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\begin{document}


\section{Ma 681}

\vspace{1pt}

\section{Lecture 4}

\vspace{1pt}

Continuity of a Complex Function

\vspace{1pt}

Note: In all that follows $f(z)$ is assumed to be a \textit{single-valued}
function unless indicated otherwise.

\vspace{1pt}

Definition: We say\textit{\ }that a complex function $f(z)$ defined in a
domain $D$ \textit{approaches the limit }$A$ \textit{\ as }$z$ \textit{%
approaches a point }$z_{0}$ in $D$ and we write

\vspace{1pt}

\begin{center}
$\lim_{z\rightarrow z_{0}}f(z)=A$

\vspace{1pt}
\end{center}

or $f(z)\rightarrow A$ as $z\rightarrow z_{0}$ if given any $\in $ $>0$ $%
\exists $ a number $\delta =\delta (\in )$ $>0$ such that

\vspace{1pt}

\begin{center}
$\left| f(z)-A\right| <\in $

\vspace{1pt}
\end{center}

for all $z$ such that

\vspace{1pt}

\begin{center}
$0<\left| z-z_{0}\right| <\delta .$

\vspace{1pt}
\end{center}

Suppose that in addition $A=f(z_{0})$, so that

\vspace{1pt}

\begin{center}
$\lim_{z\rightarrow z_{0}}f(z)=f(z_{0}).$

\vspace{1pt}
\end{center}

Then we say that $f(z)$ is \textit{continuous} \textit{at }$z_{0}.$

\vspace{1pt}

Remark: Thus $f(z)$ is said to be \textit{continuous} at $z_{0}$ if given $%
\in $ $>0$ $\exists $ a number $\delta (\in ,z_{0})>0$ such that

\vspace{1pt}

\begin{center}
$\left| f(z)-f(z_{0})\right| <$ $\in $

\vspace{1pt}
\end{center}

$\forall $ $z$ such that

\vspace{1pt}

\begin{center}
$\left| z-z_{0}\right| <\delta $

\vspace{1pt}
\end{center}

Geometrically this means that the values of $w=f(z)$ at every point of a
sufficiently small neighborhood $\left| z-z_{0}\right| <\delta $ of the
point $z_{0}$ all lie inside an arbitrarily small neighborhood $\left|
w-w_{0}\right| <\in $ of the point $w_{0}=f(z_{0}).$

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Example: The function\qquad $w=z^{n}\qquad (n=1,2,3,\ldots )$

\vspace{1pt}is continuous in the whole finite plane. In fact, setting $%
w_{0}=z_{0}^{n}$ where $z_{0}$ is any finite point, we have

\vspace{1pt}

\begin{center}
$w-w_{0}=z^{n}-z_{0}^{n}=(z-z_{0})(z^{n-1}+z^{n-2}z_{0}+\cdots
+z_{0}^{n-1}). $

\vspace{1pt}
\end{center}

$\Longrightarrow \qquad \left| w-w_{0}\right| =\left| z-z_{0}\right| \left|
z^{n-1}+z^{n-2}z_{0}+\cdots +z_{0}^{n-1}\right| \leq \left| z-z_{0}\right|
(r^{n-1}+r^{n-2}r_{0}+\cdots +r_{0}^{n-1})$

\vspace{1pt}

where $r=\left| z\right| $ and $r_{0}=\left| z_{0}\right| .$ But $\left|
z-z_{0}\right| <\delta $ $\Longrightarrow $

\vspace{1pt}

\begin{center}
$r=\left| z\right| =\left| z_{0}+(z-z_{0})\right| \leq \left| z_{0}\right|
+\left| z-z_{0}\right| <r_{0}+\delta $

\vspace{1pt}
\end{center}

Hence\qquad $\left| w-w_{0}\right| <\left| z-z_{0}\right| \lbrack
(r_{0}+\delta )^{n-1}+(r_{0}+\delta )^{n-2}r_{0}+\cdots +r_{0}^{n-1}\rbrack
<n\delta (r_{0}+\delta )^{n-1}.$

