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\centerline{{\it  Class 29} {\bf (Monday April 6)}}
 
\begin{description}
\item[SUMMARY] Classifying Singularities and Representing Functions using Series
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\item[CURRENT READING] Brown \& Curchill pages 150-154
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\item[NEXT READING] Brown \& Curchill pages 138-156
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\centerline{\bf Cauchy's Second Residue Theorem}
If a function $f(z)$ is analytic everywhere in the finite plane except for a finite number of singular points interior to a positively oriented  simple closed contour $C$, then
$${\displaystyle \oint f(z) \ dz = 2 \pi i \mbox{\bf Res} \left[ \frac{1}{z^2} f\left( \frac{1}{z} \right) \mbox{\bf ; 0} \right] }$$
In other words, instead of finding the residues of all the singularities of the given function $f(z)$ which lie inside the given contour $C$, all you need to do is find the residue at a single point, $z=0$, of the associated function ${\displaystyle \frac{1}{z^2}f(\frac{1}{z})}$. Note what's really going on involves finding the residue of the function at the point at infinity.

{\bf Exercise}\\
Evaluate ${\displaystyle \oint_{|z|=2\pi} \tan(z) \ dz}$ using Cauchy's Second Residue Theorem.
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{\sc Groupwork }\\
Evaluate ${\displaystyle \oint_{|z|=2} \frac{3z+2}{z^2+1} \ dz}$ using Cauchy's Second Residue Theorem.
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\centerline{\bf Classifying Singularities}
There are basically three types of singularities (points where $f(z)$ is not analytic) in the complex plane. They are called {\bf removable singularities}, {\bf isolated singularities} and {\bf branch singularities}. \\

{\bf Isolated Singularity}\\
An isolated singularity of a function $f(z)$ is a point $z_0$ such that $f(z)$ is analytic on the punctured disc $0<|z-z_0|<r$ but is {\it undefined} at $z=z_0$. We usually call isolated singularities {\bf poles}. An example is $z=i$ for the function $z/(z-i)$.\\

{\bf Removable Singularity}\\
A removable singularity is a point $z_0$ where the function $f(z_0)$ appears to be undefined but if we assign $f(z_0)$ the value $w_0$ with the knowledge that
${\displaystyle \lim_{z \rightarrow z_0} f(z) = w_0}$ then we can say that we have ``removed'' the singularity. An example would be the point $z=0$ for $f(z) = \sin(z)/z$.\\

{\bf Branch Singularity}\\
A branch singularity is a point $z_0$ through which all possible branch cuts of a multi-valued function can be drawn to produce a single-valued function. An example of such a point would be the point $z=0$ for $\Log(z)$.\\

There is also a special kind of isolated singularity, called an {\bf essential singularity}. 
The canonical example of an essential singularity is $z=0$ for the function $f(z) = e^{1/z}$. The easiest way to define an essential singularity of a function involves Laurent Series (see below).

{\bf Laurent series}\\
In fact, the best way to identify an essential singularity $z_0$ of a function $f(z)$ (and an alternative way to compute residues) is to look at the {\bf series representation} of the function $f(z)$ about the point $z_0$

That is,
$$f(z) = \sum_{n=0}^{\infty} a_n (z-z_0)^n + \sum_{n=1}^{\infty} \frac{b_n}{(z-z_0)^n}, \qquad R_1 < |z-z_0| < R_2$$
This formula for a Laurent series is sometimes written as
$$f(z) = \sum_{n=-\infty}^{\infty} c_n (z-z_0)^n \qquad \mbox{ where } c_n = \frac{1}{2 \pi i} \oint_C \frac{f(z)}{(z-z_0)^{n+1}} dz, \quad n = \pm 1, \pm 2, \ldots$$

This first part of this series should look somewhat familiar from your experience with real functions, since the expression is clearly a {\bf Taylor series} if $b_n=0$ for all $n$. This first part of the series representation is known as the {\bf analytic part} of the function. The second part (with the {\it negative exponents}) is called the {\bf principal part} of the function. However if $a_n$ and $b_n$ are not all identically zero this type of series is called a {\bf Laurent series} and converges to the function $f(z)$ in the annular region $R_1 < |z-z_0| < R_2$. 

{\bf Exercise}\\
Let's show why expressing the function $f(z)$ in terms of a Laurent Series is useful by proving that the value of the {\bf Res}$(f;z_0)$ is exactly equal to $b_{-1}$ (or $c_{-1}$), that is, the coefficient of the ${\displaystyle \frac{1}{z-z_0}}$ term. We can do this by integrating the Laurent series term by term on some closed contour $C$ and using the CIF.

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%Our goal is to see how series representations of functions allows us to compute integrals of functions easily by computing residues, and other means. But first we need to update our knowledge of sequences and series of real variables to the same objects using complex variables.

