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\centerline{\bf COMPLEX ANALYSIS\footnote{These lecture notes were prepared
for the instructor's personal use in teaching a half-semester course
on complex analysis at the beginning graduate level at Penn State, in
Spring 1997.  They are certainly not meant to replace a good text on the
subject, such as those listed on this page.}}
\medskip
\centerline{Douglas N.~Arnold\footnote{Department of Mathematics,
Penn State University, University Park, PA   16802.  \hfil\break
Web: http://www.math.psu.edu/dna/.}}
\vfill
References:
\smallskip
John B.~Conway, {\it Functions of One Complex Variable}, Springer-Verlag, 1978.

Lars V. Ahlfors, {\it Complex Analysis}, McGraw-Hill, 1966.

Raghavan Narasimhan, {\it Complex Analysis in One Variable}, Birkh\"auser, 1985.
\vfill
\goodbreak\head CONTENTS \endhead\nobreak
\roster
\item"I." {\smc The Complex Number System\dotfill2}
\item"II." {\smc Elementary Properties and Examples of Analytic Fns.\dotfill3}
\item""\qquad Differentiability and analyticity\dotfill4
\item""\qquad The Logarithm\dotfill6
\item""\qquad Conformality\dotfill6
\item""\qquad Cauchy--Riemann Equations\dotfill7
\item""\qquad M\"obius transformations\dotfill9
\item"III." {\smc Complex Integration and Applications to Analytic
Fns.\dotfill11}
\item""\qquad Local results and consequences\dotfill12
\item""\qquad Homotopy of paths and Cauchy's Theorem\dotfill14
\item""\qquad Winding numbers and Cauchy's Integral Formula\dotfill15
\item""\qquad Zero counting; Open Mapping Theorem\dotfill17
\item""\qquad Morera's Theorem and Goursat's Theorem\dotfill18
\item"IV." {\smc Singularities of Analytic Functions\dotfill19}
\item""\qquad Laurent series\dotfill20
\item""\qquad Residue integrals\dotfill23
\item"V." {\smc Further results on analytic functions\dotfill26}
\item""\qquad The theorems of Weierstrass, Hurwitz, and Montel\dotfill26
\item""\qquad Schwarz's Lemma\dotfill28
\item""\qquad The Riemann Mapping Theorem\dotfill29
\item""\qquad Complements on Conformal Mapping\dotfill31
\item"VI." {\smc Harmonic Functions\dotfill32}
\item""\qquad The Poisson kernel\dotfill33
\item""\qquad Subharmonic functions and the solution of the Dirichlet
Problem\dotfill36
\item""\qquad The Schwarz Reflection Principle\dotfill39
\endroster
\vfill\eject

\goodbreak\head I. The Complex Number System \endhead\nobreak

$\R$ is a field.  For $n>1$, $\R^n$ is a vectorspace over $\R$, so is
an additive group, but doesn't have a multiplication on it.  We can
endow $\R^2$ with a multiplication by
$$
(a,b)(c,d)=(ac-bd,bc+ad).
$$
Under this definition $\R^2$ becomes a field, denoted $\C$.  Note
that $(a/(a^2+b^2),-b/(a^2+b^2))$ is the multiplicative inverse
of $(a,b)$.  (Remark:
it is not possible to endow $\R^n$ with a field structure for $n>2$.)
We denote $(0,1)$ by $i$ and identify $x\in\R$ with $(x,0)$, so
$\R\subset\C$.  Thus $(a,b)=a+b i$, $a,b\in\R$.  Note that $i^2=-1$.
$\C$ is generated by adjoining $i$ to $\R$ and closing under addition
and multiplication.  It is remarkable that the addition of $i$ lets us
not only solve the equation $x^2+1=0$, but every polynomial equation.

For $a$ and $b$ real and $z=a+bi$ we define $\Re z=a$, $\Im z=b$,
$\bar z=a-bi$, and $|z|=(a^2+b^2)^{1/2}$.  Then
$$
\gather
\Re z=(z+\bar z)/2,\quad
\Im z=(z-\bar z)/(2i),\\
|z|^2=z\bar z,\quad
\frac 1 z=\frac{\bar z}{|z|^2},\\
\overline{z\pm w}=\bar z\pm\bar w,\quad
\overline{zw}=\bar z \bar w, \\
\overline{z/w}=\bar z/\bar w,\quad
|z+w|\le |z|+|w|.
\endgather
$$

The map $\theta\mapsto (\cos\theta,\sin\theta)$ defines a
$2\pi$-periodic map of the real line onto the unit circle
in $\R^2$.  In complex notation this map is $\theta\mapsto
\cis\theta:=\cos\theta+i\sin\theta$.  Every nonzero complex number can
be written as $r\cis\theta$ where $r>0$ is uniquely determined and
$\theta\in\R$ is uniquely determined modulo $2\pi$.  The number $0$
is equal to $r\cis\theta$ where $r=0$ and $\theta$ is arbitrary.  The
relation $z=r\cis\theta$ determines the relations $z\mapsto r$ which
is simply the function $r=|z|$ and $z\mapsto\theta$.  The latter is
denoted $\theta=\arg\theta$.  Note that for $z\ne 0$, $\arg\theta$
is determined modulo $2\pi$ (while $\arg 0$ is arbitrary).  
We can normalize $\arg$ by insisting that $\arg z\in(-\pi,\pi]$.  Note
that if $z_1=r\cis \theta_1$ and $z_2=r\cis\theta_2$ then $z_1 z_2=r_1
r_2\cis(\theta_1+\theta_2)$.  The latter formula just encapsulates the
formula for the sine and cosine of a sum, and gives $\arg z_1 z_2=\arg
z_1 +\arg z_2$. In particular, $i r\cis\theta=r\cis(\theta+\pi/2)$, so
multiplication by $i$ is just the operation of rotation by $\pi/2$
in the complex plane.  Multiplication by an arbitrary complex number
$r\cis\theta$ is just rotation by $\arg\theta$ followed by (or preceded
by) dilation by a factor $r$.  Further, $z^n=r^n\cis(n\theta)$.  Every
nonzero $z\in\C$ admits $n$ distinct $n$th roots: the $n$th roots
of $r\cis\theta$ are $\root n\of r \cis[(\theta+2\pi k)/n]$, $k=0,1,\ldots,n$.

Lines and circles in the plane.  Circles given by $|z-a|=r$ where $a\in\C$
is the center and $r>0$ is the radius.  If $0\ne b\in\C$ then
the line through the origin in the direction $b$ is the set of all points
of the form $tb$, $t\in\R$, or all $z$ with $\Im(z/b)=0$.  If $t\in\R$
and $c>0$ then $(t+c i)b=tb+c ib$ represents a point in the half plane
to the left of $b$ determined by the line $tb$, i.e., $\{z:\Im(z/b)>0\}$
is the equation of that half-plane.  Similarly, $\{z:\Im[(z-a)/b]>0\}$
is the translation of that half-plane by $a$, i.e., the half-plane determined
by the line through $a$ parallel to $b$ and in the direction to the left of $b$.

Stereographic projection determines a one-to-one correspondence between
the unit sphere in $\R^3$ minus the north-pole, $S$, and the complex
plane via the correspondence
$$
\gather
z\leftrightarrow \frac{x_1+i x_2}{1-x_3},
\qquad 
\\
x_1=\frac{2\Re z}{1+|z|^2},\quad x_2=\frac{2\Im z}{1+|z|^2},
\quad x_3=\frac{|z|^2-1}{|z|+1}.
\endgather
$$
If we define $\C_\infty=\C\cup\{\infty\}$, then we have a one-to-one
correspondence between $S$ and $\C_\infty$.  This allows us to define
a metric on $\C_\infty$, which is given by
$$
d(z_1,z_2)=\frac{2|z_1-z_2|}{\sqrt{(1+|z_1|^2)(1+|z_2|^2)}},
\qquad d(z,\infty)=\frac2{\sqrt{1+|z|^2}}.
$$

\goodbreak\head II. Elementary Properties and Examples of Analytic Functions \endhead\nobreak

For $z\ne 1$, $\sum_{n=0}^N z^n=(1-z^{N+1})/(1-z)$.  Therefore
the geometric series $\sum_{n=0}^\infty z^n$ converges (to $1/(1-z)$)
if $|z|<1$.  It clearly diverges, in fact its terms become unbounded,
if $|z|>1$.

\proclaim{Weierstrass M-Test}  Let $M_0,M_1,\ldots$ be positive numbers
with $\sum M_n<\infty$ and suppose that $f_n:X\to \C$ are functions on some set
$X$ satisfying $\sup_{x\in X}|f_n(x)|\le M_n$.  Then $\sum_{n=0}^\infty f_n(x)$
is absolutely and uniformly convergent.
\endproclaim

\proclaim{Theorem}  Let $a_0,a_1,\dots\in\C$ be given and define the
number $R$ by
$$
\frac1R=\lim\sup|a_n|^{1/n}.
$$
Then (1) for any $a\in\C$ the power series $\sum_{n=0}^\infty a_n(z-a)^n$
converges absolutely for all $|z-a|<R$ and it converges absolutely
and uniformly on the disk $|z-a|\le r$ for all $r<R$.  (2) The
sequence $a_n(z-a)^n$ is unbounded for all $|z-a|>R$ (and hence
the series is certainly divergent).
\endproclaim

Thus we see that the set of points where a power series converges consists
of a disk $|z-a|<R$ and possibly a subset of its boundary.  $R$ is called
the radius of convergence of its series.  The case $R=\infty$ is allowed.

\demo{Proof of theorem} For any $r<R$ we show absolute uniform convergence on
$D_r=\{|z-a|\le r\}$.  Choose $\tilde r\in(r,R)$.
Then, $1/\tilde r > \lim\sup|a_n|^{1/n}$, so $|a_n|^{1/n}<1/\tilde r$
for all $n$ sufficiently large.  For such $n$, $|a_n|<1/\tilde r^n$ and so
$$
\sup_{z\in D_r}|a_n(z-a)^n|<(r/\tilde r)^n.
$$
Since $\sum (r/\tilde r)^n<\infty$ we get the absolute uniform convergence
on $D_r$.

If $|z-a|=r>R$, take $\tilde r\in(R,r)$.  Then there exist $n$ arbitrarily
large such that $|a_n|^{1/n}\ge 1/\tilde r$.  Then,
$|a_n(z-a)^n|\ge (r/\tilde r)^n$, which can be arbitrarily large. \qed
\enddemo

\proclaim{Theorem}  If $a_0,a_1,\ldots\in\C$ and $\lim|a_n/a_{n+1}|$
exists as a finite number or infinity, then this limit is the radius
of convergence $R$ of $\sum a_n(z-a)^n$.
\endproclaim
\demo{Proof}  Without loss of generality we can suppose that $a=0$.
Suppose that $|z|>\lim|a_n/a_{n+1}|$.  Then for all $n$ sufficiently
large $|a_n|<|a_{n+1}z|$ and $|a_nz^n|<|a_{n+1}z^{n+1}|$.  Thus the series
$\sum a_nz^n$ has terms of increasing magnitude, and so cannot be convergent.
Thus $|z|\ge R$.  This shows that $\lim|a_n/a_{n+1}|\ge R$.

Similarly, suppose that $z<\lim|a_n/a_{n+1}|$.  Then for all $n$ sufficiently
large $|a_n|>|a_{n+1}z|$ and $|a_nz^n|>|a_{n+1}z^{n+1}|$.  Thus the
series has terms of decreasing magnitude, and so, by the previous theorem,
$|z|\le R$.  This shows that $\lim|a_n/a_{n+1}|\le R$.\qed
\enddemo

\remark{Remark} On the circle of convergence, many different behaviors are
possible. $\sum z^n$ diverges for all $|z|=1$.  $\sum z^n/n$ diverges
for $z=1$, else converges, but not absolutely (this follows from
the fact that the partial sums of $\sum z^n$ are bounded for $z\ne1$
and $1/n\downarrow 0$).  $\sum z^n/n^2$ converges absolutely on
$|z|\le1$.  Sierpinski gave a (complicated) example of a function
which diverges at every point of the unit circle except $z=1$.
\endremark

As an application, we see that the series
$$
\sum_{n=0}^\infty \frac{z^n}{n!}
$$
converges absolutely for all $z\in\C$ and that the convergence is uniform
on all bounded sets.  The sum is, by definition, $\exp z$.

Now suppose that $\sum_{n=0}^\infty a_n(z-a)^n$ has radius of convergence
$R$, and consider its formal derivative $\sum_{n=1}^\infty na_n(z-a)^{n-1}
=\sum_{n=0}^\infty (n+1)a_{n+1} (z-a)^n$.  Now clearly
$\sum_{n=0}a_{n+1} (z-a)^n$ has the same radius of convergence as
$\sum_{n=0}^\infty a_n(z-a)^n$ since
$$
(z-a)\sum_{n=0}^N a_{n+1} (z-a)^n=\sum_{n=0}^{N+1} a_n(z-a) -a_0,
$$
and so the partial sums on the left and right either both diverge for
a given $z$ or both converge.  This shows (in a roundabout way)
that $\lim\sup |a_{n+1}|^{1/n}=\lim\sup |a_n|^{1/n}=1/R$.  Now
$\lim (n+1)^{1/n}=1$ as is easily seen by taking logs.  Moreover,
it is easy to see that if $\lim\sup b_n=b$ and $\lim c_n=c>0$, then
$\lim\sup b_n c_n=bc$.  Thus $\lim\sup|(n+1)a_{n+1}|^{1/n}=1/R$.  This
shows that the formal derivative of a power series has the same radius
of convergence as the original power series.

\goodbreak\subhead Differentiability and analyticity \endsubhead\nobreak
Definition of differentiability at a point (assumes function
is defined in a neighborhood of the point).

Most of the consequences of differentiability are quite different in
the real and complex case, but the simplest algebraic rules are the
same, with the same proofs.  First of all, differentiability at a point
implies continuity there. If $f$ and $g$ are both differentiable at a
point $a$, then so are $f\pm g$, $f\cdot g$, and, if $g(a)\ne 0$, $f/g$,
and the usual sum, product, and quotient rules hold.  If $f$ is differentiable
at $a$ and $g$ is differentiable at $f(a)$, then $g\circ f$ is differentiable
at $a$ and the chain rule holds.  Suppose that $f$ is continuous at $a$, $g$ is continous
at $f(a)$, and $g(f(z))=z$ for all $z$ in a neighborhood of $a$.  Then
if $g'(f(a))$ exists and is non-zero, then $f'(a)$ exists and equals
$1/g'(f(a))$.

\definition{Definition} Let $f$ be a complex-valued function defined on an open
set $G$ in $\C$.  Then $f$ is said to be analytic on $G$ if $f'$ exists and is
continuous at every point of $G$.
\enddefinition

\remark{Remark} We shall prove later that if $f$ is differentiable at every
point of an open set in $\C$ it is automatically analytic; in fact, it is
automatically infinitely differentiable.  This is of course vastly
different from the real case.
\endremark

If $Q$ is an arbitrary non-empty subset of $\C$ we say $f$ is analytic
on $Q$ if it is defined and analytic on an open set containing $Q$.