\vspace{1pt}

It follows that $\left| w-w_{0}\right| $ can be made less than any
preassigned positive number $\in $ by choosing $\delta $ suitably. Hence the
function $w=z^{n}$ is continuous at every point $z_{0}\neq \infty .$

\vspace{1pt}

Remarks: Since the definition of continuity for complex functions is
formally the same as for real functions, the proofs of the familiar theorems
concerning algebraic operations on continuous functions carry over from the
real case to the complex case. Thus if $f(z)$ and $g(z)$ are both continuous
at a point $z_{0}$, then so is $f(z)\pm g(z),$ $f(z)g(z),$ and $f(z)/g(z)$
provided that $g(z_{0})\neq 0$ in the quotient. Moreover, if the function $%
f(z)$ is continuous at the point $z_{0}$ and if the function $\varphi (w)$
is continuous at the point $w_{0}=f(z_{0})$, then the composite function $%
\varphi (f(z))$ is continuous at the point $z_{0}.$

\vspace{1pt}

Definition: A complex function $f(z)$ is said to be continuous in the domain 
$D$ if it is continuous at every point of $D.$

\vspace{1pt}

Remarks: When a complex function $f(z)$ is also defined at some boundary
points of $D,$we speak of ``continuity from the interior''.

\vspace{1pt}

$f(z)=u(x,y)+iv(x,y)$ is continuous $\Longleftrightarrow $ $u(x,y)$ and $%
v(x,y)$ are continuous functions of the real variables $x$ and $y.$

\vspace{1pt}

Uniform Continuity

\vspace{1pt}

In general in the definition of continuity $\delta $ $=\delta (\in ,z_{0}),$
that is, $\delta $ depends both upon the given value of $\in $ and the point 
$z_{0}$ under consideration. However, in certain cases $\delta $ depends 
\textit{only} upon $\in .$ This situation is called \textit{uniform
continuity. }

\vspace{1pt}

Definition: A complex function $f(z)$ defined in a domain $D$ is said to be 
\textit{uniformly continuous} in $D$ if given $\in $ $>0$ $\exists $ a
number $\delta =\delta (\in )$ such that

\vspace{1pt}

\begin{center}
$\left| f(z^{\prime })-f(z^{\prime \prime })\right| <\in $

\vspace{1pt}
\end{center}

for \textit{all} points $z^{\prime },z^{\prime \prime }$ in $D$ such that

\vspace{1pt}

\begin{center}
$\left| z^{\prime }-z^{\prime \prime }\right| <\delta $

\vspace{1pt}
\end{center}

Remark: Unlike the case of ordinary continuity, it is meaningless to talk
about uniform continuity at a single point $z_{0}$. The key observation here
is that the $\delta $ in the above definition must be \textit{independent}
of the points $z^{\prime }$ and $z^{\prime \prime }$ in $D.$ If $f(z)$ is
uniformly continuous in $D$, then it is certainly continuous in $D.$
However, a function can be continuous in $D$ without being uniformly
continuous in $D.$ In the real case continuity in a closed bounded domain
implies uniform continuity there. We have a similar result in the complex
case.

\vspace{1pt}

Theorem: If $f(z)$ is continuous in a closed bounded domain $\bar{D}$, then
it is uniformly continuous in $\bar{D}.$

\vspace{1pt}

Differentiability

\vspace{1pt}

Definition: A function $w=f(z)$ defined in a domain $D$ is said to be 
\textit{differentiable} at a point $z\in D$ if

\begin{center}
$\lim_{h\rightarrow 0}\frac{f(z+h)-f(z)}{h}\qquad \qquad (1)$

\vspace{1pt}
\end{center}

exits and its value is \textit{independent }of the way in which $%
h\rightarrow 0.$ This limit is denoted by $f^{\prime }(z)$ or $\frac{dw}{dz}$
and is called the \textit{derivative} of $f(z)$ at the point $z.$

