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\begin{document}
\section{Ma 681}
\vspace{1pt}
\section{Lecture 12}
Recall:
Theorem: Laurent Expansion \ Let $0\leq r0$ be given. Then $\exists $ $\delta >0$ such that
\vspace{1pt}
\begin{center}
$\left| \left( z-z_{0}\right) ^{m}f\left( z\right) -L\right| <\in \qquad $if
$0<\left| z-z_{0}\right| <\delta $
\vspace{1pt}
\end{center}
But $\left| \left| \left( z-z_{0}\right) ^{m}f\left( z\right) \right|
-\left| -L\right| \right| \leq \left| \left( z-z_{0}\right) ^{m}f\left(
z\right) -L\right| <\in $ since $\left| \left| z\right| -\left| w\right|
\right| \leq \left| z+w\right| $ and this implies that $\left| \left|
z\right| -\left| w\right| \right| =\left| \left| z\right| -\left| -w\right|
\right| \leq \left| z-w\right| .$ Thus
\vspace{1pt}
\begin{center}
$\left| \left( z-z_{0}\right) ^{m}f\left( z\right) \right| <\left| L\right|
+\in $ \qquad if $0<\left| z-z_{0}\right| <\delta $
\vspace{1pt}
\end{center}
Taking $\left| z-z_{0}\right| =\delta $ gives
\vspace{1pt}
\begin{center}
$\left| \left( z-z_{0}\right) ^{-n-1}f\left( z\right) \right| <\left( \left|
L\right| +\in \right) \left| \left( z-z_{0}\right) ^{-n-m-1}\right| =$ $%
\left( \left| L\right| +\in \right) \delta ^{-n-m-1}$
\vspace{1pt}
\end{center}
Hence
\vspace{1pt}
\begin{center}
$\left| c_{n}\right| \leq \dfrac{1}{2\pi }\doint_{\gamma }\left| \dfrac{%
f\left( z\right) }{\left( z-z_{0}\right) ^{n+1}}\right| dz\leq \dfrac{1}{%
2\pi }\left( \left| L\right| +\in \right) \delta ^{-n-m-1}\left( 2\pi \delta
\right) =\left( \left| L\right| +\in \right) \delta ^{-n-m}$
$\qquad \qquad \qquad \qquad \qquad $
\vspace{1pt}
\end{center}
where we have used the deformation theorem to replace $\gamma $ by the
circle $\left| z-z_{0}\right| =\delta .$ If $n<-m,$ then $0<-n-m$ and we can
make $\delta ^{-n-m}$ as small as we like by choosing $\delta $ sufficiently
small. Thus $c_{n}=0$ if $n<-m.$ Also,
\vspace{1pt}
\begin{center}
$\lim_{z\rightarrow z_{0}}\left( z-z_{0}\right) ^{m}f\left( z\right)
=c_{-m}=L\neq 0.$
\vspace{1pt}
\end{center}
Thus $f$ has a pole of order $m$ at $z_{0}.$
\vspace{1pt}
Example: Let $f(z)=\dfrac{1}{z+i}.$ Then $\lim_{z\rightarrow -i}\left(
z+i\right) f\left( z\right) =1\neq 0$ so this function has a simple pole at $%
-i.$
\vspace{1pt}
Remark: Poles are intimately related to zeroes of a function, because often
a quotient has a pole where the denominator vanishes. Recall that $z_{0}$ is
called a \textit{zero} of $f$ if $f\left( z_{0}\right) =0.$ If $f$ \ is
differentiable in an open disk about $z_{0},$ then the \textit{order} of the
zero in the index of the first nonzero coefficient in the Taylor series
expansion of $f$ about $z_{0}.$ Thus $f$ has a zero of order $m$ when it is
of the form
\vspace{1pt}
\begin{center}
$f\left( z\right) =\dsum_{n=m}^{\infty }c_{n}\left( z-z_{0}\right)
^{n}=c_{m}\left( z-z_{0}\right) ^{m}+c_{m+1}\left( z-z_{0}\right)
^{m+1}+\cdots $
\vspace{1pt}
\end{center}
$c_{m}\neq 0.$ From the Taylor formula for these coefficients it follows
that this is equivalent to the condition
\vspace{1pt}
\begin{center}
$f\left( z_{0}\right) =f^{\prime }\left( z_{0}\right) =\cdots
=f^{(m-1)}\left( z_{0}\right) =0,\qquad $but $f^{\left( m\right) }\left(
z_{0}\right) \neq 0$
\vspace{1pt}
\end{center}
Thus $f\left( z\right) =\left( z+3i\right) ^{3}$ has a third order zero at $%
-2i$ since $f\left( -2i\right) =f^{\prime }\left( -2i\right) =f^{\prime
\prime }\left( 2i\right) =0$ but $f^{\left( 3\right) }\left( 3i\right) \neq
0.$
\vspace{1pt}
Theorem: Let $f\left( z\right) =\frac{h\left( z\right) }{g\left( z\right) }$
where $h$ and $g$ are differentiable in an open disk about $z_{0}.$ Suppose $%
h\left( z_{0}\right) \neq 0$ and $g$ has a zero or order $m$ at $z_{0}.$
Then $f$ has a pole of order $m$ at $z_{0}.$
\vspace{1pt}
Theorem: \textit{Poles of Products \ }Let $f$ have a pole of order $m$ at $%
z_{0}$ and let $g$ have a pole of order $n$ at $z_{0}.$ Then $fg$ has a pole
of order $m+n$ at $z_{0}.$
\vspace{1pt}
Theorem: Let $g$ have a pole of order $m$ at $z_{0}$ and $f$ have a zero or
order $n