% vim: tw=50
% 10/02/2023 10AM

\begin{flashcard}[uniform-limit-of-holomorphic-is-holomorphic]
\begin{corollary*}[Uniform convergence of holomorphic functions]
    \cloze{Let $f_n : \mathcal{U} \to \CC$ holomorphic
    functions on a domain $\mathcal{U}$, and $f_n
    \to f$ uniformly on $\mathcal{U}$ (sufficient:
    uniform convergence on compact subsets of
    $\mathcal{U}$). Then $f$ is holomorphic on
    $\mathcal{U}$, and $f'(z) = \lim_{n \to
    \infty} f_n'(z)$.}
\end{corollary*}

\begin{proof}
    \cloze{
    $\mathcal{U}$ is a union of open disks, so it
    suffices to work with $D(z, \eps) \subset
    \mathcal{U}$. Given $\gamma$ closed curve in
    $D(z, \eps)$, $\int_\gamma f_n \to \int_\gamma
    f$ (A\&T), and $\int_\gamma f_n = 0$, so $\int_\gamma
    f = 0$. Since $f$ is continuous, Morera's
    theorem applies, so $f$ is holomorphic on
    $D(z, \eps)$.

    \myskip
    Recall Taylor expansion computation: for $0 <
    \rho < \eps$,
    \[ f^{(m)}(z) = \frac{m!}{2\pi i}
    \int_{\partial D(z, \rho)}
    \frac{f(\zeta)}{(\zeta - z)^{m + 1}} \dd \zeta \]
    So $f'(z) = \frac{1}{2\pi i} \int_{\partial
    D(z, \rho)} \frac{f(\zeta)}{(\zeta - z)^2} \dd
    \zeta$.
    \begin{align*}
        |f'(z) - f_n'(z)|
        &= \frac{1}{2\pi} \left| \int_{\partial
        D(z, \rho)} \frac{f(\zeta)}{(\zeta - z)^2}
        - \frac{f_n(\zeta)}{(\zeta - z)^2} \dd
        \zeta \right| \\
        &\le \rho \cdot \frac{1}{\rho^2} \cdot
        \sup_{\zeta \in \partial D(z, \rho)}
        |f(\zeta) - f_n(\zeta)| \\
        &\to 0
    \end{align*}
    as $n \to \infty$. So $f'(z) = \lim_{n \to
    \infty} f_n'(z)$.
}
\end{proof}
\end{flashcard}

\begin{remark*}
    $f$ need not be non-constant; for example,
    $f_n(z) = z^n$ on $D(0, r)$, $0 < r < 1$. Then
    $f_n \to 0$ uniformly.
\end{remark*}

\begin{corollary*}
    If $f : \mathcal{U} \to \CC$ is continuous on a
    domain $\mathcal{U} \setminus S$ for some
    finite set $S$, then $f$ is holomorphic on
    $\mathcal{U}$.
\end{corollary*}

\begin{proof}
    If $a \in S$, find $D(a, r) \subset
    \mathcal{U}$ open disk. Cauchy's theorem in a
    disk implies $\int_\gamma = 0$ for any closed
    curve $\gamma$ in $D(a, r)$. Morera's theorem
    implies $f$ is holomorphic on $D(a, r)$, at
    $a$. So $f$ is holomorphic on $\mathcal{U}$.
\end{proof}

\subsubsection*{zeroes of holomorphic maps}

Let $f : D(a, R) \to \CC$ be holomorphic, so $f(z)
= \sum_{n \ge 0} c_n(z - a)^n$ on $D(a, R)$. If $f
\not \equiv 0$ then some $c_n$ is non-zero; let
\[ m \defeq \min\{n \in \NN \cup \{0\} : c_n \neq
0\} .\]
If $m > 0$ then we say $f$ has a \emph{zero of
order $m$ at $a$}. In this case, we can write
\[ f(z) = (z - a)^m g(z) \]
where $g(z)$ is holomorphic on $D(a, R)$, $g(a)
\neq 0$.

\begin{flashcard}[principle-of-isolated-zeroes]
\begin{theorem*}[Principle of Isolated Zeroes]
\cloze{
    If $f : D(a, R) \to \CC$ is holomorphic,
    \fcemph{not
    identically} $0$, then there exists $0 < r \le
    R$ such that $f(z) \neq 0$ on $0 < |z - a| <
    r$.
}
\end{theorem*}

\begin{proof}
\cloze{
    If $f(a) \neq 0$ then $f(z) \neq 0$ on $D(a,
    r)$ for some $0 < r \le R$ by continuity of
    $f$. If $f$ has a zero of order $m$ at $a$,
    write $f(z) = (z - a)^m g(z)$ where $g(a) \neq
    0$, $g$ holomorphic. By continuity of $g$,
    there exists $0 < r \le R$ such that $g(z)
    \neq 0$ for all $z \in D(a, r)$, so $f(z) \neq
    0$ for all $0 < |z - a| < r$.
}
\end{proof}
\end{flashcard}

\begin{remark*}
    Principle of isolated zeroes says that there is
    no accumulation point of the zero set of a
    holomorphic map inside its domain, unless
    $\equiv 0$.
\end{remark*}