\vspace{1pt}

Remarks: That is, $f(z)$ is differentiable at $z$ if $\exists $ a new number 
$f^{\prime }(z)$ associated with the point $z$ in such a way that given an
arbitrary $\in $ $>0$ $\ \exists $ a $\delta (\in ,z)$ such that

\vspace{1pt}

\begin{center}
$\left| \frac{f(z+h)-f(z)}{h}-f^{\prime }(z)\right| <\in $

\vspace{1pt}
\end{center}

whenever $z+h\in D$ and $\left| h\right| <\delta .$

\vspace{1pt}

Alternative Definition: $f(z)$ is differentiable at $z$ and has derivative $%
f^{\prime }(z)$ there, $\Longleftrightarrow $ for any sequence $\{z_{n}\}$
tending to $z,$ the sequence $\left\{ \frac{f(z_{n})-f(z)}{z_{n}-z}\right\}
\rightarrow f^{\prime }(z).$

\vspace{1pt}

\vspace{1pt}Example: let $f(z)=z^{2}.$ Then $f^{\prime }(z)=2z$ $\forall $ $%
z $ because

\vspace{1pt}

\begin{center}
$f^{\prime }(z)=\lim_{h\rightarrow 0}\frac{(z+h)^{2}-z^{2}}{h}%
=\lim_{h\rightarrow 0}\frac{2zh+h^{2}}{h}=\lim_{h\rightarrow 0}(2z+h)=2z$
\end{center}

\vspace{1pt}

Remark: Differentiability$\Longrightarrow $continuity. However, the converse
is not true.

\vspace{1pt}

Example: The function $f(z)=\left| z\right| $ is continuous, but it is not
differentiable.

\vspace{1pt}

Consider\qquad \qquad \qquad \qquad $\lim_{h\rightarrow 0}\frac{\left|
z+h\right| -\left| z\right| }{h}\qquad \qquad (2)$

\vspace{1pt}

and let $z=r(\cos \theta +i\sin \theta ).$

\vspace{1pt}

For $\theta =\theta _{0},$ a constant, let $h=\rho (\cos \theta _{0}+i\sin
\theta _{0}).$ Then $(2)$ becomes

\vspace{1pt}

\begin{center}
$\lim_{\rho \rightarrow 0}\frac{r+\rho -r}{\rho (\cos \theta _{0}+i\sin
\theta _{0})}=(\cos \theta _{0}-i\sin \theta _{0})$

\vspace{1pt}
\end{center}

This last expression is clearly \textit{not} independent of the approach to $%
z.$

\vspace{1pt}

Definition: If a complex function $w=f(z)$ is defined and differentiable at
all points of a domain $D$, we say that $f(z)$ is an \textit{analytic (or
regular or holomorphic)} function in $D.$

\vspace{1pt}

Remark: The statement, ``$f(z)$ is an analytic function'' means that $%
\exists $ \textit{some} domain of analyticity.

\vspace{1pt}

Definition: An analytic function is said to be \textit{regular at a point} $%
z_{0}$ if it is regular in a neighborhood $\left| z-z_{0}\right| <\in $ of $%
z_{0}.$

\vspace{1pt}

Example: $f(z)=\left| z\right| ^{2}$ is differentiable at the origin but it
is not regular there, since it is not differentiable anywhere else.
(Homework)

\begin{center}
\vspace{1pt}
\end{center}

Definition: $z_{0}$ is called a \textit{singular point }of $f\left( z\right) 
$ if $f\left( z\right) $ is not differentiable at $z_{0}$, but if every
neighborhood of $z_{0}$ contains points at which $f\left( z\right) $ is
differentiable.

\vspace{1pt}

Remark: $z^{n}$ ($n$ a positive integer) is a regular function of $z$ at all
finite points of the complex plane.

