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% 24/02/2023 10AM

\begin{flashcard}[casorati-weierstrass]
\begin{theorem*}[Casorati-Weierstrass]
    \cloze{
    $f : D(a, R)^\times \to \CC$ with essential
    singularity at $z = a$. Then $f$ has dense
    image on any neighbourhood of $a$; that is,
    $\forall w \in \CC$, $\forall \eps > 0$,
    $\forall \delta > 0$ then $\exists z \in D(a,
    \delta)^\times$ such that $|f(z) - w| < \eps$.}
\end{theorem*}
\end{flashcard}

\begin{proof}
    Example Sheet 2.
\end{proof}

\myskip
More difficult: ``great Picard theorem''. If $z =
a$ is an essential singularity of $f$, then
$\exists b \in \CC$ such that $\forall \eps > 0$,
$\CC \setminus \{b\} \subseteq f((D(a, \eps)^*)$.

\myskip
Exercise: $f(z) = e^z$ has an essential
singularity at $\infty$, and takes every non-zero
value on every neighbourhood of $\infty$.

\begin{remark*}
    An advantage of the Riemann sphere
    perspective: if $f : D(a, R)^* \to \CC$ has a
    pole at $z = a$, we can view $f$ as a
    continuous map $f : D(a, R) \to \CC_\infty$,
    with $f(a) = \infty$. $f$ is ``holomorphic at
    $a$'' in the $\CC_\infty$ sense since
    $\frac{1}{f}$ is holomorphic on a
    neighbourhood of $a$, with a zero of the same
    order as the pole of $f$.
\end{remark*}

\begin{flashcard}[meromorphic-function]
\begin{definition*}
    Suppose $D$ is a domain. A function $f$ is
    \emph{meromorphic} on $D$ if \cloze{$f : D \setminus
    S \to \CC$ is holomorphic, where $S$ is a set
    of \fcemph{isolated singularities} for $f$ which are
    removable or poles.}
\end{definition*}
\end{flashcard}

\begin{flashcard}[residue-of-function]
\begin{definition*}
    Let $f : D(a, R)^* \to \CC$ be holomorphic
    with Laurent expansion $f(z) = \sum_{n =
    -\infty}^\infty c_n (z - a)^n$. The
    \emph{residue} of $f$ at $z = a$ is
    \[ \cloze{\Res_{z = a} f(z) \defeq c_{-1} \in
    \CC} \]
\end{definition*}
\end{flashcard}

\begin{flashcard}[principal-part-of-function]
\begin{definition*}
    Let $f : D(a, R)^* \to \CC$ be holomorphic
    with Laurent expansion $f(z) = \sum_{n =
    -\infty}^\infty c_n(z - a)^n$. The
    \emph{principal part} of $f$ at $z = a$ is
    \[ \cloze{\sum_{n = -\infty}^{-1} c_n(z -
    a)^n} \]
\end{definition*}
\end{flashcard}

\begin{flashcard}[integral-formula-for-annulus]
\begin{proposition*}
    Let $\gamma$ be a closed curve in $D(a, R)^*$.
    Then
    \[ \int_\gamma f(z) \dd z = \cloze{2\pi i I(\gamma;
    a) \Res_{z = a} f(z)} \]
\end{proposition*}
\end{flashcard}

\begin{proof}
    Using uniform convergence of Laurent expansion
    of $f$, we have that:
    \[ \int_\gamma f(z) \dd z = \sum_{n =
    -\infty}^\infty c_n \left[ \int_\gamma (z -
    a)^n \dd z \right] \]
    Since
    \[ \int_\gamma (z - a)^n \dd z = \begin{cases}
        0 & n \neq -1 \\
        2\pi i I(\gamma; a) & n = -1
    \end{cases} \]
    the proposition is proved.
\end{proof}

\myskip
If $f$ is meromorphic on a domain $D$, and $z = a$
is a pole of $f$ in $D$, then its principal part
at $z = a$ is of the form
\[ \frac{c_{-k}}{(z - a)^k} + \frac{c_{-k + 1}}{(z
- a)^{k - 1}} + \cdots + \frac{c_{-1}}{z - a} \]
a polynomial in $\frac{1}{z - a}$, and can be
written as $\frac{p(z)}{(z - a)^k}$ for some
polynomial $p$. So the principal part of $f$ at $z
= a$ is holomorphic on $\CC \setminus \{a\}$.

