1).
\end{align}
 In this form we see that $u$ and $v$ are real differentiable at $(x_1,y_1)$ and $u_x = v_y, \ u_y=-v_x$ at $(x_1,y_1)$.
 
 Next suppose $u$ and $v$ are real differentiable at $(x_1,y_1)$ and (3) is true.  Let $a_1=u_x(x_1,y_1)=v_y(x_1,y_1), \  b_1= v_x(x_1,y_1)=-u_y(x_1,y_1)$. Let's write the definition of real differentiability as follows
 \begin{align}
 u(x,y)=&u(x_1,y_1) + a_1(x-x_1)-b_1(y-y_1)+e(x,y)\\
 v(x,y) = & v(x_1,y_1) + b_1(x-x_1)+a_1(y-y_1) +d(x,y),
 \end{align}
 where $e(x,y)/|z-z_1| \to 0, d(x,y)/|z-z_1| \to 0 $ as $z \to z_1$.  So 
 \begin{equation}
 f(z)=f(z_1) + (a_1+ib_1)(z-z_1) + \epsilon (z)(z-z_1), \text{ where } \epsilon (z)= \frac{e(z)+id(z)}{z-z_1}.
 \end{equation}
 and $\epsilon(z) \to 0$ as $z\to z_1$.
 This can be rewritten as
 \begin{equation}
 f(z)=f(z_1) + p(z)(z-z_1), 
 \end{equation}
 where $p(z) = a_1+ib_1+ \epsilon(z)$ and  $p(z) \to a_1+ib_1$ as $z\to z_1$ so $f$ is complex differentiable at $z_1$.
 \end{proof}
 

\end{document}