\vspace{1pt}

Remark: Given that there is a complete analogy between the definition $(1)$
of the complex derivative and the corresponding formula

\begin{center}
\vspace{1pt}
\end{center}

\vspace{1pt}

\begin{center}
$f^{\prime }(x)=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$

\vspace{1pt}
\end{center}

defining the definition of a real function, all of the standard formulae for
derivatives of sums, products, quotients, etc., also hold in the complex
case. Thus if $f(z)$ and $g(z)$ are differentiable at a point $z$ \ and $%
c_{1}$ and $c_{2}$ are any complex constants, then so are $c_{1}f(z)\pm
c_{2}g(z),$ $c_{1}f(z)$, $f(z)g\left( z\right) ,$ $f\left( z\right) /g\left(
z\right) ,$ (provided $g\left( z\right) \neq 0),$ and

\vspace{1pt}

\begin{center}
$\lbrack c_{1}f(z)\pm c_{2}g(z)\rbrack =c_{1}f^{\prime }\left( z\right) \pm
c_{2}g^{\prime }\left( z\right) $

\vspace{1pt}

$\lbrack c_{1}f(z)\rbrack ^{\prime }=c_{1}f^{\prime }\left( z\right) $

\vspace{1pt}

$\lbrack f(z)g\left( z\right) \rbrack ^{\prime }=f^{\prime }\left( z\right)
g(z)+f\left( z\right) g^{\prime }\left( z\right) $

\vspace{1pt}

$\left[ \frac{f(z)}{g\left( z\right) }\right] ^{\prime }=\frac{f^{\prime
}\left( z\right) g\left( z\right) -f\left( z\right) g^{\prime }\left(
z\right) }{g^{2}(z)}\qquad g\left( z\right) \neq 0$

\vspace{1pt}

\vspace{1pt}
\end{center}

Thus every \textit{polynomial} $f\left( z\right) =a_{0}+a_{1}z+\cdots
+a_{n}z^{n}$ \ (where $a_{0},a_{1},\ldots ,a_{n}$ are complex constants) is
regular in the finite complex plane, and every \textit{rational} function

\vspace{1pt}

\begin{center}
\vspace{1pt}$\frac{a_{0}+a_{1}z+\cdots +a_{n}z^{n}}{b_{0}+b_{1}z+\cdots
b_{m}z^{m}}$

\vspace{1pt}
\end{center}

(where the $a^{\prime }s$ and $b^{\prime }s$ are constants) is regular in
the finite complex plane, except at the zeroes of the denominator.

\vspace{1pt}

Example: a zero of the denominator is a singular point of a rational
function.

\vspace{1pt}

Theorem: Suppose $w=g(z)$ is an analytic function in a domain $D$ of the $%
z-plane$, and when $z\in D$ $\ w=g(z)$ lies in a domain $D^{\prime }$ of the
complex $w-plane.$ Also, suppose that $F\lbrack w\rbrack $ is an analytic
function of $w$ in $D^{\prime }.$ Then $f(z)=F\lbrack g\left( z\right)
\rbrack $ is an analytic function of $z$ in $D.$

\vspace{1pt}

In other words, an analytic function of an analytic function is again an
analytic function.

\vspace{1pt}

The derivative of $f\left( z\right) $ is given by the chain rule

\vspace{1pt}

\begin{center}
$f^{\prime }\left( z\right) $\vspace{1pt}$=F^{\prime }\left[ w\right]
g^{\prime }\left( z\right) =F^{\prime }\left[ g\left( z\right) \right]
g^{\prime }\left( z\right) $

\vspace{1pt}
\end{center}

Example: If $D$ is $\left| z\right| <1,$ and $w=g\left( z\right) =5z$ then $%
D^{\prime }$ is $\left| w\right| <5.$

If $D:\left| z\right| >0$ and $w=g\left( z\right) =\frac{1}{z}$, then $%
D^{\prime }:\left| w<\infty \right| $