\myskip
More generally, if $f$ is meromorphic on $D$, and
$\{a_1, \ldots, a_m\} \subseteq \{\text{poles of
$f$ in $D$}\}$, with $p_i(x)$ the principal part
of $f$ at $z = a_i$, then the function
\[ g(z) = f(z) - \sum_{i = 1}^m p_i(z) \]
is meromorphic on $D$, with removable
singularities at $a_1, \ldots, a_m$.

\begin{flashcard}[residue-thm]
\begin{theorem*}[Residue Theorem]
    \cloze{
    Let $f$ be \fcemph{meromorphic} on a domain $D$, and
    $\gamma$ a curve which is \fcemph{homologous
    to zero}
    in $D$. Suppose $\gamma$ \fcemph{does not contain any
    pole} of $f$, and $f$ has only \fcemph{finitely many
    poles} in $D$ with non-zero winding number for
    $\gamma$; call them $\{a_1, \ldots, a_m\}$.
    Then
    \[ \int_\gamma f(z) \dd z = 2\pi i \sum_{i =
    1}^m I(\gamma; a_i) \Res_{z = a_i} f(z) \]
}
\end{theorem*}
\end{flashcard}

\begin{proof}
    Let $P_i$ denote the principal part of $f$ at
    $z = a_i$, and write $g = f - \sum_{i =
    1}^\infty P_i$. Then by Cauchy's theorem,
    \[ \int_\gamma g = 0, \qquad \text{i.e.}
    \qquad \int_\gamma f = \sum_{i = 1}^m
    \int_\gamma P_i \]
    Each $P_i$ is holomorphic on $\CC \setminus
    \{a_i\}$ as we argued, so by the previous
    proposition we have
    \[ \int_\gamma P_i = 2\pi i I(\gamma; a_i)
    \Res_{z = a_i} P_i(z) \]
    By definition, $\Res_{z = a_i} P_i(z) =
    \Res_{z = a_i} f(z)$, so
    \[ \int_\gamma f = 2\pi i \sum_{i = 1}^\infty
    I(\gamma; a_i) \Res_{z = a_i} f(z) \qedhere \]
\end{proof}

\subsubsection*{Remarks}

\begin{enumerate}[(1)]
    \item[($*$)] If $\gamma$ is homologous to $0$
        in a domain $D$, then $\{z \in \CC :
        I(\gamma; z) \neq 0 \text{ or } z \in
        \gamma\}$ is a closed set and
        a bounded set. Notice that the winding
        number is a continuous function on $\CC
        \setminus \gamma$, taking values in a
        discrete set, then $\{z \in \CC \setminus
        \gamma : I(\gamma; z) = 0\}$ is open. So
        the complement is closed. Since the polws
        of $f$ are isolated, this closed bounded
        set contains only finitely many of them
        (Bolzano-Weierstrass).
    \item $f$ holomorphic on $D$: Residue theorem
        implies Cauchy's theorem.
    \item $f(z) = \frac{g(z)}{z - a}$. Then
        $\Res_{z = a} f(z) = g(a)$, so Residue
        theorem implies CIF.
    \item 
        \begin{flashcard}[bounds-a-domain]
        \prompt{What does it mean for a curve to
        bound a domain? \\}
        \cloze{
        We say a closed curve $\gamma$
        \emph{bounds a domain} $\mathcal{U}$ if
        \[ I(\gamma; z) = \begin{cases}
            1 & z \in \mathcal{U} \\
            0 & z \not\in \mathcal{U}
        \end{cases} \]
    }
        \end{flashcard}
        If $\gamma$ is a closed curve in a domain
        $D$ which bounds a domain $\mathcal{U}$,
        and $f$ is holomorphic on $D$, then
        $\int_\gamma f = 0$ and
        \[ \int_\gamma \frac{f(z)}{z - w} \dd z =
        2\pi i f(w) \qquad \forall w \in
        \mathcal{U} \setminus \gamma \]
        If $f$ is meromorphic on $D$ with no poles
        on $\gamma$, then
        \[ \int_\gamma f = 2\pi i \sum_{\text{$w$
        poles in $\mathcal{U}$}} \Res_{z = w} f(z) \]
\end{enumerate}