We now show that a power series is differentiable at every point in
its disk of convergence and that its derivative is given by the formal
derivative obtained by differentiating term-by-term.  Since we know
that that power series has the same radius of convergence, it follows
that a power series is analytic and infinitely differentiable in
its convergence disk.  For simplicity, and without loss of generality
we consider a power series centered at zero: $f(z)=\sum_n a_nz^n$.  Suppose
that the radius of convergence is $R$ and that $|z_0|<R$.  We must
show that for any $\epsilon>0$, the inequality
$$
\left| \frac{f(z)-f(z_0)}{z-z_0}-g(z_0)\right|\le \epsilon
$$
is satisfied for all $z$ sufficiently close to $z_0$,
where $g(z)=\sum_{n=1}^\infty n a_n z^{n-1}$.
Let $s_N(z)=\sum_{n=0}^N a_nz^n$, $R_N(z)=\sum_{n=N+1}^\infty a_n z^n$.
Then 
$$
\multline
\left|\frac{f(z)-f(z_0)}{z-z_0}-g(z_0)\right|
\le\left|\frac{s_N(z) -s_N(z_0)}{z-z_0} -s_N'(z_0)\right|
\\
+\left|s_N'(z_0)-g(z_0)\right| + \left|\frac{R_N(z) -R_N(z_0)}{z-z_0}\right|
=:T_1+T_2+T_3.
\endmultline
$$
Now $s_N'(z_0)$ is just a partial sum for $g(z_0)$, so for $N$ sufficiently
large (and all $z$), $T_2\le\epsilon/3$.  Also,
$$
\frac{R_N(z) -R_n(z_0)}{z-z_0}=\sum_{n=N+1}^\infty a_n\frac{z^n-z_0^n}{z-z_0}.
$$
Now $|z_0|<r<R$ for some $r$, and if we restrict to $|z|<r$, we have
$$
\left|a_n\frac{z^n-z_0^n}{z-z_0}\right|
=|a_n||z^{n-1}+z^{n-2}z_0+\cdots+z_0^{n-1}|\le a_nnr^{n-1}.
$$
Since $\sum a_n nr^{n-1}$ is convergent, we have for $N$ sufficiently large
and all $|z|<r$ then $T_3<\epsilon/3$.  Now fix a value of $N$ which 
is sufficiently large by both criteria.  Then the differentiability of
the polynomial $s_N$ shows that $T_1\le \epsilon/3$ for all $z$ sufficiently
close to $z_0$.

We thus know that if $f(z)=\sum a_n z^n$, then, within the disk of convergence,
$f'(z)=\sum na_nz^{n-1}$, and by induction, $f''(z)=\sum n(n-1)a_nz^{n-2}$,
etc.  Thus $a_0=f(0)$, $a_1=f'(0)$, $a_2=f''(0)/2$, $a_3=f'''(0)/3!$, etc.
This shows that any convergent power series is the sum of its
Taylor series in the disk of convergence:
$$
f(z)=\sum\frac{f^{n}(a)}{n!}(z-a)^n.
$$

In particular, $\exp'=\exp$.

\proclaim{Lemma} A function with vanishing derivative on a region (connected
open set) is constant. 
\endproclaim
Indeed, it is constant on disks, since we can
restrict to segments and use the real result.  Then we can use connectedness
to see that the set of points where it takes a given value is open and closed.

Use this to show that $\exp(z)\exp(a-z)\equiv \exp a$.  Define $\cos z$
and $\sin z$, get $\cos^2+\sin^2\equiv1$, $\exp z =\cos z+i\sin z$,
$\cis\theta=\exp(i\theta)$,
$\exp z=\exp(\Re z)\cis(\Im z)$.

\goodbreak\subhead The Logarithm\endsubhead\nobreak
If $x\ne0$, the most general solution of $\exp z=w$ is $z=\log|w|+i\arg w+2\pi i n$,
$n\in\Z$.  There is no solution to $\exp z=0$.

\definition{Definition} If $G$ is an open set and $f:G\to\C$ is a continuous
function satisfying $\exp(f(z))=z$, then $f$ is called a branch of the
logarithm on $G$.
\enddefinition

Examples: $\C\setminus\R_-$ (principal branch), $\C\setminus\R_+$, a spiral strip.

By the formula for the derivative of an inverse, a branch of the logarithm is
analytic with derivative $1/z$.

A branch of the logarithm gives a branch of $z^b:=\exp(b\log z)$ (understood
to be the principal branch if not otherwise noted).  Note that $e^b=\exp b$.

Note that different branches of the logarithm assign values to $\log z$
that differ by addition of $2\pi i n$.  Consequently different branches
of $z^b$ differ by factors of $\exp(2\pi i n b)$.  If $b$ is an integer,
all values agree.  If $b$ is a rational number with denominator $d$
there are $d$ values.

\goodbreak\subhead Conformality\endsubhead\nobreak
Consider the angle between two line segments with a common vertex
in the complex plane.  We wish to consider how this angle is
transformed under a complex map.  The image will be two smooth
curves meeting at a common point, so we have to define the
angle between two such curves.

Define a path in the complex plane as a continuous map $\gamma:[a,b]\to\C$.
If $\gamma'(t)\in\C$ exists at some point and is not zero, then it is a vector
tangent to the curve at $\gamma(t)$, and hence its argument is the
angle between the (tangent to the) curve and the horizontal.
(Note: if $\gamma'(t)=0$, the curve may not have a tangent; e.g.,
$\gamma(t)=t^2-it^3$.)  Let $\gamma:[a,b]\to\C$ and
$\tilde\gamma:[\tilde a,\tilde b]\to\C$ be two smooth curves, such that
$\gamma(a)=\tilde\gamma(\tilde a)=z$, and $\gamma'(a),\tilde\gamma'(\tilde
a)\ne0$.
Then $\arg\tilde\gamma'(\tilde a)-\arg\gamma'(a)$ measures the angle
between the curves.  Now consider the images under $f$, e.g., of
$\rho(t)=f(\gamma(t))$.  Then $\rho'(a)=f'(z)\gamma'(a)$,
$\tilde\rho'(\tilde a)=f'(z)\tilde\gamma'(\tilde a)$.  Therefore
$\arg\tilde\rho'(\tilde a)-\arg\rho'(a)=\arg\tilde\gamma'(\tilde a)
+\arg f'(z)-\arg\gamma'(a)-\arg f'(z)$, i.e., the angle is invariant
in both magnitude and sign.  This says that an analytic mapping is
conformal, whenever its derivative is not zero.  (Not true if the
derivative is zero.  E.g., $z^2$ doubles angles at the origin.)

The conformality of an analytic map arises from the fact that in a
neighborhood of a point $z_0$ such a map behaves to first order like
multiplication by $f'(z_0)$.  Since $f'(z_0)=r\exp i\theta$, near $z_0$
$f$ simply behave like rotation by $\theta$ followed by dilation.  By
contrast, a smooth map from $\R^2\to\R^2$ can behave like an arbitrary
linear operator in a neighborhood of a point.

At this point, show the computer graphics square1.mps and square2.mps.

\goodbreak\subhead Cauchy--Riemann Equations\endsubhead\nobreak
Let $f:\Omega\to\C$ with $\Omega\subset\C$ open.  Abuse notation slightly
by writing $f(x,y)$ as an alternative for $f(x+iy)$.  If $f'(z)$
exists for some $z=x+iy\in\Omega$, then
$$
f'(z)=\lim\Sb h\in\R\\h\to 0\endSb \frac{f(z+h)-f(z)}{h}
=\lim\Sb h\in\R\\h\to 0\endSb\frac{f(x+h,y)-f(x,y)}{h}
=\frac{\partial f}{\partial x},
$$
and
$$
f'(z)=\lim\Sb h\in\R\\h\to 0\endSb \frac{f(z+ih)-f(z)}{ih}
=\lim\Sb h\in\R\\h\to 0\endSb-i\frac{f(x,y+h)-f(x,y)}{h}
=-i\frac{\partial f}{\partial y}.
$$
Thus complex-differentiability of $f$ at $z$ implies not only
that the partial derivatives of $f$ exist there, but also that they
satisfy the Cauchy--Riemann equation
$$
\frac{\partial f}{\partial x}=-i\frac{\partial f}{\partial y}.
$$
If $f=u+iv$, then this equation is equivalent to the system
$$
\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y},
\qquad
\frac{\partial v}{\partial x}=-\frac{\partial u}{\partial y}.
$$

Another convenient notation is to introduce
$$
\frac{\partial f}{\partial z}:=\frac12\left(\frac{\partial f}{\partial x}-i\frac{\partial
f}{\partial y}\right),
\qquad
\frac{\partial f}{\partial \bar z}:=\frac12\left(\frac{\partial f}{\partial x}+i\frac{\partial
f}{\partial y}\right).
$$
(These are motivated by the equations $x=(z+\bar z)/2$,
$y=(z-\bar z)/(2i)$, which, if $z$ and $\bar z$ were independent variables,
would give $\partial x/\partial z=1/2$, $\partial y/\partial z=-i/2$, etc.)

In terms of these, the Cauchy--Riemann equations are exactly equivalent to
$$
\frac{\partial f}{\partial \bar z}=0,
$$
which is also equivalent to
$$
\frac{\partial f}{\partial z}=\frac{\partial f}{\partial x}.
$$

Thus, if $f$ is analytic on $\Omega$, then the partial derivaties of $f$
exist and are continuous on $\Omega$, and the Cauchy-Riemann equations
are satisfied there.  The converse is true as well:

\proclaim{Theorem} If the partial derivatives of $f$ exist and are
continuous on $\Omega$, and the Cauchy-Riemann equations
are satisfied there, then $f$ is analytic on $\Omega$.
\endproclaim
\demo{Proof} Let $z=x+iy\in\Omega$.  We must show that $f'(z)$ exists.
Let $r$ be small enough that the disk of radius $r$ around $z$ belongs
to $\Omega$ and choose $h=s+it$ with $0<|s+it|<r$.  Then, by the mean
value theorem,
$$
\align
&\frac{f(z+h)-f(z)}{h}=\frac{f(x+s,y+t)-f(x,y)}{s+it}
\\
&\qquad=\frac{f(x+s,y+t)-f(x,y+t)}{s}\frac{s}{s+it}
+\frac{f(x,y+t)-f(x,y)}{t}\frac{t}{s+it}
\\
&\qquad=\frac{\partial f}{\partial x}(x+s^*,y+t)\frac{s}{s+it}
+\frac{\partial f}{\partial y}(x,y+t^*)\frac{t}{s+it},
\endalign
$$
where $|s^*|<s$ and $|t^*|<t$.  Note that
$$
\frac{\partial f}{\partial x}(x+s^*,y+t)-\frac{\partial f}{\partial x}(x,y)
\to 0
$$
as $h\to0$, and similarly for the second partial derivative.  Moreover
$|s/(s+it)|$ stays bounded (by $1$).  Thus
$$
\frac{f(z+h)-f(z)}{h}=\frac{\partial f}{\partial x}(x,y)\frac{s}{s+it}
+\frac{\partial f}{\partial y}(x,y)\frac{t}{s+it} +R(h),
$$
where $\lim_{h\to0} R(h)=0$.  Now if the Cauchy--Riemann equations
hold, then the the first two terms on the right hand side sum to
$\partial f/\partial x (x,y)$, which is independent of $h$, so the limit
exists and is equal to $\partial f/\partial x (x,y)$.\qed
\enddemo

\remark{Remark} The theorem can be weakened to say that if $f$ is
continuous on $\Omega$ and the partial derivatives exist and
satisfy the Cauchy--Riemann equations there (without assuming
that the partial derivatives are continuous), then the
complex derivative of $f$ exists on $\Omega$ (which is equivalent
to $f$ being analytic on $\Omega$.  This is the Looman--Menchoff
Theorem.  See Narasimhan, {\it Complex Analysis in One Variable},
for a proof.  We do need at least continuity, since otherwise we
could take $f$ to be the characteristic function of the coordinate
axes.
\endremark

Note that
$$
\frac{\partial}{\partial z}\frac{\partial}{\partial \bar z}
=\frac{\partial}{\partial\bar z}\frac{\partial}{\partial z}
=\frac14\Delta.
$$

This shows that any analytic function is harmonic (equivalently,
its real and imaginary parts are harmonic).  It also shows
that the conjugate of an analytic function, while not analytic,
is harmonic.

If $u(x,y)=\Re f(x+iy)$, then $u$ is harmonic.  Conversely, if $u$
is harmonic on a region $\Omega$, does there exist a {\it harmonic
conjugate}, i.e., a function $v$ on $\Omega$ such that $f=u+iv$
is analytic?  Clearly, if there exists $v$ it is determined up to
addition of a real constant (or $f$ is determined up to addition of
an imaginary constant).  The book gives an elementary construction of
the harmonic conjugate in a disk and in the whole plane.  We sketch
the idea of a more general proof.  We require that the domain satisfy
the condition that if $\gamma$ is any piecewise simple smooth closed
curve in $\Omega$, then $\gamma$ is the boundary of a subset $G$ of
$\Omega$.  In other words, $\Omega$ has no holes (is simply-connected).
Our proof follows directly from the following result, which holds on
simply-connected domains: let $f:\Omega\to\R^2$ be a $C^1$ vectorfield.
Then there exists $v:\Omega\to\R$ such that $f=\nabla v$ if and only
if ${\partial f_1}/{\partial y}={\partial f_2}/{\partial x}$.  The ``only
if'' part is obvious.  To prove the ``if'' part, fix a point $(x_0,y_0)$
in $\Omega$ and for any $(x,y)\in\Omega$, let $\gamma_{(x,y)}$ be a
piecewise smooth path in $\Omega$ from $(x_0,y_0)$ to $(x,y)$, and let
$\tau$ be the unit tangent to the path pointing from $(x_0,y_0)$ towards
$(x,y)$.  Define
$$
v(x,y)=\int_{\gamma_{(x,y)}} f\cdot\tau\,ds.
$$
It is essential that this quantity doesn't depend on the choice of path,
or, equivalently, that
$$
\int_\gamma f\cdot\tau\,ds=0
$$
for all piecewise smooth simple closed paths $\gamma$.  But
$$
\int_\gamma f\cdot\tau\,ds=\int_\gamma (f_2,-f_1)\cdot n\,ds
=\iint \nabla\cdot(f_2,-f_1)\,dx\,dy =0,
$$
by the divergence theorem.  It is easy to check that $\nabla v=f$.

For a non-simply connected region, there may exist no harmonic
conugate.  E.g., if $u=\log|z|$ on $\C\setminus\{0\}$, then
$u+i\arg z$ is analytic on $C\setminus\{z\le0\}$, but cannot
be extended to an analytic function on $\C\setminus\{0\}$.

\goodbreak\subhead M\"obius transformations \endsubhead\nobreak
\definition{Definition} Let $a$, $b$, $c$, $d$ be complex numbers
with $ad\ne bc$.  Then the mapping
$$
S(z)=\frac{az+b}{cz+d}, \qquad z\in\C,\quad z\ne -d/c,
$$
is a M\"obius transformation.
\enddefinition

\remark{Remarks}
1) Note that if $ad=bc$ the same expression would yield a constant.
2) The coefficents aren't unique, since we can multiply them all
by any nonzero complex constant. To each MT
we associate the nonsingular matrix $\pmatrix a&b\\c&d\endpmatrix$ of
its coefficients, which is determined up to a non-zero multiple.
3) The linear (but non-constant) polynomials are MTs, namely the ones for
which $c=0$.
4) If $z\ne -d/c$ then $S(z)\in\C$ and $S'(z)=(ad-bc)/(cz+d)^2\ne0$.
5) We can define $S(-d/c)=\infty$ and $S(\infty)=a/c$, so $S$ can
be viewed of as a map from $\C_\infty$ into itself (which is,
as we will now see, 1-1 and onto).
\endremark

If $S$ and $T$ are MTs, then so is $S\circ T$,
its coefficient matrix being the product of the coefficient matrices
of $S$ and $T$.  Any MTs admits an inverse,
namely the transform with the inverse coefficient matrix.  With the
understanding that $S(\infty)=a/c$ and $S(-d/c)=\infty$ (or, if $c=0$,
then $S(\infty)=\infty$, the MT maps $\C_\infty$
1-1 onto itself.  This is evident from the existence of its inverse
$S^{-1}(z)=(dz-b)/(-cz+a)$.  Moreover, the composition of M\"obius
transformations is again one, so they form a group under composition.
If $S$ is a non-linear MT we can write it as
$$
\frac{az+b}{z+d}=a+\frac{b-ad}{z+d}.
$$
This shows that any MT can be written as a composition
of translations, rotations and dilations (multiplication by a complex number),
and inversion (reciprocals).