\vspace{1pt}

\subsection{The Cauchy-Riemann Equations}

\vspace{1pt}

Suppose that $f\left( z\right) =f\left( x+iy\right) =u\left( x,y\right)
+iv\left( x,y\right) $ is differentiable at a point $z.$ We shall show that
differentiability implies a simple but characteristic property of $u\left(
x,y\right) $ and $v\left( x,y\right) .$

\vspace{1pt}

Recall that when $f^{\prime }\left( z\right) $ exists then the value of

\vspace{1pt}

\begin{center}
$f^{\prime }\left( z\right) =\lim_{h\rightarrow 0}\frac{f\left( z+h\right)
-f\left( z\right) }{h}$

\vspace{1pt}
\end{center}

is independent of the way in which $h\rightarrow 0.$ Let $h=\Delta x+i\Delta
y$ and consider the difference quotient

\vspace{1pt}

\begin{center}
$\frac{f\left( z+h\right) -f\left( z\right) }{h}=\frac{\lbrack u(x+\Delta x,%
\text{ }y+\Delta y)+iv(x+\Delta x,\text{ }y+\Delta y)\rbrack -\lbrack
u\left( x,y\right) +iv(x,y)\rbrack }{\Delta x+i\Delta y}.\qquad \qquad (1)$

\vspace{1pt}
\end{center}

We shall let $h=\Delta x+i\Delta y$ go to zero in two different ways:

\vspace{1pt}

1) Let $\Delta y=0,$ $\Delta x>0,$ and $\Delta x\rightarrow 0.$ Then $(1)$
becomes

\vspace{1pt}

\begin{center}
$\frac{f\left( z+h\right) -f\left( z\right) }{h}=\frac{\lbrack u(x+\Delta
x,y)+iv(x+\Delta x,y)\rbrack -\lbrack u\left( x,y\right) +iv(x,y)\rbrack }{%
\Delta x}=\frac{u\left( x+\Delta x,y\right) -u\left( x,y\right) }{\Delta x}+i%
\frac{v\left( x+\Delta x,y\right) -v\left( x,y\right) }{\Delta x}\qquad (2)$

\vspace{1pt}
\end{center}

Since the limit of the left hand side exist and equals $f^{\prime }\left(
z\right) $ then the limit of the right hand side must also exist, and
therefore

\vspace{1pt}

$\lim_{\Delta x\rightarrow 0}\frac{u\left( x+\Delta x,y\right) -u\left(
x,y\right) }{\Delta x}$ exists and $=$ $\func{Re}\lbrack f^{\prime
}(z)\rbrack $ and \ $\lim_{\Delta x\rightarrow 0}\frac{v\left( x+\Delta
x,y\right) -v\left( x,y\right) }{\Delta x}$ also exists and $=\func{Im}%
\lbrack f^{\prime }\left( z\right) \rbrack $. \ Hence $\frac{\partial u}{%
\partial x}$ and $\frac{\partial v}{\partial x}$ exist at $z=x+iy$ and

\vspace{1pt}

\begin{center}
$f^{\prime }\left( z\right) =u_{x}+iv_{x}\qquad \qquad (3)$

\vspace{1pt}
\end{center}

2) Let $\Delta x=0,$ $\Delta y>0$ and $\Delta y\rightarrow 0.$ Then $(1)$
becomes

\vspace{1pt}

\begin{center}
$\frac{f\left( z+h\right) -f\left( z\right) }{h}=\frac{u\left( x,y+i\Delta
y\right) -u\left( x,y\right) }{i\Delta y}+i\frac{v\left( x,y+i\Delta
y\right) -v\left( x,y\right) }{i\Delta y}\qquad \qquad (4)$

\vspace{1pt}
\end{center}

Let $\Delta y\rightarrow 0.$ $\Longrightarrow u_{y}$ and $v_{y}$ exist at
the point $z$ and

\vspace{1pt}

\begin{center}
$f^{\prime }\left( z\right) =v_{y}-iu_{y}\qquad \qquad (5)$

\vspace{1pt}
\end{center}

From $(3)$ and $\left( 5\right) $ we have that

\vspace{1pt}

\begin{center}
$u_{x}=v_{y}\qquad $and\qquad $u_{y}=-v_{x}\qquad \qquad (6)$

\vspace{1pt}
\end{center}

These are called the Cauchy-Riemann equations.

