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% 20/02/2023 10AM

\begin{flashcard}[laurent-expansion]
\begin{theorem*}[Laurent expansion]
    \cloze{
    Let $f$ be holomorphic on an annulus $A = \{z
    \in \CC : r < |z - a| < R\}$, where $0 \le r <
    R \le \infty$. Then:
    \begin{enumerate}[(i)]
        \item $f$ has a (unique) convergent
            expansion on $A$:
            \[ f(z) = \sum_{n = -\infty}^\infty
            c_n(z - a)^n \]
            ``Laurent series''
        \item For any $r < \rho < R$, we have
            \[ c_n = \frac{1}{2\pi i}
            \int_{\partial D(a, \rho)}
            \frac{f(z)}{(z - a)^{n + 1}} \dd z \]
        \item If $r < \rho' \le \rho < R$, the
            Laurent series converges uniformly on
            $\{z \in \CC : \rho' \le |z - a| \le
            \rho\}$.
    \end{enumerate}
}
\end{theorem*}
\end{flashcard}

\begin{flashcard}[laurent-expansion-proof]
\prompt{Laurent expansion proof? \\}
\begin{proof}
    \cloze{
    Fix $w \in A$, and choose $r < \rho_1 < |w -
    a| < \rho_2 < R$. Define two closed curves
    $\gamma_1, \gamma_2$ by a diameter of the
    annulus, labelled such that $I(\gamma_1; w) =
    1$, $I(\gamma_2; w) = 0$.
    \fcscrap{
    \begin{center}
        \includegraphics[width=0.6\linewidth]
        {images/db860fdab10911ed.png}
    \end{center}
    }
    $\gamma_1, \gamma_2$ are both homologous to
    zero in $A$, so by the generalised CIF we have
    \[ f(w) = \frac{1}{2\pi i} \int_{\gamma_1}
    \frac{f(z)}{z - w} \dd z = \frac{1}{2\pi i}
    \int_{\gamma_1 + \gamma_2} \frac{f(z)}{z - w}
    \dd z \]
    Travelling $\gamma_1 + \gamma_2$ is the same
    as travelling $\partial D(a, \rho_2) -
    \partial D(a, \rho_1)$. So
    \[ f(w) = \ub{\frac{1}{2\pi i} \int_{|z - a| =
    \rho_1} \frac{f(z)}{z - w} \dd z}_{I_2} -
    \ub{\frac{1}{2\pi i} \int_{|z - a| = \rho_1}
    \frac{f(z)}{z - w} \dd z}_{I_1} \]
    Using the same geometric series for
    $\frac{1}{1 - \frac{w - a}{z - a}}$ to
    compute $I_2$ as in the Taylor series case
    gives
    \[ I_2 = \sum_{n = 0}^\infty c_n (w - a)^n \]
    where
    \[ c_n = \frac{1}{2\pi i} \int_{|z - a| =
    \rho_2} \frac{f(z)}{(z - a)^{n + 1}} \dd z \]
    for $n \ge 0$. For $I_1$, using the expansion
    (since $|z - a| < |w - a|$)
    \[ -\frac{1}{z - w} = \frac{\frac{1}{w -
    a}}{1 - \frac{z - a}{w - a}} = \sum_{m =
    1}^\infty \frac{(z - a)^{m - 1}}{(w -a)^m} ,\]
    gives
    \[ I_1 = \sum_{m = 1}^\infty d_m (w - a)^{-m} \]
    where
    \[ d_m = \frac{1}{2\pi i} \int_{|z - a| =
    \rho_1} \frac{f(z)}{(z - a)^{-m + 1}} \dd z
    \qquad \forall m \ge 1 \]
    Reindex with $n = -m$, we obtain the Laurent
    expansion for $f$.

    \myskip
    To show (ii), (iii), suppose $f(z) = \sum_{n =
    -\infty}^\infty c_n(z - a)^n$ on $A$, and let
    $r < \rho' \le \rho < R$. The non-negative
    power series $\sum_{n = 0}^\infty c_n(z -
    a)^n$ has Radius of Convergence $\ge R$, so
    converges uniformly on $D(a, \rho)$.
    Similarly, if $u = \frac{1}{z - a}$ the
    negative part of the Laurent expansion,
    $\sum_{n = 1}^\infty c_{-n} u^n$ has Radius of
    Convergence $\ge \frac{1}{r}$, so converges
    uniformly on $\rho' \le |z - a| \le \rho$, so
    we can integrate term by term:
    \begin{align*}
        \frac{1}{2\pi i} \int_{\partial D(a,
        \rho)} \frac{f(z)}{(z - a)^{n + 1}} \dd z
        &= \frac{1}{2\pi i} \sum_{n =
        -\infty}^\infty c_n \int_{\partial D(a,
        \rho)} (z - a)^{n - m - 1} \dd z \\
        &= c_m
    \end{align*}
    since this integral $=0$ unless $n - m - 1 =
    -1$, in which case it is $2\pi i$.
}
\end{proof}
\end{flashcard}