Since the geometric action of translation, dilation, and rotation are
quite evident, let's consider inversion.  It takes the unit disk $D$
to $\C_\infty\setminus \bar D$.  We now show that inversion maps
lines and circles to other lines and circles.

Consider the equation $|z-p|=k|z-q|$ where $k>0$, $p,q\in\C$.
This is
$$
(x-a)^2+(y-b)^2=k^2[(x-c)^2+(y-d)^2]
$$
which is clearly the equation of a circle if $k\ne1$ and the
equation of a line if $k=1$.  In either case we have a circle
in the Riemann sphere, which we call $C_1$.
Substituting $1/z$ for $z$ and doing some simple manipulations gives
the equation $|z-1/p|=(k|q|/|p|) |z-1/q|$, which is another such
circle, $C_2$.  Thus we have $1/z\in C_1\iff z\in C_2$, showing
that inversion maps circles in the sphere onto other circles in the
sphere.  Since the same property is evident for translation, dilation,
and rotation, we conclude that this property holds for all MT.

Some further analysis, which will be omitted, establishes
two further properties.  
1) A MT is orientiation preserving in the sense that, if we traverse a
circle in the order of three distinct points on it, $z_1$, $z_2$, $z_3$,
the region to the left of the circle will map to the region to the left
of the image circle, with respect to the image orientation.  2) A MT
preserves the property of two points being symmetric with respect to a
circle, i.e., lying on the same ray from the center, and such that the
geometric mean of their distances from the center equals the radius.

Let $z_2$, $z_3$, $z_4$ be distinct points in $\C$.  Then it is easy to
see that there is a unique MT taking these
points to $1$, $0$, and $\infty$, respectively, namely
$$
Sz=\frac{z-z_3}{z-z_4}\frac{z_2-z_4}{z_2-z_3}.
$$
We can also handle, as a special case, the possibility that one of the
$z_i$ is $\infty$.  We infer from this and the invertibility of the MTs
that given two ordered triples of distinct points in $\C_\infty$, there
is a unique MT taking the first onto the second.

If we have any two disks or halfplanes, or one of each, we can
map one to the other by a MT, and can arrange that the image of
any three points on the boundary of one go to three points on the boundary of
the other.

The book makes heavy use of the the notation $(z_1,z_2,z_3,z_4)$ (cross
ratio) for the image of $z_1$ under the transformation taking $z_2$,
$z_3$, $z_4$ to $1$, $0$, and $\infty$, respectively, so
$$
(z_1,z_2,z_3,z_4)=\frac{z_1-z_3}{z_1-z_4}\frac{z_2-z_4}{z_2-z_3}.
$$
From the definition, $(Tz_1,Tz_2,Tz_3,Tz_4)=(z_1,z_2,z_3,z_4)$ for any
MT $T$.

Example: $z\mapsto (1-z)/(1+z)$ takes the right half plane onto the unit
disk.

\goodbreak\head III. Complex Integration and Applications to Analytic Functions \endhead\nobreak

If $\gamma:[a,b]\to\C$ is piecewise smooth, and $f$ is a continuous
complex-valued function defined on the image of $\gamma$, define
$$
\int_\gamma f(z)\,dz=\int_a^b f(\gamma(t))\gamma'(t)\,dt.
$$
This is simply the line integral of $f$ along $\gamma$.  It can be
defined analogously to the Riemann integral as the limit of sums of
the form $\sum (f\circ\gamma)(\tau_k)[\gamma(t_k)-\gamma(t_{k-1})]$,
so is the {\it Riemann--Stieltjes integral}\/ of $f\circ\gamma$ with
respect to $\gamma$.  Using this definition it can be extended to
rectifiable paths, i.e., ones for which $\gamma$ is only of bounded
variation.

If $\phi$ is a piecewise smooth increasing function from $[c,d]$ onto
$[a,b]$ and we let $\rho=\gamma\circ \phi$ (so $\rho$ is the same
path as $\gamma$ but with a different parametrization), then
$\int_\rho f(z)\,dz = \int_\gamma f(z)\,dz$.

From the fundamental theorem of calculus we see that if
$F$ is any analytic function on a neighborhood of $\gamma$,
then $\int_\gamma F'(z)\,dz=
F(\beta)-F(\alpha)$ where $\alpha=f(a)$, $\beta=f(b)$.  If $\gamma$
is a closed path, this is zero.

As a consequence, if $\gamma=r e^{i\theta}$, then
$$
\int_{\gamma}z^n\,dz=0\text{\qquad for all $n\in\Z$, $n\ne-1$}.
$$
By direct calculation we have $\int_{\gamma}z^{-1}\,dz=2\pi i$.

More generally, let $|z_0|<r$ and consider
$\int_{\gamma}\frac{1}{z-z_0}\,dz$.  Let $z^\pm_\epsilon$ be the point
on the circle $|z-z_0|=r$ with $\arg(z^\pm_\epsilon-z_0)=\pm(\pi-\epsilon)$.
Clearly, as $\epsilon\to0$, $z^\pm_\epsilon\to z_1$, the point on the
circle with the same imaginary part as $z_0$ and negative real part.
Now let $\gamma_\epsilon$
be the curve running from $z^-_\epsilon$ to $z^+_\epsilon$ on the circle.
In a neighborhood of $\gamma_\epsilon$ the function $\Log(z-z_0)$ is
a primitive of $1/(z-z_0)$.  So
$$
\int_{\gamma_\epsilon}\frac1{z-z_0}\,dz=\Log(z^+_\epsilon-z_0)-\Log(z^-_\epsilon-z_0)
=\log r+i(\pi-\epsilon)-\log r+i(\pi-\epsilon).
$$
Taking the limit as $\epsilon\downarrow0$ gives
$$
\int_{\gamma}\frac{1}{z-z_0}\,dz=2\pi i.
$$

\goodbreak\subhead Local results and consequences\endsubhead\nobreak
\proclaim{Theorem (Cauchy's integral formula for a disk)} Let $f$ be analytic on
a closed disk and let $\gamma$ be a path around the boundary of the disk.
Then
$$
f(z)=\frac1{2\pi i}\int_\gamma \frac{f(w)}{w-z}\,dw
$$
for all $z$ in the open disk.
\endproclaim
\demo{Proof} Consider the function
$$
g(t)=\int_\gamma \frac{f(z+t(w-z))}{w-z}\,dw.
$$
Then
$$
g'(t)=\int_\gamma f'(z+t(w-z))\,dw,
$$
and, since  $F(w)=f(z+t(w-z))/t$ is a primitive for $f'(z+t(w-z))$,
$g'(t)\equiv0$.  Therefore,
$$
\int_\gamma \frac{f(w)}{w-z}\,dw=g(1)=g(0)=f(z)\int_\gamma \frac1{w-z}\,dw
=2\pi i f(z).\qed
$$
\enddemo

As an easy corollary, we have:
\proclaim{Cauchy's Theorem for a disk}  Let $f$ be analytic on
a closed disk and let $\gamma$ be a path around the boundary of the disk.
Then
$$
\int_\gamma f(w)\,dw =0.
$$
\endproclaim
\demo{Proof} Fix $z$ in the open disk and apply Cauchy's formula
to the function $w\mapsto f(w)(w-z)$.\qed
\enddemo

We can now show that an analytic (i.e., continuously complex differentiable)
function has a power series expansion around each point in its domain.
(And we already know the converse is true.)

\proclaim{Theorem} Let $f$ be analytic in $B(a;R)$.  Then $f(z)=\sum a_n(z-a)^n$
for $|z-a|<R$, for some $a_n\in\C$.
\endproclaim
\demo{Proof}  Fix $z$ and choose $r<R$ with $|z-a|<r$.  Let $\gamma$ be the path
around $|w-a|=r$.  Then $|(z-a)/(w-a)|<1$, so the Weierstrass M-test
implies that the series
$$
\sum\frac1{(w-a)^{n+1}}(z-a)^n=\frac1{w-z},
$$
converges absolutely and uniformly in $w$ for $|w-a|=r$.  Since $f$ is
bounded on the circle, the same sort of convergence holds for
$$
\sum\frac{f(w)}{(w-a)^{n+1}}(z-a)^n=\frac{f(w)}{w-z}.
$$
Therefore we can interchange summation and integraton to get
$$
f(z)=\frac1{2\pi i}\int_\gamma \frac{f(w)}{w-z}\,dw=\sum a_n (z-a)^n,
$$
where
$$
a_n=\int_\gamma\frac{f(w)}{(w-a)^{n+1}}\,dw.\qed
$$
\enddemo

\proclaim{Corollary (of proof)} Under the same hypotheses
$$
f^{(n)}(a)=\frac{n!}{2\pi i}\int_\gamma\frac{f(w)}{(w-a)^{n+1}}\,dw.
$$
\endproclaim

\proclaim{Corollary (Cauchy's estimate)}  If $f$ is analytic in
$B(a;R)$, then
$$
|f^{(n)}(a)|\le \frac{n!\max_{B(a;R)}|f|}{R^n}.
$$
\endproclaim
\demo{Proof} Evident if $R$ is replaced by $r<R$. Then pass to limit.\qed
\enddemo


\proclaim{Corollary (Maximum Modulus Principle)}  An analytic function on
a region which achieves a maximum modulus in the region is constant.
\endproclaim
\demo{Proof} By the Cauchy estimate for $n=0$, if the function achieves
its maximum modulus at a point, it achieves it everywhere on a disk
containing the point.  Thus, the set of points were the maximum modulus
is achieved is open.  It is also clearly closed. Hence the function has
constant modulus.  It follows from the Cauchy--Riemann equations that
the function itself is constant.\qed
\enddemo

\proclaim{Corollary (Liouville's Theorem)} A bounded entire function
is constant.
\endproclaim
\demo{Proof} Cauchy's estimate for $n=1$ implies $f'\equiv0$.\qed
\enddemo

\proclaim{Corollary (Fundamental Theorem of Algebra)} Every nonconstant
polynomial has a root in $\C$. \endproclaim

\proclaim{Theorem} If $f$ is analytic on a region then there does
not exist a point $a$ such that $f^{(n)}(a)=0$ for all $n$, unless
$f$ is identically zero on the region.
\endproclaim
\demo{Proof}  The set of such $a$ is closed, and using the Taylor
expansion, is open.\qed
\enddemo

\proclaim{Corollary} If $f$ is analytic on a region $\Omega$ and not
identically zero, then for each $a\in\Omega$ there exists a unique
integer $p\in\N$ and a function $g$, analytic in a neighborhood of $a$
and nonzero at $a$, such that
$$
f(z)=(z-a)^p g(z).
$$
\endproclaim
\demo{Proof} $f(z)=\sum c_n(z-a)^n$ with $c_n=f^{(n)}(a)/n!$.  Let $p$
be the least integer with $c_p\ne0$ (which exists by the preceeding
theorem), and $g(z)=\sum_{n=p}^\infty c_n(z-a)^{n-p}$.\qed
\enddemo

The number $p$ is the {\it multiplicity}\/ of $a$ as a root of $f$.

\proclaim{Corollary} If $f$ is analytic on a region and not
identically zero, then the roots of $f$ are isolated.
\endproclaim

\proclaim{Corollary} An analytic function on a region is completely
determined by its value on any set of points containing a limit point.
\endproclaim

\goodbreak\subhead Homotopy of paths and Cauchy's Theorem \endsubhead\nobreak
%
\definition{Definition} Let $\gamma_0,\gamma_1:[0,1]\to G$ be two
closed paths in a region $G\subset\C$.  We say that the paths
are {\it homotopic in $G$}\/ if there exists a continuous
function $\Gamma:[0,1]\x[0,1]\to G$ such that for each $s$
$\Gamma(\,\cdot\,,s)$ is a closed path with $\Gamma(\,\cdot\,,0)=\gamma_0$
and $\Gamma(\,\cdot\,,1)=\gamma_1$.
\enddefinition
This is an equivalence relation, written $\gamma_0\sim\gamma_1$ in $G$.

\proclaim{Theorem}  Suppose that $\gamma_0$ and $\gamma_1$ are two homotopic
piecewise smooth closed curves in $G$ and that $f$ is analytic on $G$.
Then
$$
\int_{\gamma_0} f(z)\,dz = \int_{\gamma_1} f(z)\,dz.
$$
\endproclaim

If we assume that the homotopy $\Gamma$ is $C^2$, the result is
easy: let
$$
g(s)=\int_0^1 f(\Gamma(t,s))\frac{\partial\Gamma}{\partial t}(t,s)\,dt.
$$
Then
$$
g'(s)=\int_0^1\frac{\partial}{\partial
s}[f(\Gamma(t,s))\frac{\partial\Gamma}{\partial t}(t,s)]\,dt
=\int_0^1\frac{\partial}{\partial
t}[f(\Gamma(t,s))\frac{\partial\Gamma}{\partial s}(t,s)]\,dt
=0,
$$
since both integrands equal
$$
f'(\Gamma(s,t))\frac{\partial\Gamma}{\partial t}
\frac{\partial\Gamma}{\partial s}
+f(\Gamma(s,t))\frac{\partial^2\Gamma}{\partial t\partial s}.
$$

The proof without assuming smoothness is based on Cauchy's theorem
for a disk and a polygonal approximation argument.
\demo{Sketch of proof}  Since $\Gamma$ is continuous on the compact set
$[0,1]\x[0,1]$, it is uniformly continuous.  Moreover, there exists
$\epsilon>0$ so that $G$ contains a closed disk of radius $\epsilon$ about
each point in the range of $\Gamma$.  Therefore, we can find
an integer $n$, so that $\Gamma$ maps each of the squares
$[j/n,(j+1)/n]\x[k/n,(k+1)/n]$ into the disk of radius $\epsilon$
about $\Gamma(j/n,k/n)$.  Let $\rho_k$ be the closed polygonal path
determined by the points $\Gamma(j/n,k/n)$, $j=0,1,\ldots,n$.  It
is easy to conclude from Cauchy's theorem for a disk that
$$
\int_{\rho_k} f(z)\,dz=\int_{\rho_{k+1}} f(z)\,dz,
$$
and also that
$$
\int_{\gamma_0} f(z)\,dz=\int_{\rho_0} f(z)\,dz,\qquad
\int_{\gamma_1} f(z)\,dz=\int_{\rho_n} f(z)\,dz. \qed
$$
\enddemo

\definition{Definition} A closed path in $G$ is {\it homotopic to 0}\/
in $G$ if it is homotopic to a constant path.
\enddefinition

\proclaim{Corollary} If $\gamma$
is a piecewise smooth path which is homotopic to $0$ in $G$, and $f$ is
analytic in $G$, then
$$
\int_\gamma f(z)\,dz=0.
$$
\endproclaim

\definition{Definition} A region $G$ is simply-connected if every closed
path in $G$ is homotopic to $0$.
\enddefinition

As an immediate consequence of the theorem above, we have Cauchy's theorem
(previously proved only for $G$ a disk).