\vspace{1pt}

Theorem: A necessary condition for $f\left( z\right) $ to be differentiable
at $z$ is that the four partial derivatives with respect to $x$ and $y$
exist at $z=x+iy$ and are related there by Equations $(6).$

\vspace{1pt}

Theorem: If $f\left( z\right) $ is regular in a domain $D,$ and its
derivative is zero at all points of $D$, then $f\left( z\right) $ \ equals a
constant in $D.$ Equivalently, Two functions which are regular in the same
domain $D$ \ and whose derivatives coincide at all points of $D$ differ in $%
D $ only by a constant.

\vspace{1pt}

Proof: $f^{\prime }\left( z\right) =0\Longrightarrow
u_{x}=u_{y}=v_{x}=v_{y}=0$ throughout $D.$ Hence $u\left( x,y\right) $ and $%
v\left( x,y\right) $ and thus $f\left( z\right) $ are constant throughout $%
D. $

\vspace{1pt}

Example: $\ f(z)=\func{Re}\lbrack z\rbrack =x.$ here $u=x$ and $v=0$ so that 
$u_{x}=1,u_{y}=v_{x}=v_{y}=0.$ Thus the C-R Equations are satisfied nowhere
so that $f(z)=x$ is differentiable nowhere.

\vspace{1pt}

Example: $f(z)=\left| z\right| ^{2}=x^{2}+y^{2}.$ Here $u=x^{2}+y^{2}$ and $%
v=0,$ so that $u_{x}=2x,$ $u_{y}=2y$ and $v_{x}=v_{y}=0.$ Therefore the C-R
Equations hold only at the origin. Thus $z=0$ is the only possible point
where $\left| z\right| ^{2}$ might be differentiable. That this is indeed
the case may be determined by a separate computation.

\vspace{1pt}

Remark: Satisfaction of the Cauchy-Riemann equations is a necessary
condition for $f(z)=u+iv$ to be differentiable at a point. however, it is
not sufficient as the example below shows.

\vspace{1pt}

Example: Let $f(z)=\sqrt{\left| xy\right| }.$ The $u=\sqrt{\left| xy\right| }
$ and $v=0.$ Using the definition it is not hard to show that at $(0,0)$ $\
u_{x}=u_{y}=0$, whereas it is obvious that $v_{x}=v_{y}=0$ everywhere$.$
Hence the C-R Equations hold at the origin $z=0.$ However, $f\left( z\right) 
$ is not differentiable at $z=0$ \ since the difference quotient

\vspace{1pt}

\begin{center}
$\frac{f\left( 0+h\right) -f\left( 0\right) }{h}=\frac{f\left( h\right) }{h}=%
\frac{\sqrt{\Delta x\Delta y}}{\Delta x+i\Delta y}$

\vspace{1pt}
\end{center}

for $h=\Delta x+i\Delta y.$ If we let $\Delta x=\alpha r$ and $\Delta
y=\beta r,$ where $\alpha $ and $\beta $ are real and constant, and $r$ is a
real parameter (thus $h\rightarrow 0$ along any straight line), then

\vspace{1pt}

\begin{center}
$\lim_{h\rightarrow 0}\frac{f\left( h\right) }{h}=\lim_{\Delta x,\Delta
y\rightarrow 0}\frac{\sqrt{\Delta x\Delta y}}{\Delta x+i\Delta y}%
=\lim_{r\rightarrow 0}\frac{r\sqrt{\left| \alpha \beta \right| }}{r(\alpha
+i\beta )}=$ $\frac{\sqrt{\left| \alpha \beta \right| }}{(\alpha +i\beta )}$

\vspace{1pt}
\end{center}

Hence the limit is not independent of the path of approach, and therefore $%
f(z)$ is not differentiable at $z=0.$

\vspace{1pt}

Theorem: Suppose that the four partial derivatives of first order of $u$ and 
$v$ with respect to $x$ and $y$ \ exist and are continuous throughout a
domain $D.$ Then for $f\left( z\right) =u\left( x,y\right) +iv\left(
x,y\right) $ to be regular in $D,$ it is necessary and sufficient that the
Cauchy-Riemann equations $(6)$ hold throughout $D.$

\vspace{1pt}

Proof: Necessity - done.