\begin{remark*}
    Proof shows $f = f_1 + f_2$, $f_1$ holomorphic
    on $D(a, R)$, and $f_2$ holomorphic on $|z -
    a| > r$. Applying when $r = 0$, we have three
    possibilities on a punctured disk domain i.e.
    an isolated singularity at $z = a$.
    \begin{enumerate}[(1)]
        \item $c_n = 0$ $\forall n < 0$. Then $f$
            is the restriction to $D(a, R)^\times$
            of a function holomorphic on $D(a,
            R)$. We say $f$ has a \emph{removable
            singularity} at $a$. For example,
            $f(z) = \frac{\sin z}{z}$ at $a = 0$.
        \item $\exists k < 0$ such that $c_k \neq
            0$ but $c_n = 0$ for all $n < k$> We
            have $(z - a)^{-k} f(z)$ holomorphic
            and non-zero at $a$. We say $f$ has a
            \emph{pole of order $-k$ at $a$}. For
            example, $g(z) = \frac{1}{z^6}$ at $a
            = 0$.
        \item $c_n \neq 0$ for infinitely many $n
            < 0$. $f$ has an essential singularity
            at $a$. For example, $e^{\frac{1}{z}}$
            at $a = 0$.
    \end{enumerate}
\end{remark*}

\begin{flashcard}[removable-singularity-iff]
\begin{proposition*}
    An isolated singularity at $z = a$ for $f$ is
    removable if and only if \cloze{$\lim_{z \to a} (z -
    a) f(z) = 0$.}
\end{proposition*}

\begin{proof}
    \cloze{
    Forwards direction is clear. For backwards
    direction, consider
    \[ g(z) = \begin{cases}
        (z - a)^2 f(z) & z \neq a \\
        0 & z = a
    \end{cases} \]
    $g'(a) = \lim_{z \to a} (z - a) f(z) = 0$, so
    $g$ is holomorphic at $a$, with $g(a) = 0$. So
    $g(z) = \sum_{n = 2}^\infty c_n(z - a)^n$. So
    $f(z) = \sum_{n = 0}^\infty c_{n + 2} (z -
    a)^n$, so is holomorphic at $a$.
}
\end{proof}
\end{flashcard}

\begin{flashcard}[pole-iff]
\begin{proposition*}
    An isolated singularity at $z = a$ for $f$ is
    a pole $\iff$ \cloze{$|f(z)| \to \infty$ as $z \to
    a$.} \\
    Furthermore, the following are equivalent:
    \begin{enumerate}[(1)]
        \item $f$ has a pole of order $k$ at $z =
            a$.
        \item \cloze{$f(z) = (z - a)^{-k} g(z)$, where $g$
            is holomorphic and nonzero at $a$.}
        \item \cloze{$f(z) = \frac{1}{h(z)}$ where $h$ is
            holomorphic at $a$ with a zero of
            order $k$ at $a$.}
    \end{enumerate}
\end{proposition*}
\end{flashcard}

\begin{proof}
    (1) $\iff$ (2) is immediate using Laurent
    expansion. (2) $\iff$ (3) since $g$ is
    holomorphic, nonzero at $a$ $\iff$
    $\frac{1}{g}$ is holomorphic at $a$.

    \myskip
    If $f$ has a pole of order $k$ at $z = a$,
    then $f(z) = (z - a)^{-k} g(z)$, so $|f(z)| \to
    \infty$ as $z \to a$. Convsersely if $|f| \to
    \infty$ as $z \to a$, then there exists $r >
    0$ such that $f(z) \neq 0$ for all $0 < |z -
    a| < r$. So $\frac{1}{f}$ is holomorphic on
    $D(a, r)^\times$, and $\left| \frac{1}{f}
    \right| \to 0$ as $z \to a$, so the
    singularity for $\frac{1}{f}$ is removable,
    and $\frac{1}{f(z)} = h(z)$, holomorphic $h$
    on $D(a, r)$. $h$ has a zero of order $k$, so
    $h(z) = (z - a)^k l(z)$ for $l$ holomorphic
    and nonzero at $a$, so $f(z) = (z - a)^{-k}
    g(z)$, i.e. $f$ has a pole of order $k$. at $z
    = a$.
\end{proof}

\begin{flashcard}[essential-singularity-iff]
\begin{corollary*}
    An isolated singularity at $z = a$ is
    essential $\iff$ \cloze{$|f|$ does not approach a
    limit in $\RR \cup \{\infty\}$ as $z \to a$.}
\end{corollary*}
\end{flashcard}