\proclaim{Cauchy's Theorem} If $G$ is
is a simply-connected region in the complex plane then
$$
\int_\gamma f(z)\,dz=0
$$
for every function $f$ analytic in $G$ and every piecewise smooth
closed curve $\gamma$ in $G$.
\endproclaim

\goodbreak\subhead Winding numbers and Cauchy's Integral Formula \endsubhead\nobreak
%
Intuitive meaning of winding number: if a pole is planted in the
plane at a point $a$ and a closed curve is drawn in the plane
not meeting $a$, if the curve were made of well-lubricated rubber
and were to be contracted as small as possible, if it contracts
to a point other than $a$ its winding number is 0.  If it contracts
around the pole planted at $a$, its winding number is the number of
times it circles the pole (signed by orientation).

\definition{Definition} Let $\gamma$ be a piecewise smooth closed
curve in $\C$ and $a\in\C\setminus\gamma$.  Then the {\it index}\/
of $\gamma$ with respect to $a$, or the {\it winding number}\/ of
$\gamma$ about $a$ is defined to be
$$
n(\gamma;a)=\frac1{2\pi i}\int_\gamma\frac 1{z-a}dz.
$$
\enddefinition

Demonstrate how the winding number can be calculated by drawing a negatively
directed ray from $a$ and integrating over pieces of the curve between
successive intersections of the curve with the ray.

\proclaim{Proposition} The index $n(\gamma;a)$ is an integer for
all $a\in\C\setminus\gamma$.  It is constant on each connected
component of $\C\setminus\gamma$.
\endproclaim
\demo{Proof}  Assuming $\gamma$ is parametrized by $[0,1]$, define
$$
g(t)=\int_0^t \frac{\gamma'(s)}{\gamma(s)-a}\,ds.
$$
An easy direct calculation shows that
$$
\frac{d}{dt}\exp[-g(t)][\gamma(t)-a]=0,
$$
so, indeed, $\exp[-g(t)][\gamma(t)-a]=\gamma(0)-a$.  Taking $t=1$,
$$
\exp[-2\pi i n(\gamma;a)][\gamma(1)-a] = \gamma(0)-a,
$$
or, $\exp[-2\pi i n(\gamma;a)]=1$, which implies that $n(\gamma;a)$
is indeed integral.

The function is obviously continuous, and being integer, it is constant
on connected components.  Clearly also, it tends to $0$ as $a$ tends
to $\infty$, so it is identically $0$ on the unbounded component.\qed
\enddemo

The next result is an immediate corollary of the invariance of path
integrals of analytic functions under homotopy.

\proclaim{Proposition} If $\gamma_0$ and $\gamma_1$ are paths which are
homotopic in $\C\setminus\{a\}$ for some point $a$, 
then $n(\gamma_0;a)=n(\gamma_1;a)$.
\endproclaim

\proclaim{Cauchy's Integral Formula} Let $\gamma$ a piecewise smooth curve
in a region $G$ which is null homotopic there, and let $f$ be an analytic function
on $G$.  Then
$$
n(\gamma;a)f(a)=\frac1{2\pi i}\int_\gamma \frac{f(z)}{z-a}\,dw,\quad
a\in G\setminus\gamma.
$$
\endproclaim
\demo{Proof} For $a\in G\setminus\gamma$ fixed, let
$$
g(z)=\cases
\displaystyle\frac{f(z)-f(a)}{z-a},&z\in G\setminus\{a\},
\\
f'(a),& z=a.
\endcases
$$
It is easy to show that $g$ is continuous (even analytic!) on $G$ (show
the continuity as a homework exercise).  Applying Cauchy's theorem to
$g$ gives the desired result immediately.
\enddemo

\remark{Remark} This presentation is quite a bit different from the
book's.  The book proves Cauchy's formula under the assumption that
$\gamma$ is {\it homologous to 0}, i.e., that $n(\gamma;a)=0$ for all
$a\in\C\setminus G$.  It is easy to see that homotopic to 0 implies
homologous to 0, but the reverse is not true, so the book's result is
a bit stronger.  It is true that all piecewise smooth closed curves
in region are homologous to zero iff $G$ is
simply-connected (see Theorem 2.2 of Chapter 8).

The book also goes on to consider chains, which are unions of curves
and proves that
$$
f(a)\sum_{j=1}^k n(\gamma_j;a) = \frac1{2\pi i}\sum_{j=1}^k\int_{\gamma_j}
\frac{f(w)}{w-z}\,dz,
$$
if the chain $\gamma_1+\cdots+\gamma_n$ is homologous to zero, i.e.,
$\sum_{j=1}^k n(\gamma_j;z)=0$ for all $z\in \C\setminus G$.
This sort of result is useful, e.g., in the case of an annulus.
However we can deal with this solution in other ways (by adding
a line connecting the circles).
\endremark

\goodbreak\subhead Zero counting; Open Mapping Theorem \endsubhead\nobreak
%
\proclaim{Theorem} Let $G$ be a region, $\gamma$ a curve homotopic
to $0$ in $G$ and $f$ an analytic function on $G$ with zeros
$a_1$, $a_2$, \dots, $a_m$ repeated according to multiplicity.
Then
$$
\frac1{2\pi i}\int_\gamma \frac{f'(z)}{f(z)}\,dz=\sum_{k=1}^m n(\gamma;a_k).
$$
\endproclaim
\demo{Proof} We can write $f(z)=(z-a_1)(z-a_2)\ldots(z-a_m)g(z)$ where
$g(z)$ is analytic and nonzero on $G$.  Then
$$
\frac{f'(z)}{f(z)}=\frac1{z-a_1}+\frac1{z-a_2}+\cdots\frac1{z-a_m}
+\frac{g'(z)}{g(z)},
$$
and the theorem follows.\qed
\enddemo

If the $a_i$ are the solutions of the equation $f(z)=\alpha$, then
$$
\frac1{2\pi i}\int_\gamma \frac{f'(z)}{f(z)-\alpha}\,dz
=\sum_{k=1}^m n(\gamma;a_k).
$$
Note that we may as well write the integral as
$$
\frac1{2\pi i}\int_{f\circ\gamma}\frac1{w-\alpha}{dw}.
$$

\proclaim{Theorem} If $f$ is analytic at $a$ and $f(a)=\alpha$ with
finite multiplicity $m\ge 1$, then there exists $\epsilon,\delta>0$ such that
for all $0<|\zeta-\alpha|<\delta$, the equation $f(z)=\zeta$ has precisely
$m$ roots in $|z-a|<\epsilon$ and all are simple.
\endproclaim
\demo{Proof} That the number of roots is $m$ comes from the continuity
of the integral as long as $\zeta$ remains in the same connected component
of $\C\setminus f\circ\gamma$ as $\alpha$, and the fact that it is integer
valued.  We can insure that the roots are simple by taking $\epsilon$
small enough to avoid a root of $f'$.\qed
\enddemo

\proclaim{Open Mapping Theorem} A non-constant analytic function is
open.
\endproclaim
\demo{Proof} This is an immediate corollary.\enddemo

\goodbreak\subhead Morera's Theorem and Goursat's Theorem \endsubhead\nobreak
%
\proclaim{Morera's Theorem} If $G$ is a region and $f:G\to\C$ a continuous
function such that $\int_T f\,dz=0$ for each triangular path
$T$ in $G$, then $f$ is analytic.
\endproclaim
\demo{Proof} Since it is enough to show $f$ is analytic on disks, we can
assume $G$ is a disk.  Let $a$ be the center of the disk, and define
$F(z)=\int_{[a,z]}f$.  By hypothesis
$$
\frac{F(z)-F(z_0)}{z-z_0}=\frac1{z-z_0}\int_{[z,z_0]}f(w)\,dw
=\int_0^1 f(z_0+t(z-z_0))dt.
$$
Continuity then implies that $F'(z_0)=f(z_0)$, so $f$ has a primitive,
so is analytic.
\enddemo

Before proceeding to Goursat's Theorem, we consider another result
that uses an argument similar to Morera's Theorem.

\proclaim{Theorem} Let $G$ be a region and $f:G\to\C$ a continuous function.
Then $f$ admits a primitive on $G$ if and only if $\int_\gamma f=0$
for all piecewise smooth closed curves $\gamma$ in $G$.
\endproclaim
\demo{Proof} Only if is clear.  For the if direction, fix $a\in G$ and
define $F(z)=\int_{\gamma(a,z)}f$ where $\gamma(a,z)$ is any piecewise smooth
path from $a$ to $z$.  Reasoning as in Morera's theorem $F'=f$.\qed
\enddemo

In particular, if $G$ is a simply-connected region then every analytic
function admits a primitive.  If $0\notin G$, then $1/z$ admits a primitive.
Adjusting the primitive by a constant we can assure that it agrees with
a logarithm of $z_0$ at some point $z_0$.  It is then easy to check
by differentiation that $\exp(F(z))/z\equiv1$.  Thus

\proclaim{Proposition} If $G$ is a simply connected domain, $0\in G$,
then a branch of the logarithm may be defined there.
\endproclaim

(As an interesting example, take $G$ to be the complement of the spiral
$r=\theta$.)

We can at long last show that a function which is complex differentiable in
a region (but without assuming it is continuously differentiable) is analytic.

\proclaim{Goursat's Theorem} Let $G$ be a region and $f:G\to\C$ complex
differentiable at each point of $G$.  Then $f$ is analytic.
\endproclaim
\demo{Proof} We will show that the integral of $f$ vanishes over every
triangular path.  Let $\Delta_0$ be a closed triangle with boundary $T_0$.  Using
repeated quadrisection we obtain triangles
$$
\Delta_0\subset\Delta_1\subset\ldots
$$
with boundaries $T_n$ such that
$$
|\int_{T_0}f|\le 4^n|\int_{T_n}f|.
$$
Let $d=\operatorname{diam}T_0$, $p=\operatorname{perim} T_0$.
Then $\operatorname{diam}T_n=d/2^n$, $\operatorname{perim} T_n=p/2^n$.
By compactness $\bigcap\Delta_n\ne\emptyset$.  Let $z_0$ be in the intersection
(it's unique).

From the differentiability of $f$ at $z_0$ we have: given $\epsilon>0$
$$
|f(z)-f(z_0)-f'(z_0)(z-z_0)|\le\epsilon|z-z_0|
$$
for all $z$ in some disk around $z_0$, therefore for all $z\in\Delta_n$,
some $n$.  Since $\int_{T_n}f(z)\,dz=\int_{T_n}z\,dz=0$, 
$$
\align
|\int_{T_n}f|&=|\int_{T_n}[f(z)-f(z_0)-f'(z_0)(z-z_0)]dz|
\\
&\le 
\epsilon\operatorname{diam}T_n\operatorname{perim} T_n=\epsilon d p /4^n.
\endalign
$$
Therefore $|\int_{T_0}f|\le\epsilon d p$, and, since $\epsilon$ was arbitrary,
it must vanish.\qed
\enddemo

\goodbreak\head IV. Singularities of Analytic Functions \endhead\nobreak

\definition{Definition} If $f$ is analytic in $B(a;R)\setminus\{a\}$,
for some $a\in\C$, $R>0$, but not in $B(a;R)$.  We say $f$ has an
{\it isolated singularity}\/ at $a$.  If there exists $g$ analytic
on some disk centered at $a$ which agrees with $f$ except at $a$,
we say $f$ has a {\it removable}\/ singularity.
\enddefinition

\proclaim{Theorem} Suppose $f$ has an isolated singularity at $a$.  It
is removable if and only if $\lim_{z\to a} (z-a)f(z)=0$.
\endproclaim
\demo{Proof} ``Only if'' is clear.  For ``if'' let $g(z)=(z-a)f(z)$,
$z\ne a$, $g(a)=0$.  Using Morera's theorem (separating the cases
$a$ inside the triangle, on the boundary, or in the interior), it is
easy to see that $g$ is analytic.  Therefore $g(z)=(z-a)h(z)$ for $h$
analytic, and $h$ is the desired extension of $f$.\qed
\enddemo

\proclaim{Corollary} If $\lim_{z\to a}f(z)\in\C$, or even $f(z)$ is bounded
in a punctured neighborhood of $a$, or even $|f(z)|\le C|z-a|^{-(1-\epsilon)}$
for some $\epsilon>0$, some $C$, then $f$ has a removable singularity.
\endproclaim

Now either $\lim_{z\to a}|f(z)|$ exists as a finite real number,
equals $+\infty$, or does not exist as a finite number or infinity.
The first case is that of a removable singularity.  The second is
called a {\it pole}, the third, an {\it essential singularity}.

Note that if $f$ has a pole at $a$, then $1/f(z)$ has a removable
singularity there and extends to $a$ with value $0$.  Using this we see
that $f$ has a pole at $a$ if and only if $f(z)=(z-a)^{-m}g(z)$ for some
integer $m>0$ and $g$ analytic in a neighborhood of $a$.  Thus
$$
f(z)=\sum_{n=-m}^\infty a_n(z-a)^n
$$
in a punctured neighborhood of $a$.

The function $e^{1/z}$ has an essential singularity since
$$
\lim_{z\downarrow 0}e^{1/z}=+\infty,\quad
\lim_{z\uparrow 0}e^{1/z}=0.
$$

\proclaim{Theorem (Casorati-Weierstrass)} If $f$ has an essential
singularity at $a$, then the image under $f$ of any punctured disk
around $a$ is dense in $\C$.
\endproclaim
\demo{Proof}  WLOG, assume $a=0$.  If the conclusion is false,
there is a punctured disk around $0$ in which $f(z)$ stays a fixed
positive distance $\epsilon$ away from some number $c\in\C$.
Consider the function $g(z)=[f(z)-c]/z$.  It tends to $\infty$
as $z\to 0$, so it has a pole at $0$.  Therefore $z^m[f(z)-c]\to0$
as $z\to 0$ for sufficiently large $m$.  Therefore, $z^mf(z)\to 0$,
which is not possible if $f$ has an essential singularity.
\enddemo

\goodbreak\subhead Laurent series \endsubhead\nobreak

\definition{Definition} Suppose $f$ is analytic on $\{z:|z|>R\}$
for some $R$.  Then we say that $f$ has an isolated singularity 
at $\infty$.  It is removable if $f(1/z)$ has a removable
singularity at $0$.
\enddefinition

We easily see that $f$ has a removable singularity at $\infty$
$\iff$ $\exists\lim_{z\to\infty}f(z)\in\C$ $\iff$ $f$ is bounded
in a neighborhood of infinity.

Note that if $f$ has a removable singularity at $\infty$ then we
can expand $f$ in a power series in $1/z$, convergent for $|z|>R$
uniformly and absolutely for $|z|\ge\rho>R$.

We say that a series of the form $\sum_{n=-\infty}^\infty a_n(z-a)^n$
converges if both the series $\sum_{n=1}^\infty a_{-n}(z-a)^{-n}$
and $\sum_{n=0}^\infty a_{n}(z-a)^{n}$ converge (and their sum is the
value of the doubly infinite summation).