\vspace{1pt}

Sufficiency: Since $u_{x},$ $u_{y,}$ $v_{x},$ and $v_{y}$ are continuous
throughout $D$, then for $\left( x,y\right) \in D$ and $\left( x+\Delta
x,y+\Delta y\right) $ in a neighborhood of $\left( x,y\right) $ which lies
in $D,$

\vspace{1pt}

$(7)\qquad \left\{ 
\begin{array}{l}
u(x+\Delta x,y+\Delta y)-u\left( x,y\right) =\left[ u_{x}\left( x,y\right)
+\in _{1}\right] \Delta x+\left[ u_{y}\left( x,y\right) +\in _{2}\right]
\Delta y \\ 
\\ 
v(x+\Delta x,y+\Delta y)-v\left( x,y\right) =\left[ v_{x}\left( x,y\right)
+\in _{3}\right] \Delta x+\left[ v_{y}\left( x,y\right) +\in _{4}\right]
\Delta y%
\end{array}
\right. $

\vspace{1pt}

where $\in _{i}=\in _{i}\left( x,y,\Delta x,\Delta y\right) \rightarrow 0$
as $h=\Delta x+i\Delta y\rightarrow 0\qquad (i=1,2,3,4).$

\vspace{1pt}We must show that the limit of

\begin{center}
$\frac{\Delta f}{h}=\frac{f\left( z+h\right) -f(z)}{h}=\frac{u(x+\Delta
x,y+\Delta y)-u\left( x,y\right) }{\Delta x+i\Delta y}+i\frac{v(x+\Delta
x,y+\Delta y)-v\left( x,y\right) }{\Delta x+i\Delta y}$
\end{center}

\vspace{1pt}

exists as $h\rightarrow 0.$ From $(7)$

\vspace{1pt}

\begin{center}
$\frac{\Delta f}{h}=\frac{\left[ u_{x}\left( x,y\right) +\in _{1}\right]
\Delta x+\left[ u_{y}\left( x,y\right) +\in _{2}\right] \Delta y+i\{\left[
v_{x}\left( x,y\right) +\in _{3}\right] \Delta x+\left[ v_{y}\left(
x,y\right) +\in _{4}\right] \Delta y\}}{\Delta x+i\Delta y}$

\vspace{1pt}

$\frac{\Delta f}{h}=\frac{(u_{x}+iv_{x})\Delta x+i(v_{y}-iu_{y})\Delta y+\in
_{1}\Delta x+\in _{3}\Delta y+i(\in _{2}\Delta y+\in _{4}\Delta y)}{\Delta
x+i\Delta y}$

\vspace{1pt}

$\frac{\Delta f}{h}=\frac{(u_{x}+iv_{x})(\Delta x+i\Delta y)+\in _{1}\Delta
x+\in _{3}\Delta x+i(\in _{2}\Delta y+\in _{4}\Delta y)}{\Delta x+i\Delta y}$

\vspace{1pt}
\end{center}

by the Cauchy-Riemann equations. Thus

\vspace{1pt}

\begin{center}
$\frac{\Delta f}{h}=(u_{x}+iv_{x})+\frac{\in _{1}\Delta x+\in _{3}\Delta
x+i(\in _{2}\Delta y+\in _{4}\Delta y)}{\Delta x+i\Delta y}$