\begin{remark*}
    It \emph{is} possible for the zeroes of a
    holomorphic map to accumulate ouside its
    domain:
    \[ \sin z \defeq \frac{e^{iz} - e^{-iz}}{2i} =
    0 \iff e^{-z} = e^{-iz} \]
    i.e. $e^{2iz} = 1$, which holds for all $z =
    n\pi$, $n \in \ZZ$. So $\sin \left(
    \frac{1}{z} \right)$ has zeroes accumulating
    at $0$, on the boundary of its domain $\CC^* =
    \CC \setminus \{0\}$.
\end{remark*}

\begin{remark*}
    Another application of Principle of Isolated
    Zeroes: since $\cos^2 z + \sin^2 z = 1$ holds
    for all $z \in \RR$, then $\cos^2 z + \sin^2 z
    - 1$ is entire with $\RR \subset \{\text{zero
    set}\}$. So by PIZ, $\cos^2 z + \sin^2 z = 1$
    for all $z \in \CC$.
\end{remark*}

\begin{flashcard}[identity-theorem-for-holomorphic-functions]
\begin{proposition*}[Identity theorem for holomorphic functions]
    \cloze{
    Let $f, g : \mathcal{U} \to \CC$ be
    holomorphic on a domain $\mathcal{U}$. Let $S
    \defeq \{z \in \mathcal{U} : f(z) = g(z)\}$.
    If $S$ has a \fcemph{non-isolated} point
    \fcemph{(i.e. there
    exists $w \in S$ such that for all $\eps > 0$,
    $D(w, \eps) \setminus \{w\} \cap S \neq
    \emptyset$)} then $f(z) = g(z)$ for all $z \in
    \mathcal{U}$.
}
\end{proposition*}

\begin{proof}
\cloze{
    Define $h(z) = f(z) - g(z)$, holomorphic on
    $\mathcal{U}$, and suppose $w$ is non-isolated
    in $S$. Then for $\eps > 0$ with $D(w, \eps)
    \subseteq \mathcal{U}$, PIZ implies $h \equiv
    0$ on $D(w, \eps)$.

    \myskip
    Given $z \in \mathcal{U}$, let $\gamma : [0,
    1] \to \mathcal{U}$ be a path with $\gamma(0)
    = w$, $\gamma(1) = z$. Consider the set
    \[ T \defeq \{t \in [0, 1] :
    h^{(n)}(\gamma(t)) = 0 ~\forall n \ge 0\} \]
    Note that $T$ is closed by definition. Since
    $h \equiv 0$ on $D(w, \eps)$, Taylor expansion
    implies $T$ is non-empty, since $0 \in T$.
    Define $t_0 \defeq \sup\{t' \in [0, 1] : t \in
    T ~\forall t \le t'\}$. Then $T$ closed and
    non-empty so $t_0 \in T$. Since
    $h^{(n)}(\gamma(t_0)) = 0$ for all $n \ge 0$,
    $h \equiv 0$ on a neighbourhood of
    $\gamma(t_0)$, contradicting the maximality of
    $t_0$, unless $t_0 = 1$. So $h(\gamma(1)) =
    0$, i.e. $h(z) = 0$ as claimed.
}
\end{proof}
\end{flashcard}

\begin{flashcard}[analytic-continuation]
\begin{definition*}[Analytic Continuation]
\cloze{
    Let $\mathcal{U} \subseteq V \subseteq \CC$ be
    domains and $f : \mathcal{U} \to \CC$ is
    holomorphic. $g : V \to \CC$ is an
    \emph{analytic continuation} of $f$ if:
    \begin{enumerate}[(1)]
        \item $g$ is holomorphic on $V$
        \item $g|_{\mathcal{U}} = f$.
    \end{enumerate}
}
\end{definition*}
\end{flashcard}

\begin{example*}
    \begin{enumerate}[(1)]
        \item The series $\sum_{n \ge 1}
            \frac{(-1)^{n + 1}}{n} z^n$ converges
            on $D(0, 1)$, and takes the value
            $\Log(1 + z)$ on $D(0, 1)$. So $\Log(1
            + z)$ is an analytic continuation of
            this series to the domain $\CC
            \setminus (-\infty, -1]$.
        \item $\sum_{n \ge 0} z^n$ has radius of
            convergence 1 about $a = 0$, and on
            $D(0, 1)$, we have $\frac{1}{1 - z}
            \sum_{n \ge 0} z^n$. So $\frac{1}{1 -
            z}$ is an analytic continuation to
            $\CC \setminus \{1\}$.
            \begin{center}
                \includegraphics[width=0.6\linewidth]
                {images/7daec5fca93211ed.png}
            \end{center}
        \item Considering $f(z) = \sum_{n \ge 0}
            z^{2^n}$, $f$ converges on $D(0, 1)$
            and cannot be analytically continued
            to any larger domain. We say $\partial
            D(0, 1)$ is \emph{natural boundary}
            for $f(z)$.
    \end{enumerate}
\end{example*}