\proclaim{Lemma}  Let $\gamma$ be a piecewise smooth curve
and $g$ a continuous complex-valued function on $\gamma$.
Define
$$
f(z)=\int_\gamma\frac{g(w)}{z-w}dw, \qquad z\notin\gamma.
$$
The $f$ is analytic on $\C\setminus\gamma$.
\endproclaim
\demo{Proof}  Fix $z\notin\gamma$.  Then
$$
\frac{\frac{g(w)}{\zeta-w}-\frac{g(w)}{z-w}}{\ssize\zeta-z}
=-\frac{g(w)}{(\zeta-w)(z-w)}\to -\frac{g(w)}{(z-w)^2}
$$
as $\zeta\to z$ with the convergence uniform in $w\in\gamma$.
Thus $f$ is differentiable at each $z\in\C\setminus\gamma$.\qed
\enddemo

Let $f$ be defined and analytic in an annulus $R_1<|z|<R_2$.
Define
$$
f_2(z)=\frac1{2\pi i}\int_{|w|=r}\frac{f(w)}{w-z}dw,
$$
where $R_2>r>|z|$ (it doesn't matter which such $r$ we take,
by Cauchy's theorem).  Then $f_2$ is analytic in $|z|<R_2$ and so
has a convergent power series there:
$$
f_2(z)=\sum_{n=0}^\infty a_n z^n.
$$

Also, let
$$
f_1(z)=-\frac1{2\pi i}\int_{|w|=r}\frac{f(w)}{w-z}dw
$$
where now $r$ is taken with $R_1<r<|z|$.  Then this is analytic in
$|z|>R_1$ and tends to 0 at $\infty$.  Thus
$$
f_1(z)=\sum_{n=1}^\infty a_{-n}z^{-n}.
$$
Now, it is easy to use Cauchy's integral formula to see that
$f=f_1+f_2$ in the annulus.  This shows that
$$
f(z)=\sum_{n=-\infty}^\infty a_nz^n,
$$
with the convergence absolute and uniform on compact subsets
of the annulus.  Note that such an expansion is unique, since we
can easily recover the $a_n$:
$$
a_m=\frac1{2\pi i}\sum_{n\in\Z}{a_n}\int_{|z|=r}\frac{z^n}{z^{m+1}}dz
=\frac1{2\pi i}\int_{|z|=r}\frac{f(z)}{z^{m+1}}dz
$$
where $r$ is any number in $(R_1,R_2)$ (result is independent of choice
of $r$ since two different circles are homotopic in the annulus).

Summarizing:
\proclaim{Theorem (Laurent Expansion)} If $f$ is an analytic function on the
annulus $G=\ann(0;R_1,R_2)$ for some $0\le R_1<R_2\le\infty$, then
$$
f(z)=\sum_{n=-\infty}^{\infty} a_n z^n
$$
for all $z$ in the annulus.  The convergence is uniform and absolute
on compact subsets of the annulus and the coefficients $a_n$ are
uniquely determined by the formula above.
\endproclaim

If $f$ has an isolated singularity at $a$, it has a Laurent
expansion in a punctured disk at $a$:
$$
f(z)=\sum_{n=-\infty}^\infty a_n (z-a)^n, \quad 0<|z-a|<R.
$$
We define
$$
\ord_a(f)=\inf\{n\in \Z:a_n\ne0\}.
$$
Thus $f$ has a removable singularity $\iff$ $\ord_a(f)\ge 0$,
and it extend to $a$ as $0$ $\iff$ $\ord_a(f)>0$, in which case this is
the order of the $0$.  $\ord_a(f)=+\infty$ $\iff$ $f$ is identically
0 near $a$.  $\ord_a(f)=-\infty$ $\iff$ $f$ has an essential singularity
at $a$.  $\ord_a(f)<0$ but finite $\iff$ $f$ has a pole, in which case
$-\ord_a(f)$ is the order of the pole.

We call $a_{-1}$ the {\it residue} of $f$ at $a$, denoted $\Res(f;a)$.
Note that, if $f$ has an isolated singularity at $a$, then for
$r$ sufficiently small,
$$
\Res(f;a)=\frac1{2\pi i}\int_{|z-a|=r}f(z)\,dz.
$$

Give pictoral proof that if $G$ is simply-connected domain, $f$
analytic on $G$ except for isolated singularies, and $\gamma$
is a simple closed curve that encloses singularities at
$a_1,a_2,\ldots,a_m$ (necessarily finite), then
$$
\frac1{2\pi i}\int_\gamma f(z)\,dz = \sum _{j=1}^m \Res(f;a_j).
$$
a simple closed curve in $G$ .

More precisely:
\proclaim{Theorem (Residue Theorem)} Let $G$ be an open set, $E$ a discrete subset
of $G$, and $\gamma$ a null-homotopic piecewise smooth closed curve
in $G$ which doesn't intersect $E$.  Then $\{a\in E|n(\gamma;a)\ne0\}$
is finite and
$$
\frac1{2\pi i}\int_\gamma f(z)\,dz = \sum _{a\in E} n(\gamma;a)\Res(f;a).
$$
for all functions $f$ which are analytic in $G\setminus E$.
\endproclaim
\demo{Proof} Let $F:[0,1]\x[0,1]\to G$ be a homotopy from $\gamma$
to a constant.  The image $K$ of $F$ is compact, so has a finite intersection
with $E$.  If $a\in E$ does not belong to this intersection, the homotopy
takes place in the the complement of $a$, showing that the $n(\gamma;a)=0$.

Let $a_1,\dots,a_m$ be the points in the intersection, and let $g_i$ be
the singular part of $f$ at $a_i$.  The $f-\sum g_i$ is analytic
on $K$ (after removing the removable singularities at the $a_i$),
so $\int_\gamma f-\sum g_i =0$.  The result follows easily.\qed
\enddemo

We will proceed to two easy but useful consequences of the residue
theorem, the argument principle and Rouch\'e's theorem.  First, however,
we consider the application of the residue theorem to the computation of
definite integrals.

\goodbreak\subhead Residue integrals \endsubhead\nobreak

First, how to practically compute residues.  For essential singularities
it can be very difficult, but for poles it is usually easy.
If $f$ has a simple pole
at $a$, then $f(z)=g(z)/(z-a)$ with $g$ analytic at $a$ and $\Res(f;a)=g(a)$
(and thus Cauchy's therem may be viewed as the special case of
the residue theorem where $f$ has a simple pole).
If $f$ has a double pole at $a$, then
$$
f(z)=g(z)/(z-a)^2=\frac{g(a)}{(z-a)^2}+\frac{g'(a)}{z-a}+\cdots,
$$
so $\Res(f;a)=g'(a)$. Similarly, if $f$ has a pole of order $m$
with $f(z)=g(z)/(z-a)^m$, we expand $g$ as a Taylor series around $a$
and can read off $\Res(f;a)$.  (It is $\frac1{(m-1)!}g^{(m-1)}(a)$,
but it is easier to remember the procedure than the formula.)

$$
\text{Example 1: evaluate\qquad\qquad}
\int_{-\infty}^\infty \frac 1{(x^2+a^2)(x^2+b^2)}dx,\qquad 0<a<b.
$$
Use a semicircle centered at the origin in upper half plane and let
radius tend to infinity to get the answer of $\pi/[a b (a+b)]$.  Same
technique works for all rational functions of degree $\le-2$.

$$
\text{Example 2: evaluate\qquad\qquad}
\int_0^{2\pi} \sin^4(x)\,dx.
$$
Write as $\int_\gamma \frac1{i z} \left(\frac{z+1/z}{2i}\right)^4\,dz$,
with $\gamma$ the unit circle, to get the answer $3\pi/4$.
This approach applies to rational functions in $\sin$ and $\cos$.

$$
\text{Example 3: evaluate\qquad\qquad}
\int_0^\infty \frac1{t^{1/2}+t^{3/2}}dt.
$$
Use a branch of the $z^{-1/2}$ with the {\it positive} real axis as
the branch cut, and integrate around this cut and over a large circle
centered at the origin (avoid origin with a small circle).  The answer
is $\pi$.  This approach applies to integrals of the form $\int_0^\infty
t^\alpha r(t) dt$ where $\alpha\in(0,1)$ and $r(t)$ is a rational
function of degree $\le-2$ with at worst a simple pole at $0$.

$$
\text{Example 4: evaluate\qquad\qquad}\int_0^\infty \frac{\sin x}{x}\,dx
$$
We will evaluate
$$
\lim\Sb X\to+\infty\\ \delta\downarrow 0\endSb
\int_{-X}^{-\delta}+\int_{\delta}^X \frac{e^{iz}}{z}\,dz.
$$
The imaginary part of the quantity inside the limit is
$2\int_\delta^X (\sin x)/x\,dx$.  Note: it is essential
that when taking the limit we excise a symmetric interval about zero.
(The ordinary integral doesn't exist: $e^{ix}/x$ has a non-integrable
singularity near the origin, even though its imaginary part, $(\sin x)/x$
doesn't.)  On the other hand, it is {\it not}\/ necessary that we cut
off the integrand at symmetric points $-X$ and $X$ (we could have used
$-X_1$ and $X_2$ and let them go independently to zero).

We use a somewhat fancy path: starting at $-X$ we follow the $x$-axis to
$-\delta$, then a semicircle in the upper-half plane to $\delta$, then
along the $x$-axis to $X$, then along a vertical segment to $X+iX$, then
back along a horizontal segment to $-X+iX$, and then down a vertical
segment back to $-X$.  The integrand $e^z/z$ has no poles inside this
curve, so its contour integral is $0$.  It is easy to see that the
integral is bounded by $1/X$ on each of the vertical segments and by $2e^{-X}$
on the horizontal segment.  On the other hand, $e^{iz}/z=1/z+\text{analytic}$
and the integral of $1/z$ over the semicircle (integrated clockwise) is
easily seen to be $-\pi i$.  We conclude that the limit above equals $\pi i$,
whence
$$
\int_0^\infty \frac{\sin x}{x}\,dx=\frac{\pi}2.
$$

$$
\text{Example 5: evaluate\qquad\qquad}\sum_{n=-\infty}^\infty \frac1{(n-a)^2},
\quad a\in\R\setminus\Z.
$$
Use a rectangular path with opposite corners $\pm(n+1/2 + n i)$,
$n$ a positive integer and evaluate $\int_\gamma (z-a)^{-2} \cot \pi z\,dz$,
$a\notin\Z$.  The poles of the integrand are at $a$ and the integers
and the residue is $-\pi/\sin^2(\pi a)$ and $1/[\pi(n-a)^2]$.  Since
$\sin x+iy = \sin x \cosh y + i \cos x \sinh y$ and $\cos x+iy =
\cos x \cosh y - i \sin x \sinh y$, we get
$$
|\cot x+iy|^2 = \frac{\cos^2 x + \sinh^2 y}{\sin^2 x + \sinh^2 y},
$$
so the $|\cot|$ equals $1$ on the vertical segments and, if $n$ is
large, is bounded by $2$ on the horizontal segments.  It follows easily
that the integral goes to 0 with $n$, so
$$
\sum_{n=-\infty}^\infty \frac1{(n-a)^2} = \frac {\pi^2}{\sin^2 \pi a}.
$$

To state the argument principle we define:
\definition{Definition} A complex function on an open set $G$ is called
{\it meromorphic} if it is analytic on $G$ except for a set of poles.
\enddefinition

Note that the pole set of a meromorphic function is
discrete (but may be infinite).  A meromorphic function is continuous
when viewed as taking values in $\C_\infty$.

Suppose $f$ is meromorphic at $a$ (i.e., on a neighborhood of $a$).
Then $\Res(f/f';a) = \ord_a(f)$.  Indeed, $f(z)=(z-a)^mg(z)$ with
$m=\ord_a(f)$ and $g(a)\ne0$, and $f'/f=m/(z-a)+g'/g$.  The residue
theorem then immediately gives
\proclaim{Theorem (Argument Principle)} Let $G$ be an open set, $f$ a
meromorphic function on $G$, and $\gamma$ a null-homotopic piecewise smooth
closed curve in $G$ which doesn't intersect either the zero set $Z$ or
the pole set $P$ of $f$.  Then
$$
\frac1{2\pi i}\int_\gamma\frac{f'}{f} = \sum_a n(\gamma;a)\ord_a(f)
$$
where the sum is over $Z\cup P$ and contains only finitely many terms.
\endproclaim

Note that the left hand side is nothing other than $n(f\circ\gamma;0)$,
which is behind the name (how much does the argument change as we traverse
this curve).  We saw this theorem previously in the case where $f$ was
analytic, in which case the integral counts the zeros of $f$.  For meromorphic
$f$ we count the poles as negative zeros.

Our final application of the residue theorem is Rouch\'e's theorem.
\proclaim{Theorem (Rouch\'e)} Let $G$ be an open set, $f$ and $g$
meromorphic functions on $G$, and $\gamma$ a null-homotopic piecewise smooth
closed curve in $G$ which doesn't intersect $Z_f\cup P_f\cup Z_g \cup P_g$.
If
$$
|f(z)+g(z)|<|f(z)|+|g(z)|,\quad z\in\gamma,
$$
then
$$
\sum_{a\in Z_f\cup P_f} n(\gamma;a)\ord_a(f)
=\sum_{a\in Z_g\cup P_g} n(\gamma;a)\ord_a(g).
$$
\endproclaim

\demo{Proof} The hypothesis means that on $\gamma$ the quotient $f(z)/g(z)$ is not
a non-negative real number.  Let $l$ be a branch of the logarithm
on $\C\setminus[0,\infty)$.  Then $l(f/g)$ is a primitive of
$(f/g)'/(f/g)=f'/f-g'/g$ which is analytic in a neighborhood of $\gamma$.
Hence, $\int_\gamma f'/f=\int_\gamma g'/g$ and the theorem follows
from the argument principle.\qed
\enddemo

\proclaim{Corollary (Classical statement of Rouch\'e's theorem)} Suppose
$f$ and $\phi$ are analytic on a simply-connected region $G$ and
that $\gamma$ is a simple closed piecewise smooth curve in $G$.
If $|\phi|<|f|$ on $\gamma$, then $f+\phi$ has the same
number of zeros as $f$ in the region bounded by $\gamma$.
\endproclaim
\demo{Proof} $|(f+\phi)+(-f)|=|\phi|<|f|\le|f+\phi|+|-f|$.\qed\enddemo

Example: If $\lambda>1$, then the equation $z\exp(\lambda-z)=1$ has exactly
one solution in the unit disk (take $f=z\exp(\lambda-z)$, $\phi=-1$).

\goodbreak\head V. Further results on analytic functions \endhead\nobreak

\goodbreak\subhead The theorems of Weierstrass, Hurwitz, and Montel \endsubhead\nobreak
%
Consider the space of continuous functions on a region in the complex
plane, endowed with the topology of uniform convergence on compact
subsets.  (In fact this is a metric space topology: we can define
a metric such that a sequence of continuous functions converge to
another continuous function in this metric, if and only if the sequence
converges uniformly on compact subsets.)  Weierstrass's theorems
states that the space of analytic functions on the region is a closed
subspace and differentiation acts continuously on it.  Montel's theorem
identifies the precompact subsets of the space of analytic functions.

\proclaim{Theorem (Weierstrass)} Let $G\subset\C$ be open, $f_n:G\to\C$,
$n=1,2,\ldots$, analytic, $f:G\to\C$.  If $f_n\to f$ uniformly on
compact subsets of $G$, then $f$ is analytic.  Moreover
$f_n'$ converges to $f'$ uniformly on compact subsets of $G$.
\endproclaim
\demo{Proof} If $T$ is a triangular path,
$$
\int_T f = \lim_{n\to\infty}\int_T f_n =0,
$$
so $f$ is analytic by Morera's theorem.