\vspace{1pt}
\end{center}

But $\left| \in _{1}\Delta x+\in _{3}\Delta x+i(\in _{2}\Delta y+\in
_{4}\Delta y\right| \leq \left| \in _{1}+\in _{3}\right| \left| \Delta
x\right| +\left| \in _{2}+\in _{4}\right| \left| \Delta y\right| $ and

\vspace{1pt}

\begin{center}
$\frac{|\Delta x|}{\left| \Delta x+i\Delta y\right| }\leq 1\qquad $and $%
\frac{|\Delta y|}{\left| \Delta x+i\Delta y\right| }\leq 1$

\vspace{1pt}

$\left| \frac{\in _{1}\Delta x+\in _{3}\Delta x+i(\in _{2}\Delta y+\in
_{4}\Delta y)}{\Delta x+i\Delta y}\right| \leq \left| \in _{1}\right|
+\left| \in _{2}\right| +\left| \in _{3}\right| +\left| \in _{4}\right|
\rightarrow 0\qquad $as $h\rightarrow 0$

\vspace{1pt}
\end{center}

Thus $\lim_{h\rightarrow 0}\frac{\Delta f}{h}=u_{x}+iv_{x}$ \ and is
independent of the method of approach by $h.$

\vspace{1pt}

Note: \ $f^{\prime }(z)=u_{x}+iv_{x}=v_{y}-iu_{y}$ \ by the C-R Equations.

\vspace{1pt}

\subsection{Harmonic Functions}

\vspace{1pt}

If we assume further the existence and continuity in $D$ of the second order
partials derivatives of $u$ and $v$ with respect to $x$ and $y$ (which we
will see later is automatically true), then fro the Cauchy-Riemann Equations
we have

\vspace{1pt}

\begin{center}
$\frac{\partial ^{2}u}{\partial x^{2}}=\frac{\partial ^{2}v}{\partial
x\partial y}\qquad \frac{\partial ^{2}v}{\partial y\partial x}=-\frac{%
\partial ^{2}u}{\partial y^{2}}$

\vspace{1pt}
\end{center}

$\Longrightarrow \qquad \qquad \qquad \qquad \frac{\partial ^{2}u}{\partial
x^{2}}+\frac{\partial ^{2}u}{\partial y^{2}}=0\qquad \qquad $and \qquad $%
\frac{\partial ^{2}v}{\partial x^{2}}+\frac{\partial ^{2}v}{\partial y^{2}}%
=0 $

\vspace{1pt}

Thus $u$ and $v$ both satisfy Laplace's equation (or the potential equation)
in two dimensions of the form

\vspace{1pt}

\begin{center}
$\nabla ^{2}\phi =\phi _{xx}+\phi _{yy}=0$

\vspace{1pt}
\end{center}

Definition: A function that has second order partial derivatives in a domain 
$D$ \ and satisfies Laplace's equation in $D$ is called a \textit{harmonic
(or potential)} function.

\vspace{1pt}

Theorem: Let $u(x,y)$ be harmonic in $D$ and let $N(z_{0})$ be any
neighborhood of $z_{0}=x_{0}+iy_{0}$ that is contained in $D.$ Then $u\left(
x,y\right) $ is the real part of an analytic function that is regular in $%
N\left( z_{0}\right) .$

\vspace{1pt}

Proof: See Hille, page 85

\vspace{1pt}

Remark: The three dimensional version of Laplace's equation is

\vspace{1pt}

\begin{center}
$\phi _{xx}+\phi _{yy}+\phi _{zz}=0$

\vspace{1pt}
\end{center}

A solution of this three dimensional equation is called a \textit{%
logarithmic (or Newtonian) potential.}

\end{document}

%%%%%%%%%%%%%%%%%%%%%% End /document/Lecture4.tex %%%%%%%%%%%%%%%%%%%%%

%%%%%%%%%%%%%%%%%%%%%%% Start /document/6-5b.bmp %%%%%%%%%%%%%%%%%%%%%%

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%%%%%%%%%%%%%%%%%%%%%%%% End /document/6-5b.bmp %%%%%%%%%%%%%%%%%%%%%%%