\newpage
{\bf Review of Sequences and Series}\\
Recall that an infinite {\bf sequence} $\lbrace z_n \rbrace$ converges to $z$ if for each $\epsilon>0$ there exists an $N$ such that if $n > N$ then $|z_n-z|< \epsilon$

The  sequence $z_1, z_2, z_3, \ldots, z_n, \dots$ converges to the value $z=x+iy$ if and only if the sequence $x_1, x_2, x_3, \ldots$ converges to $x$ and $y_1, y_2, y_3, \ldots$ converges to $y$.

In other words 
${\displaystyle \lim_{n \rightarrow \infty} z_n = z \Leftrightarrow \lim_{n \rightarrow \infty} x_n = x  \mbox{ and }  \lim_{n \rightarrow \infty} y_n = y}$ \\

An infinite {\bf series} ${\displaystyle \sum_{n=1}^{\infty} z_n = z_1+z_2+z_3+ \cdots +z_n+ \cdots }$ converges to $S$ if the sequence $S_N$ of {\bf partial sums}  where 
$S_N = z_1 + z_2 + z_3 + z_4 + \cdots + z_N \qquad (N=1,2,3,\ldots)$
converges to $S.$ Then we say that ${\displaystyle \sum_{n=1}^{\infty} z_n = S}.$\\

As with sequences, series can be split up into real and imaginary parts.
Suppose $z_n = x_n + i y_n$ and ${\displaystyle \sum_{n=1}^{\infty} z_n = Z,  \sum_{n=1}^{\infty} x_n = X  \mbox{ and } Y = \sum_{n=1}^{\infty} y_n}$ then $Z = X + iY$.\\

{\bf Taylor series}
Suppose a function $f$ is analytic throughout an open disk $|z-z_0|< R_0$ centered at $z_0$ with radius $R_0.$ Then at each point $z$ in this disk $f(z)$ has the series representation $$f(z) = {\displaystyle \sum_{n=0}^{\infty} a_n (z-z_0)^n} \qquad \mbox{ where } a_n = \frac{f^{(n)}(z_0)}{n!} \mbox{ for } (n=0,1,2,\ldots)$$
In other words the function $f(z)$ can be represented exactly by the infinite series in the disk $|z-z_0|<R$

When $z_0 = 0$ the series is known as a {\bf Maclaurin series}.

Here are some examples of well known Maclaurin series you should know.
$$
\begin{array}{lcccl}
 e^z & = & {\displaystyle  1 +  z + \frac{z^2}{2!} +\frac{z^3}{3!} +\dots } & = & {\displaystyle \sum_{k=0}^\infty \frac{z^k}{k!} } \nonumber \qquad |z|<\infty  \\
\sin(z) & = & {\displaystyle  z-\frac{z^3}{3!}+\frac{z^5}{5!} -\dots }& = & {\displaystyle \sum_{k=0}^\infty \frac{(-1)^{k+1} z^{2k+1}}{(2k+1)!} } \nonumber \qquad |z|<\infty \\
\cos(z) & = & {\displaystyle  1-\frac{z^2}{2!}+\frac{z^4}{4!} -\dots } & = & {\displaystyle \sum_{k=0}^\infty \frac{(-1)^{k+1} z^{2k}}{(2k)!} } \qquad |z|<\infty \nonumber \\
\sinh(z) & = & {\displaystyle  z +\frac{z^3}{3!}+\frac{z^5}{5!} +\dots }& = & {\displaystyle \sum_{k=0}^\infty \frac{z^{2k+1}}{(2k+1)!} } \nonumber \qquad |z|<\infty \\
\cosh(z) & = & {\displaystyle  1 +\frac{z^2}{2!}+\frac{z^4}{4!} +\dots }& = & {\displaystyle \sum_{k=0}^\infty \frac{z^{2k}}{(2k)!} } \nonumber \qquad |z|<\infty \\
{\displaystyle \frac{1}{1-z}} & =&  1+z+z^2+z^3+\dots & = & {\displaystyle \sum_{k=0}^\infty z^k
 } \qquad |z|<1 \nonumber \\
 \ln(1+z) & = & {\displaystyle  z - \frac{z^2}{2}+\frac{z^3}{3} +\dots } & = & {
\displaystyle  \sum_{k=1}^\infty \frac{z^k}{k} } \qquad |z|<1  \nonumber \\
 (1+z)^p & = & {\displaystyle  1 + p z + \frac{p(p-1)z^2}{2!}+\dots \frac{p(p-1)\dots (p-n+1)}{n!} \dots } &  &  \qquad |z|<1 \nonumber \\
\tan(z) & = & {\displaystyle  z+\frac{z^3}{3}+\frac{2 z^5}{15} +\dots }&  &  \nonumber \qquad |z|< \frac{pi}{2}\\
\end{array}
$$
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\underline{\sc Groupwork}\\
{\bf 1.} Write down the MacLaurin series for $f(z) = e^{1/z}$.

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{\bf 2.} What is the value of Res$(e^{1/z},0)$?
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{\bf 3.} Evaluate ${\displaystyle \oint_{|z|=1} e^{1/z} dz}$ {\bf two different ways}.
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