Let $\gamma$ be the boundary of a disk around $z$ contained in $G$.
$$
\lim f_n'(z)=\lim \frac1{2\pi i} \int_\gamma \frac{f_n(w)}{(w-z)^2}dz
=\frac1{2\pi i} \int_\gamma \frac{f(w)}{(w-z)^2}dz=f'(z).\qed
$$
\enddemo

Combining Weierstrass's Theorem and the Argument Principle
we get:
\proclaim{Theorem (Hurwitz)} Let $G$ be a domain in $\C$, $f_n$, $f$
analytic functions on $G$ with $f_n\to f$ uniformly on compact
subsets.  If each of the $f_n$ is nowhere-vanishing, then $f$ is either
nowhere-vanishing or identically zero.
\endproclaim
\demo{Proof} If $f$ is not identically zero, but $f(a)=0$, we
can choose $r>0$ so that $\bar B(a,r)\subset G$, and $a$ is
the only zero of $f$ on this disk.  Letting $\gamma$ be the boundary
of the disk, the argument principle then tells us that
$\int_\gamma f_n'/f_n =0$, but $\int_\gamma f'/f=2\pi i \ord_a(f)\ne0$.
But, in view of Weierstrass's Theorem the hypothesis tells us
that $f_n'/f_n$ converges to $f'/f$ uniformly on $\gamma$, so we have
a contradiction.
\qed
\enddemo

\proclaim{Corollary}  Let $G$ be a domain in $\C$, $f_n$, $f$
analytic functions on $G$ with $f_n\to f$ uniformly on compact
subsets.  If each of the $f_n$ is injective, then $f$ is either
injective or identically constant.
\endproclaim
\demo{Proof} If $f$ is not identically constant, but there exists
distinct $a$, $b$ with $f(a)=f(b)$, then we
can choose $r>0$ so that $B(a,r)\cup B(b,r)\subset G$, and
$B(a,r)\cap B(b,r)=\emptyset$.  Now $f_n(z)-f(a)$ converges
to $f(z)-f(a)$ uniformly on compact subsets of $B(a,r)$ and the latter
function has a zero but is not identically zero there, so for all sufficiently
large $n$, $f_n(z)=f(a)$ for some $z\in B(a,r)$, and, by the same argument,
for some $z\in B(b,r)$. This contradicts the injectivity of the $f_n$.
\qed
\enddemo

\proclaim{Theorem (Montel)} Let $G\subset\C$ be open, $\Cal F$ a
family of analytic functions on $G$.  Suppose that the functions
$f_1,f_2,\ldots$ are uniformly bounded on each compact subset of $G$.
Then there exists a subsequence which converges uniformly on compact
subsets.
\endproclaim
\demo{Proof}
Fix $a\in G$ and let $d(a)=\operatorname{dist}(a,\partial G)/2$.  First we show
that we can find a subsequence that converges uniformly on $B(a,d(a)/2)$.
Now
$$
f_m(z)=\sum_{n=0}^\infty c_n(f_m)(z-a)^n, \qquad z\in \bar B(a,d(a)).
$$
where $c_n(f)=f^{(n)}(a)/n!$.  By the Cauchy estimates
$$
|c_n(f_m)|\le M/d(a)^n
$$
where $M$ is the uniform bound for the $f_m$ on $\bar B(a,d(a))$.

Since $c_1(f_m)$ is uniformly bounded in $\C$, we can find a subsequence
$f_{m^1(j)}$ for which $c_1(f_{m^1(j)})$ converges as $j\to\infty$.
Then we can find a further subsequence so that $c_2(f_{m^2(j)})$
converges, as well, etc.  For the diagonal sequence $g_j=f_{m^j(j)}$
$c_n(g_j)$ converges for all $n$.

Claim: $g_j$ converges uniformly on $B(a,d(a)/2)$.  To prove this,
we must show that for any $\epsilon>0$
$$
|g_j-g_k|\le \epsilon\qquad\text{on $B(a,d(a)/2)$.}
$$
Now, for any $N\in\N$, $z\in B(a,d(a)/2)$,
$$
\align
|g_j(z)-g_k(z)| &\le \sum_{n=1}^N |c_n(g_j)-c_n(g_k)||z-a|^n +
 \sum_{n=N+1}^\infty \frac{2M}{d(a)^n}|z-a|^n
\\
 &\le \sum_{n=1}^N |c_n(g_j)-c_n(g_k)|[d(a)/2]^n + \frac{2M}{2^N}.
\endalign
$$
Given $\epsilon>0$ we can choose $N$ large enough that
$\frac{2M}{2^N}$ and then choose $j$ and $k$ large enough
that $|c_n(g_j)-c_n(g_k)|\le \epsilon/[2(N+1)|[d(a)/2]^n]$
for $0\le n\le N$.  This proves the claim.

Fix $a\in G$ and let $d(a)=\operatorname{dist}(a,\partial G)/2$.  First we show
that we can find a subsequence that converges uniformly on $B(a,d(a)/2)$.
Now
$$
f_m(z)=\sum_{n=0}^\infty \frac{f_m^{(n)}(a)}{n!}(z-a)^n,
\qquad z\in \bar B(a,d(a)).
$$
By the Cauchy estimates
$$
|f_m^{(n)}(a)|\le M n!/d(a)^n
$$
where $M$ is the uniform bound for the $f_m$ on $\bar B(a,d(a))$.
In particular, $\sup_{m} |f_m^{(n)}(a)|<\infty$ for each $n$.

Thus we can find a subsequence
$f_{m^0(j)}$ for which $f_{m^0(j)}(a)$ converges as $j\to\infty$.
Then we can find a further subsequence so that $f_{m^2(j)}'(a)$
converges as well, etc.  For the diagonal sequence $g_j=f_{m^j(j)}$
$g_j^{(n)}(a)$ converges for all $n$.

Claim: $g_j$ converges uniformly on $B(a,d(a)/2)$.  To prove this,
we must show that for any $\epsilon>0$
$$
|g_j-g_k|\le \epsilon\qquad\text{on $B(a,d(a)/2)$.}
$$
Now, for any $N\in\N$, $z\in B(a,d(a)/2)$,
$$
\align
|g_j(z)-g_k(z)| &\le \sum_{n=0}^N |g_j^{(n)}(a)-g_k^{(n)}(a)||z-a|^n/n! +
 \sum_{n=N+1}^\infty \frac{2M}{d(a)^n}|z-a|^n
\\
 &\le \sum_{n=1}^N |g_j^{(n)}(a)-g_k^{(n)}(a)||z-a|^n/n! + \frac{2M}{2^N}.
\endalign
$$
Given $\epsilon>0$ we can choose $N$ large enough that
$2M/2^N\le\epsilon/2$ and can then choose $j$ and $k$ large enough
that $|g_j^{(n)}(a)-g_k^{(n)}(a)|\le n!\epsilon/\{2(N+1)|[d(a)/2]^n\}$
for $0\le n\le N$.  This proves the claim and shows that we can find
a subsequence which converges uniformly on $B(a,d(a)/2)$.

Now we may choose a countable sequence of points $a_i\in G$ such
that $G=\bigcup_{i=1}^\infty B(a_i,d(a_i)/2)$ (e.g., all points
in $G$ with rational real and imaginary parts).  Given the
sequence of functions $f_m$ uniformly bounded on compact sets,
we may find a subsequence $f_{n^1(j)}$ which converges uniformly
on $B(a_1,d(a_1)/2)$.  Then we may take a subsequence of that sequence
which converges uniformly on $B(a_2,d(a_2)/2)$, etc.  The diagonal
subsequence $h_j$ converges uniformly on each $B(a_i,d(a_i)/2)$.

We complete the proof by showing that $h_j$ converges uniformly on
all compact subsets.  Indeed this is immediate, because any compact
subset of $G$ is contained in a finite union of the $B(a_i,d(a_i)/2)$.
\qed
\enddemo

\goodbreak\subhead Schwarz's Lemma \endsubhead\nobreak
%
Schwarz's lemma is an easy, but very useful, consequence of the maximum
modulus principle.  The most difficult part is to remember that the
German analyst Herman~Schwarz (actually Karl Herman Amandus Schwarz,
1843-1921), inventor of the ``Schwarz inequality,'' ``Schwarz lemma,''
and ``Schwarz alternating method,'' spells his name without a ``t'',
while the French analyst, Laurent~Schwartz, 1915--, inventor of the theory of
distributions, uses a ``t''.

Suppose that an analytic map $f$ maps a disk of radius $\epsilon$ about
$a$ into the disk of radius $\delta$ about $f(a)$.  We can
use this to estimate the derivative $f'(a)$ and the stretching
$|f(b)-f(a)|/|b-a|$.  To keep the statement simple, we translate
in the domain and range space so that $a=f(a)=0$ and dilate in the
domain and range so that $\epsilon=\delta=1$.  We then get:

\proclaim{Theorem (Schwarz's Lemma)}  Suppose $f$ maps the open unit disk
into itself leaving the origin fixed.  Then $|f(z)|\le |z|$ for all
$z$ in the disk, and if equality holds for any nonzero point, then
$f(z)=cz$ for some $c$ of modulus 1.  Also $|f'(0)|\le 1$ and if
equality holds, then $f(z)=cz$ for some $c$ of modulus 1.
\endproclaim
\demo{Proof} The function $g(z)=f(z)/z$ has a removable singularity at
the origin, and extends analytically to the disk with $g(0)=f'(0)$.  It
satisfies $|g(z)|\le 1/r$ on the circle of radius $r<1$, so by the
maximum modulus theorem it satisfies this condition on the disk of
radius $r$.  Letting $r$ tend to one gives the result.  Equality implies
$g$ is constant.\qed
\enddemo

From a homework exercise (\S III .3, no.~ 10), we know that the most
general M\"obius transformations of the open unit disk $D$ into itself
is given by
$$
z\mapsto c\frac{z-a}{1-\bar a z}
$$
with $a$ in the disk and $|c|=1$.  We can use Schwarz's lemma to show that
this is most general one-to-one analytic map of the disk onto itself.

First consider the special case $f(0)=0$.  Then Schwarz shows that $|f'(0)|\le1$
with equality if and only if $f(z)=cz$ some $|c|=1$.  Applying Schwarz to
$f^{-1}$ gives $|1/f'(0)|=|(f^{-1})'(0)|\le 1$.  Thus equality does
indeed hold.

Returning to the general case, say $f(a)=0$.  Then we can apply the
special case above to $f\circ \phi_{-a}$ to get $f(\phi_{-a}(z))=cz$ or
$f(z)=c\phi_a(z)$.

\goodbreak\subhead The Riemann Mapping Theorem \endsubhead\nobreak
%
We begin by deriving some easy consequences of simple-connectivity.
Recall that every analytic function on a simply-connected region admits a
primitive.  It follows that if $f$ is a nowhere-vanishing function on
a simply-connected domain, then there exists an analytic function $F$
on that domain such that $f=\exp(F)$.  Indeed, let $g$ be primitive of
$f'/f$.  It follows immediately that $f\exp(-F)$ is constant, and so
adjusting $g$ by an additive constant, $f=\exp(F)$.  Another consequence
is the {\it square-root property}: if $f$ is a nowhere-vanishing
analytic function on the domain, then there exists an analytic function
$g$ with $[g(z)]^2=f(z)$.  Indeed, we can just take $g=\exp(F/2)$.
For future use we remark that if there exists an analytic isomorphism of
a region $G_1$ onto a region $G_2$, then $G_1$ has the square-root
property if and only if $G_2$ does. 

We now turn to a lemma of Koebe.
\proclaim{Lemma} Let $D=B(0,1)$ and let $G\subsetneq D$ be a region
containing the origin and having the square root property.  Then
there exists an injective analytic map $r:G\to D$ such that $r(0)=0$,
$|r(z)|>|z|$, $z\in G\setminus\{0\}$.
\endproclaim
\demo{Proof}  Let $a\in D\setminus G$ and let $b$ be one of the
square roots of $-a$.  Define
$$
q(z)=\phi_{-a}([\phi_{-b}(z)]^2)
$$
(where $\phi_\alpha(z):=(z-\alpha)/(1-\bar\alpha z)$).
Clearly $q(0)=0$, and $q$ is not just multiplication by a constant (in fact,
it is not 1-1 on $D$), so Schwarz's lemma implies
$$
|q(z)|<|z|,\qquad 0\ne z\in D.
$$
Next, note that $\phi_a$ is an analytic map of $G$ into $D$ which
never vanishes, so admits an analytic square root $g(z)$, which we
may determine uniquely by insisting that $g(0)=b$.  Defining
$$
r=\phi_b\circ g,
$$
we have $r(0)=0$, and $q(r(z))\equiv z$.  Thus if $0\ne z\in G$,
$r(z)\ne0$ and $|z|=|q(r(z))|<|r(z)|$.
\qed
\enddemo

Two regions in the complex plane are called conformally equivalent
if they are analytically isomorphic, that is, if there exists a
one-to-one analytic mapping of the first domain onto the second
(whose inverse is then automatically analytic).

\proclaim{Riemann Mapping Theorem}  Let $G\subsetneq \C$ be simply-connected.
Then $G$ is conformally equivalent to the open unit disk.
\endproclaim
\demo{Proof} The only consequence of simple-connectivity that we shall
use in the proof is that $G$ has the square-root property.

The structure of the proof is as follows: first we use the hypotheses
to exhibit an injective analytic map of $h_0$ of $G$ onto a domain
$G_0$ which satisfies the hypotheses of Koebe's lemma.  Then we use an
extremal problem to define an injective analytic map $f$ of $G_0$ onto a
domain $G_1$ contained in the disk.  If $G_1$ fills the entire disk we
are done; otherwise, $G_1$ satisfies the hypotheses of Koebe's lemma and
we use the lemma to contradict the extremality of $f$.

Let $a\in\C\setminus G$ and let $g$ be an analytic square root of $z-a$.
It is easy to see that $g$ is injective and that if $w\in g(G)$ then
$-w\notin g(G)$.  Pick $w_0\in g(G)$ and then $r>0$ such that $B(w_0,r)\subset
g(G)$ (Open Mapping Theorem).  Therefore $B(-w_0,r)\cap g(G)=\emptyset$.
Set $h(z)=[g(z)+w_0]^{-1}$ so that $h:G\to \bar B(0,1/r)$ is analytic and injective.
Then set $h_0(z)=r[h(z)-h(z_0)]/3$ where $z_0\in G$.  This map is an analytic
isomorphism of $G$ onto a connected open set $G_0\subsetneq D$ containing $0$.

Let $\Cal F$ be the set of all injective analytic functions $f:G_0\to D$
with $f(0)=0$.  Fix $0\ne w_0\in G_0$, and let $\alpha=\sup_{f\in\Cal
F} |f(w_0)|$.  Note $\alpha\in(0,1]$.  We claim that the supremum is
achieved.  Indeed, let $f_n\in\Cal F$, $|f_n(w_0)|\to \alpha$.  Since the
$f_n$ are uniformly bounded (by $1$), Montel's theorem assures us that
there is a subsequence $f_{n_k}$ which converges uniformly on compact
subsets to some analytic function $f$ on $G_0$.  Obviously $f(0)=0$ and
$|f(w_0)|=\alpha>0$, so $f$ is not constant, and since the $f_n$ are injective,
so is $f$.  Clearly $|f|\le 1$ on $G_0$, and by the maximum modulus principle,
$|f|<1$.  Thus $f\in\Cal F$ and achieves the supremum.

We complete the theorem by showing that $f$ is an isomorphism of $G_0$ onto
$D$ (so $f\circ h_0$ is the desired analytic isomorphism of $G$ onto $D$).
We already know that $f$ is analytic and injective, so we need only
show that $f$ maps onto $D$.  If, to the contrary, $G_1:=f(G_0)\subsetneq D$
then (since $G_1$ inherits the square-root property from $G_0$ which inherits
it from $G$) Koebe's lemma gives an injective analytic function $r:G_1\to D$ with
$r(0)=0$, $|r(z)|>|z|$ for $z\ne 0$.  Then $r\circ f\in \Cal F$, but
$|r(f(w_0))|>|f(w_0)|$, a contradiction.
\qed
\enddemo

We now note that the square-root property is equivalent to
simple-connectivity.  Indeed we have seen that simple-connectivity
implies the square-root property.  On the other hand, if $G$ satisfies
the square root property, then either $G=\C$, which is simply-connected,
or $G$ is isomorphic to the disk (by the proof of the Riemann mapping
theorem above), and so is simply-connected.

If $G\subsetneq\C$ is a simply-connected region and $z_0\in G$ is arbitrary
we may take any analytic isomorphism of $G$ onto the disk and follow
it with a suitably chosen M\"obius transformation
($z\mapsto c(a-z)/(1-\bar a z)$, $|a|<1$, $|c|=1$, to obtain an analytic
isomorphism $f$ of $G$ onto $D$ satisfying
$$
f(z_0)=0,\quad f'(z_0)>0.
$$
If $f_1$ and $f_2$ were two such analytic isomorphisms, then $g=f_2\circ f_1^{-1}$
is an analytic isomorphism of the disk onto itself with $g(0)=0$, $g'(0)>0$,
and it easily follows that $g$ is the identity and $f_1=f_2$.

\goodbreak\subhead Complements on Conformal Mapping\endsubhead\nobreak
%
There are many interesting questions in conformal mapping which we will
not have time to investigate, but I will quickly state some key results.

{\it Extension to boundary.}  If the boundary of a simply-connected region
is a simple closed curve, then it can be proved that a conformal map of
the region onto the open disk extends to a topological homeomorphism
of the closure of the region on to the closed disk.

{\it Constructive conformal mapping.}  Conformal is used to obtain
explicit solutions to problems involving analytic and harmonic functions.
For example, Joukowski used the conformal mapping $f\circ g$ where
$$
f(z)=2\frac{1+z}{1-z},\qquad g(z)=\left(\frac{z-1}{z+1}\right)^2,
$$
applied to a circle passing through $-1$ and containing $1$ in its
interior.  The resulting region, called a Joukowski airfoil, looks like
a cross-section of a wing, and by adjusting the circle, one can
adjust the exact shape of the airfoil.  Using the explicit conformal map
to the disk and explicit formulas for the solution of Laplace's equation
on (the exterior of) a disk (similar to those that will be derived in
the next section, one can calculate things like the drag and lift of the
airfoil, and so this was a useful approach to airfoil design.

An important class of regions are the polygons.  The
Schwarz--Christoffel formula is an explicit formula for the conformal
map of a given polygon onto the disk.  The formula involves a few
complex parameters which are determined by the vertices of the
polygon.  Given the vertices, one cannot usually compute the parameters
analytically, but they are not difficult to compute numerically (with a
computer).

{\it Multiply connected regions}.  For lack of time we will not study
conformal mapping of multiply connected regions, but we just state,
without proof, the result for doubly connected regions (regions whose
complement in the Riemann sphere consists of two connected components,
i.e., roughly regions in the plane with one hole).  It can be shown
(with basically the tools at our disposal) that any such region is
conformally equivalent to the annulus $1<|z|<R$ for some $R>1$.  The
value of $R$ is uniquely determined.  In particular, two such annuli
with different values of $R_1$ are {\it not}\/ conformally equivalent.

\goodbreak\head VI. Harmonic Functions \endhead\nobreak

A real-valued $C^2$ function on an open subset of $\C$ is called {\it
harmonic}\/ if its Laplacian vanishes.  We have seen previously that the
real part of an analytic function is harmonic and that if the domain is
simply-connected every real-valued harmonic function $u$ admits a harmonic
conjugate, i.e., a real-valued function $v$, for which $u+i v$ is analytic.
In particular, we see that every harmonic function is $C^\infty$.

Suppose $u$ is harmonic on $G$, $a\in G$.  If $r>0$ is such that
$\bar B(a,r)$, then we can choose a disk $D$ with $\bar B(a,r)\subset D\subset
G$, and since $D$ is simply-connected $u=\Re f$ on $D$ for some analytic
function $f$ on $D$.  Cauchy's integral formula then gives
$$
f(a)=\frac1{2\pi i}\int_{|z-a|=r}\frac{f(z)}{z-a}dz=\frac1{2\pi}\int_0^{2\pi}
f(a+re^{i\theta})d\theta.
$$
Taking real parts we get:
\proclaim{Mean Value Theorem} If $u:G\to\R$ is harmonic and $\bar B(a,r)\subset G$, then
$$
u(a)=\frac1{2\pi}\int_0^{2\pi}
u(a+re^{i\theta})d\theta.
\tag{MVP}
$$
\endproclaim

A harmonic function satisfies the {\it Maximum Principle}: it
does not assume its maximum on a region unless it is constant
there.  Indeed the conclusion is true for any continuous function
with the mean-value property (MVP), or even just satisfying the inequality
$u(a)\le(2\pi)^{-1}\int_0^{2\pi} u(a+re^{i\theta})d\theta$ (since this
implies that the set of points were the maximum is achieved is open, and
it is obviously closed).  The reverse inequality similarly shows that a
harmonic function satisfies the Minimum Principle.

Given a bounded domain $G$ and a function $f:\partial G\to\R$, the {\it
Dirichlet problem}\/ consists of finding a function $u:\bar G\to\R$
such that $u$ is harmonic on $G$, continuous on $\bar G$, and coincident
with $f$ on $\partial G$.  An important consequence of the maximum and minimum principles
is uniqueness for the Dirichlet problem.  If $u_1$ and $u_2$ are both harmonic
on $G$, continuous on $\bar G$, and agree on $\partial G$, then $u_1-u_2$
is harmonic and vanishes on the boundary, but it takes its maximum and minimum
on the boundary, so it is identically zero, i.e., $u_1=u_2$.

When applying the maximum principle to functions which are not known
to extend continuously to the boundary this easy-to-prove consequence
is useful:  If $u$ satisfies the maximum principle on a bounded open
set and there is a number $K$ such that $\limsup_{z\to z_0} u(z) \le
K$, then $u\le K$ on $G$.  [Proof: Let $M=\sup_{z\in G}u(z)$ and take
$z_n\in G$ with $u(z_n)\to M$.  Pass to a subsequence with $z_n\to
z_0\in\bar G$.  If $z_0\in G$, then $u$ takes its maximum there, so is
constant and obviously $M\le K$.  Otherwise $z_0\in \partial G$, so
$K\ge \limsup_{z\to z_0}u(z)\ge\lim_n u(z_n)=M$.]

\goodbreak\subhead The Poisson kernel\endsubhead\nobreak
%
For $0\le r<1$, $\theta\in\R$ let
$$
P_r(\theta)=\Re\left(\frac{1+re^{i\theta}}{1-re^{i\theta}}\right)
=\frac{1-r^2}{1-2r\cos\theta+r^2},
$$
define the Poisson kernel.  Note that
$$
P_r(\theta)=\Re(1+2\sum_{n=1}^\infty r^ne^{in\theta})=\sum_{n=-\infty}^\infty
r^{|n|}e^{in\theta}.
$$

\proclaim{Lemma}  For $0\le r<1$, $P_r(\theta)$ is a smooth, positive,
even, $2\pi$-periodic function of $\theta$ with mean value $1$.  Moreover
if $\theta>0$, then
$$
\lim_{r\uparrow 1}P_r(\theta)=0, \qquad \theta\in\R\setminus2\pi\Z,
$$
and if $\delta>0$, then the convergence is uniform over $\theta\in\R$
such that $|\theta-2\pi n|\ge\delta$ for all $n\in\Z$.
\endproclaim
\demo{Proof} Elementary (use the power series expansion for the integral
and the formula
$$
P_r(\theta)=\frac{1-r^2}{(1-r)^2+2r(1-\cos\theta)}
$$
for the limits).\qed
\enddemo

Now let $D$ be the open unit disk and $\partial D$ its boundary.  The
next theorem shows how to solve the Dirichlet problem for $D$ using
the Poisson kernel:
\proclaim{Theorem} Given $f:\partial D\to\R$ continuous, define
$$
\cases u(z)=f(z), & z\in\partial D,\\
u(re^{i\theta})=\frac1{2\pi}\int_0^{2\pi} P_r(\theta-t)f(e^{it})\,dt,
& 0\le r<1, \quad \theta\in\R.\endcases
$$
The $u$ is continuous on $\bar D$ and harmonic on $D$.
\endproclaim
\demo{Proof}
$$
\align
u(re^{i\theta})&=\Re\left\{\frac1{2\pi}\int_0^{2\pi}f(e^{it})
\left[\frac{1+re^{i(\theta-t)}}{1-re^{i(\theta-t)}}\right]dt\right\}
\\
&=\Re\left\{\frac1{2\pi}\int_0^{2\pi}f(e^{it})
\left[\frac{e^{it}+re^{i\theta}}{e^{it}-re^{i\theta}}\right]dt\right\}
\endalign
$$
or
$$
u(z)=\Re\left\{\frac1{2\pi}\int_0^{2\pi}f(e^{it})
\left[\frac{e^{it}+z}{e^{it}-z}\right]\right\}, \quad z\in D.
$$
The integral defines an analytic function of $z$ (since the integrand
is continuous in $t$ and $z$ and analytic in $z$), so $u$, being the real
part of an analytic function, is harmonic.

It remains to show that $\dsize\lim\Sb z\in D\\ z\to e^{it_0}\endSb u(z)=f(e^{it_0})$
for $t_0\in\R$.  Now
$$
u(re^{i\theta})-f(e^{it_0})
=\frac1{2\pi}\int_{t_0-\pi}^{t_0+\pi}P_r(\theta-t)[f(e^{it})-f(e^{it_0})]dt,
$$
so it suffices to show that the last integral can be made arbitrarily
small by taking $\theta$ sufficiently close to $t_0$ and $r<1$
sufficiently close to $1$.  Now given $\epsilon>0$ we can choose
$\delta>0$ so that $|f(e^{it})-f(e^{it_0})|\le\epsilon$ if $|t-t_0|\le\delta$.
Split the interval of integration as
$$
\{t:|t-t_0|\le\pi\}=I_1\cup
I_2:=\{t:|t-t_0|\le\delta\}\cup\{t:\delta\le|t-t_0|\le\pi\}.
$$
On $I_1$, the integrand is bounded by $P_r(\theta-t)\epsilon$, so
the integral is bounded by
$$
\frac{\epsilon}{2\pi}\int_{I_1}P_r(\theta-t)dt\le\frac{\epsilon}{2\pi}\int_0^{2\pi}P_r(\theta-t)dt\le
\epsilon.
$$
Now suppose that $\theta$ is taken sufficiently near $t_0$, namely
$|\theta-t_0|\le\delta/2$.  Then for $t\in I_2$,
$$
|\theta-t|\ge|t-t_0|-|\theta-t_0|\ge\delta/2,
\qquad
|\theta-t|\le|t-t_0|+|\theta-t_0|\le\pi+\delta/2
$$
Thus $P_r(\theta-t)\to0$ uniformly for $t\in I_2$ as $r\uparrow1$. Since $f$ is bounded,
the entire integrand does so as well.  Thus if $|\theta-t_0|\le\delta/2$
we may take $r$ sufficiently close to $1$ that the integral over $I_2$ is
bounded by $\epsilon$.
\qed
\enddemo

It is now easy to deduce that the mean value property characterizes
harmonic functions.

\proclaim{Theorem}
If $u:G\to\R$ is a continuous function satisfying the mean
value property, then it is harmonic.
\endproclaim
\demo{Proof}
It is enough to show that $u$ is harmonic on each disk with closure
in $G$.  From the Poisson kernel construction we can construct
a harmonic function with the same boundary values as $u$ on this
disk.  The difference between this function and $u$ satisfies the
mean value property and is zero on the boundary of the disk, so
it vanishes on the disk.\qed
\enddemo

\proclaim{Corollary} If a sequence of harmonic functions converges
uniformly on compact subsets, then the limit function is harmonic.
\endproclaim

From the expressions
$$
P_r(\theta)=\frac{1-r^2}{(1+r)^2-2r(1+\cos\theta)}
=\frac{1-r^2}{(1-r)^2+2r(1-\cos\theta)}
$$
we have
$$
\frac{1-r}{1+r}\le P_r(\theta)\le\frac{1+r}{1-r}.
$$

Now suppose $u$ is harmonic on $B(0,R)$ and non-negative, and
$re^{i\theta}$ is an arbitrary point of $B(0,R)$.  Choose $\epsilon>0$ small
enough that $r<R-\epsilon$, and set
$U(z)=u((R-\epsilon)z)$.  $U$ is harmonic on a neighborhood of $\bar B(0,1)$,
so
$$
u(re^{i\theta})=U([r/(R-\epsilon)]e^{i\theta})
=\frac1{2\pi}\int_0^{2\pi}P_{r/(R-\epsilon)}(\theta-t)U(e^{it})\,dt.
$$
Using the non-negativity of $u$, the mean value property, and the upper bound
for $P_r$, we get
$$
u(re^{i\theta})\le \frac{1+r/(R-\epsilon)}{1-r/(R-\epsilon)}u(0).
$$
Letting $\epsilon\downarrow0$ this gives
$$
u(re^{i\theta})\le \frac{R+r}{R-r}u(0).
$$
We obtain a lower bound in a similar fashion.  Of course we can
translate the disk so it is centered at an arbitrary point $a$.
Thus we have proven:
\proclaim{Theorem (Harnack's Inequalities)} If $u$ is non-negative
and harmonic on $B(a,R)$, then
$$
\frac{R-r}{R+r}u(a)\le u(a+re^{i\theta})\le \frac{R+r}{R-r}u(a)
$$
for $0\le r< R$.
\endproclaim

For example, if $u$ is harmonic and non-negative on the disk of radius
$R$ about $a$ then $u(z)\in[u(a)/3,3u(a)]$ on the disk of radius $R/2$
about $a$.

\proclaim{Theorem (Harnack)} Let $0\le u_1\le u_2\le\ldots$
be a sequence of harmonic functions on a region $G$.  Then either
$\lim_n u_n(z)=+\infty$ uniformly on compact subsets of $G$
or there is an harmonic function $u$ with
$\lim_n u_n(z)=u(z)$ uniformly on compact subsets of $G$.
\endproclaim
\demo{Proof}  For any point $a\in G$ let $D$ be a disk
centered at $a$ contained in $G$.  Then there exists a constant $C>1$
with
$$
u_n(z)\le C u_n(a), \quad u_n(a)\le C u_n(z), \qquad z\in D.
$$
Consequently, the numbers $u_n(a)$ remain bounded as $n\to\infty$ if and only
if $u_n(z)$ remains bounded for all $z\in D$.  This
shows that the set of points where $u_n$  stays bounded is both
open and closed, so it is either the empty set or all of $G$.
In the first case, $u_n(z)\to+\infty$ for all $z\in G$, and the above
estimate shows that the convergence is uniform on a covering set
of discs, hence on all compact subsets.

In the second case, $\lim_n u_n(a)\in\C$ for all $a$, and since
$$
u_n(z)-u_m(z) \le C [u_n(a)-u_m(a)], \quad n\ge m, z\in D,
$$
the sequence is uniformly Cauchy on $D$ so converges uniformly
on $D$.  Again, uniform convergence on compact sets follows.
\qed\enddemo

\goodbreak\subhead Subharmonic functions and the solution of the Dirichlet problem\endsubhead\nobreak
\definition{Definition} Let $u$ be a continuous real-valued function
on a region $G$.  If
$$
u(a)\le \frac1{2\pi}\int_0^{2\pi} u(a+re^{i\theta})\,d\theta
$$
whenever $\bar B(a,r)\subset G$, we say that $u$ is {\it subharmonic}.
\enddefinition

As mentioned earlier, a subharmonic function satisfies the maximum
principle.  Other simple properties are:
\roster
\item"$\bullet$" harmonic functions are subharmonic
\item"$\bullet$" the sum of two subharmonic functions is subharmonic
\item"$\bullet$" the pointwise maximum of two subharmonic functions is
subharmonic
\endroster

The next lemma states that subharmonicity is a local property:
\proclaim{Lemma} Suppose that $u$ is subharmonic in a neighborhood
of each point of a region $G$.  Then $u$ is subharmonic on $G$.
\endproclaim
\demo{Proof} If $\bar D=\bar B(a,r)\subset G$, let $\tilde u$ be the continuous
function on $\bar D$ which agrees with $u$ on $\partial D$ and is harmonic
on $D$. From the Poisson kernel representation, we have
$$
\tilde u(a)=\frac1{2\pi}\int_0^{2\pi} u(a+re^{i\theta})\,d\theta.
$$
Now $u-\tilde u$ is subharmonic in a neighborhood of each point of $D$,
and it follows by the usual connectedness argument that the points
where it achieves its maximum on $D$ is either empty or all of $D$.
Hence its maximum on $\bar D$ is achieved on the boundary, so is
$0$.  This shows that $u(a)\le \tilde u(a)$ as desired.
\qed
\enddemo

One more property we shall use:
\proclaim{Lemma} Suppose that $u$ is subharmonic in a region $G$ and
$\bar D=\bar B(a,r)\subset G$.  Define $\tilde u$ on $\bar D$ as the
continous function on $\bar D$ which agrees with $u$ on $\partial D$ and
is harmonic on $D$, and set $\tilde u=u$ on $G\setminus \bar D$.  Then
$\tilde u$ is subharmonic on $G$.
\endproclaim
\demo{Proof} Clearly it is continuous and subharmonic in $D$ and
$G\setminus\bar D$.  So it suffices to show that for each point $a$
of $\partial D$, $\tilde u$ is bounded by its mean on any circle around $a$.
Now $u-\tilde u$ is subharmonic on $D$ and $0$ on $\partial D$, so
$u\le \tilde u$ on $D$, and therefore on $G$.  Thus
$$
\tilde u(a)=u(a)\le \frac1{2\pi}\int_0^{2\pi} u(a+re^{i\theta})\,d\theta
\le \frac1{2\pi}\int_0^{2\pi} \tilde u(a+re^{i\theta})\,d\theta.
$$
\qed
\enddemo

Note that 

Let $G$ be a bounded region.  If $u$ is a harmonic function on $G$ continuous
up to the boundary, and $v$ is any subharmonic function that is less than
$u$ on the boundary, then the maximum principle implies that $v\le u$ in
$G$.  Hence it is reasonable to try to solve the Dirichlet problem
by seeking the largest subharmonic function that doesn't exceed the given
boundary values.  This is the approach of O.~Perron.

Let $f$ be a bounded (at this point not necessarily continuous)
real-valued function on $\partial G$.  Define the Perron family $\Cal
P(f,G)$ as the set of subharmonic functions $v$ on $G$ for which
$$
\limsup_{z\to a}v(z)\le f(a) \quad \text{for all $a\in\partial G$}.
\tag{$\star$}
$$
Note that $\Cal P(f,G)$ contains all constants not exceeding $\min f$,
so it is not empty.

From the maximum principle we obtain:
\proclaim{Lemma} If $v\in\Cal P(f,G)$ then $v\le \sup f$ on $G$.
\endproclaim

Now define the {\it Perron function}
$$
u(z)=\cases \dsize\sup_{v\in\Cal P(f,G)}v(z),& z\in G, \\
f(z), & z\in \partial G.
\endcases
$$
We will show that (1) the Perron function is harmonic on $G$, and (2)
under mild restrictions on $G$, it extends continuously to $\bar G$ with
value $f$ on $\partial G$.  Thus we will have exhibited a solution to
the Dirichlet problem.

\demo{Proof that the Perron function is harmonic} Let $D$ be any
disk with closure in $G$ and let $a$ be any point in the disk.
Choose functions $v_n\in \Cal P(f,G)$ with $v_n(a)\to u(a)$.  Let
$V_n=\max(v_1,\ldots,v_n)$, and then define $\tilde V_n$ to be $V_n$
off $D$ and to be harmonic on $D$ and continuous on $\bar D$.  We know
that $V_n$ and $\tilde V_n$ are subharmonic, and clearly the $V_n$
and thus the $\tilde V_n$ are non-decreasing.  All belong to $\Cal P(f,G)$
so are bounded by $u$, but $v_n\le V_n \le \tilde V_n$, so $\tilde
V_n(a)\uparrow u(a)$.  By Harnack's Theorem, $\tilde V_n$ converges to
an harmonic function $V$ on $D$.  We know that $V(a)=u(a)$.  We will now
show that if $b$ is any other point of $D$ then $V(b)=u(b)$.  This will
show that $u$ is harmonic in $D$ as desired.

Choose functions $\phi_n\in \Cal P(f,G)$ with $\phi_n(b)\to u(b)$, and
set $w_n=\max(\phi_n,v_n)\in \Cal P(f,G)$.  Construct $W_n$ and $\tilde
W_n$ from $w_n$ in analogy with $V_n$ and $\tilde V_n$.  Then $w_n\ge
v_n$, $W_n\ge V_n$, $\tilde W_n\ge \tilde V_n$, and $W_n$ converges to
a harmonic function $W\ge V$ with $W(b)=u(b)$.  Also $W(a)=u(a)=V(a)$,
so $V-W$ is a non-positive harmonic function on $D$ which achieves its
maximum at $a$.  Therefore $V\equiv W$ on $D$, so indeed $V(b)=u(b)$.
\qed
\enddemo

We now turn to the continuity of the Perron solution at the boundary.
\definition{Definition} Let $G$ be a bounded region and $z_0\in\partial
G$.  A continuous function $\phi$ on $\bar G$ which is harmonic on $G$
is called a {\it barrier function}\/ for $G$ at $z_0$ if it is positive on
$\bar G\setminus\{z_0\}$ and zero at $z_0$.
\enddefinition

For example, suppose that there is a line through $z_0$ such that
$\bar G\setminus\{z_0\}$ lies inside one of the open half-planes determined
by the line.  We may write the half-plane as $\Im[(z-z_0)/b]>0$ for some
$b\in\C$, and so $\Im[(z-z_0)/b]$ is a barrier function for $G$ at $z_0$.

Next consider the function $\sqrt{1-1/z}$ where we use the principal
branch of the square root, defined on $\C\setminus(-\infty,0]$.  This
function is analytic on $\C\setminus[0,1]$ and extends continuously
to $1$ with value $0$, and its real part is everywhere positive
on $\C\setminus[0,1]$.  Thus if $z_0=1$ belongs to $\partial G$, and
is the only point of intersection of the interval $[0,1]$ with $\bar G$,
we can use $\Re \sqrt{1-1/z}$ as a barrier function for $G$ at $z_0$.
Similarly, if $z_0$ is any boundary point for which there exists a line
segment which intersects $\bar G$ only at $z_0$, then there is a barrier
function for $G$ at $z_0$.  Thus barrier functions exist at all boundary
points of smooth domains, polygonal domains, even on a disk minus a
segment, and on many other domains.  A standard example of a domain
without a barrier at some boundary point is a punctured disk.

\proclaim{Theorem} Let $G$ be a bounded domain, $f$ a bounded function
on $\partial G$, and $u$ the corresponding Perron function.  If $z_0$
is a point in $\partial G$ possessing a barrier function and $f$ is
continuous at $z_0$, then $\lim_{z\to z_0}u(z)=f(z_0)$.
\endproclaim
\demo{Proof} First we show that for any $\epsilon>0$, there is a
harmonic function $w$ on $G$, continuous on $\bar G$, which satisfies
$w>f$ on $\partial G$ and $w(z_0)= f(z_0)+\epsilon$.  Indeed,
let $D$ be a disk about $z_0$ such that $|f(z)-f(z_0)|<\epsilon$ for
$z\in\partial G\cap D$.  Let $M$ denote the maximum of $|f|$ on $\partial
G$ and $m$ denote to the minimum of the barrier function $\phi$ on
$G\setminus D$, so $m>0$.  Set
$$
w(z)=\frac{\phi(z)}{m}[M-f(z_0)] + f(z_0) +\epsilon =
\left[\frac{\phi(z)}{m}-1\right][M-f(z_0)]+M+\epsilon.
$$
Then for $z\in\partial G\cap D$,
$$
w(z)\ge f(z_0)+\epsilon > f(z),
$$
while for $z\in\partial G\setminus D$,
$$
w(z)> M\ge f(z).
$$
So $w> f$ on the boundary as claimed, and obviously $w(z_0)=f(z_0)+\epsilon$.

It follows that if $v\in\Cal P(f,G)$, then $v\le w$ on $G$.  Since this
is true of all such $v$, we have $u\le w$ on $G$, and so
$\limsup_{z\to z_0} u(z)\le w(z_0)=f(z_0)+\epsilon$.  Since $\epsilon$
was arbitrary, $\limsup_{z\to z_0} u(z)\le f(z_0)$.

Next let
$$
v(z)=f(z_0)-\epsilon -\frac{\phi(z)}{m}[M+f(z_0)]
=\left[1-\frac{\phi(z)}{m}\right][M+f(z_0)]-M-\epsilon.
$$
Now we get $v< f$ on $\partial G$ and $v(z_0)=f(z_0)-\epsilon$.
Since $v\in \Cal P(f,G)$, $u\ge v$ on $G$, and so $\liminf_{z\to z_0} u(z)\ge
f(z_0)-\epsilon$.  Letting $\epsilon$ tend to zero we have
$\liminf_{z\to z_0} u(z)\ge f(z_0)$.
\qed
\enddemo

\proclaim{Corollary} Let $G$ be a bounded region in $\C$ which possesses
a barrier at each point of its boundary, and let $f$ be a continuous
function on $\partial G$.  Then the Perron function for $f$ on $G$ solves
the Dirichlet problem.
\endproclaim

\goodbreak\subhead The Schwarz Reflection Principle\endsubhead\nobreak
To state the reflection principles, we introduce the following
terminology:
a region $G$ is symmetric with respect to the real axis if $z\in G$
implies $\bar z\in G$.

\proclaim{Reflection Principle for Harmonic Functions} Let $G$ be a
region which is symmetric with respect to the real axis and define
$G_+$, $G_-$, and $G_0$ as the intersection of $G$ with the upper
half-plane, lower half-plane, and real axis, respectively.  If $u$ is
a continuous real-valued function on $G_+\cup G_0$, which is harmonic
on $G_+$ and zero on $G_0$, then $u$ admits a unique extension to a
harmonic function on all of $G$.  The extension is given by $u(z)=-u(\bar z)$
for $z\in G_-$.
\endproclaim
\demo{Proof}  If such an extension exists it is certainly unique, so
it suffices to show that the stated extension defines a harmonic function
in $G$.  It certainly defines a continuous extension which is harmonic
in $G_+\cup G_-$, so it suffices to show that it is harmonic in a neighborhood
of $z_0\in G_0$.  By translating and extending we can assume that $z_0=0$
and that $G$ contains the closed unit disk, $\bar D$.  Define $U$ on $\bar D$
as the solution of the Dirichlet problem with boundary data $u|_{\partial D}$
(where $u$ has been extended to $G_-$ by the formula above).  From
the Poisson kernel representation we have $U=0$ on the real interval
$(-1,1)$.  Thus $U=u$ on the entire boundary of the upper half-disk,
and the entire boundary of the lower half-disk.  Since both $U$ and $u$
are harmonic in the half-disks, they coincide, and thus $u$, like $U$
is harmonic in the whole disk.\qed
\enddemo

\proclaim{Reflection Principle for Analytic Functions}  Let $G$,
$G_+$, $G_-$, and $G_0$ be as above.  If $f$ is a continuous complex-valued
function on $G_+\cup G_0$, which is analytic on $G_+$ and real on $G_0$, then
$f$ admits
a unique extension to a harmonic function on all of $G$.  The extension
is given by $f(z)=\overline{f(\bar z)}$ for $z\in G_-$.
\endproclaim
\demo{Proof} We could base a proof on the previous result applied to $\Im f$
and harmonic conjugates, but it is also easy to verify this
directly using Morera's theorem.\qed
\enddemo

Of course the line of symmetry could be any line, not just the real
axis.  Thus if $G$ is symmetric with respect to any line, defined and
analytic on one side of the line, and real on the line, it can be
extended to be analytic on the whole domain.  The proof can be found
by translating and rotating the domain to the standard case.

We can also translate and rotate the image, so it is not necessary
that $f$ be real on the symmetry line, it is sufficient that it
map it into some other line.  An important application is to
analytic continuation.  For example, if $f$ maps a rectangle into
itself analytically and takes the edges into the edges, we can
apply the reflection principle repeatedly to extend $f$ to an
entire function.

The reflection principles can be applied for symmetries with respect
to circles as well as for lines.

\proclaim{Theorem}  Let $f$ be analytic on the unit disk, continuous
on its closure, and real on the boundary.  Then $f$ admits a unique
extension to the entire plane.  The extension is given by
$f(z)=\overline{f(1/\bar z)}$ for $|z|>1$.
\endproclaim
\demo{Proof}  Let $g(z)=f((1+iz)/(1-iz))$, so $g$ maps the upper half plane
into $\C$ and is real on the real axis.  Applying the reflection principle
to $g$ gives an extension of $f$ with
$$
f\left(\frac{1+iz}{1-iz}\right)=\overline{f\left(\frac{1+i\bar z}{1-i\bar z}\right)}
$$
for $z$ in the lower half plane, or, equivalently,
$$
f(w)=\overline{f(1/w)}
$$
for $w$ in the exterior of the disk.\qed
\enddemo

As an application of the reflection principle consider an analytic
function on the unit disk which extends continuously to the closure
taking the unit circle into itself.  We may extend this function to
$\C_\infty$ by $f(z)=1/\overline{f(1/\bar z)}$ for $|z|>1$.  (We obtain this
by applying the standard reflection principle to $g=\phi^{-1}\circ
f\circ \phi$ where $\phi(z)=(1+iz)/(1-iz)$.)  The extended function
is analytic everywhere except at the points symmetric to the zeros of
$f$, where it has a pole.  Infinity is either a removable singularity
or a pole according to whether $f$ is 0 at 0 or not.  Thus the extended
function is meromorphic on $\C^\infty$ with a finite number of poles.
It follows that $f$ is rational.


\end

