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\begin{flashcard}[general-CIF-proof]
\prompt{General CIF proof? \\}
\begin{proof}
    \cloze{
    Notice applying the first equality to $g(z) =
    f(z)(z - w)$ gives $\int_\gamma f = 0$. So
    suffices to prove the first statement. We have
    by previous lemma that
    \[ I(\gamma, w) f(w) = \frac{1}{2\pi i}
    \int_\gamma \frac{f(w)}{z - w} \dd z \]
    so we want to show that
    \[ \frac{1}{2\pi i} \int_\gamma \frac{f(z) -
    f(w)}{z - w} \dd z = 0 \forall w \in U
    \setminus \gamma \]
    Consider the function
    \[ g(z, w) = \begin{cases}
        \frac{f(z) - f(w)}{z - w} & z \neq w \\
        f'(w) & z = w
    \end{cases} \]
    This is a continuous function on $\mathcal{U}
    \times \mathcal{U}$. Want to show that
    \[ \int_\gamma g(z, w) \dd z = 0 \forall w \in
    \mathcal{U} \setminus \gamma \]
    Consider the auxiliary function $h$ on $\CC$:
    \[ h(w) \defeq \begin{cases}
        \int_\gamma g(\zeta, w) \dd \zeta & w \in
        \mathcal{U} \\
        \int_\gamma \frac{f(\zeta)}{\zeta - w} \dd
        \zeta & \ub{\{w \in \CC \setminus \gamma :
        I(\gamma, w) = 0\}}_{\eqdef V}
    \end{cases} \]
    If $w \in \mathcal{U} \cap V$, then
    \[ \int_\gamma g(\zeta, w) \dd \zeta =
    \int_\gamma \frac{f(\zeta) - f(w)}{\zeta - w}
    \dd \zeta = \int_\gamma \frac{f(\zeta)}{\zeta
    - w} \dd \zeta \]
    so $h$ is well-defined. For any disk $D(0, R)$
    with $\gamma \subseteq D(0, R)$, we have
    $I(\gamma; w) = 0$ for all $w \not\in D(0,
    R)$. In fact, $\gamma$ homologous to zero in
    $\mathcal{U}$, so $\mathcal{U} \cup V = \CC$.
    For $w \not\in D(0, R)$, we have
    \[ |h(w)| = \left| \int_\gamma
    \frac{f(\zeta)}{\zeta - w} \dd \zeta \right|
    \le \frac{\length(\gamma) \cdot \sup_{\zeta \in
    \gamma} |f(\zeta)|}{|w| - R} \to 0 \]
    as $|w| \to \infty$.

    \myskip
    Claim: $h$ is holomorphic on $\CC$. If so, $h$
    is bounded since $|h(w)| \to 0$ as $|w| \to
    \infty$. Then by Liouville's is constant,
    taking the value $0$ on $\CC$, concluding the
    proof.\prompt{ (See separate sublemma).}

\fcscrap{
\begin{lemma*}
    Let $\mathcal{U} \subseteq \CC$ be open and
    $\phi : \mathcal{U} \times [a, b] \to \CC$
    continuous with $z \mapsto \phi(z, s)$
    holomorphic on $\mathcal{U}$ for every $s \in
    [a, b]$. Then
    \[ g(z) \defeq \int_a^b \phi(z, s) \dd s \]
    is holomorphic on $\mathcal{U}$.
\end{lemma*}

\begin{proof}
    Idea: Morera. WLOG, $\mathcal{U}$ is a disk.
    For any closed curve $\gamma : [0, 1] \to
    \mathcal{U}$ we have
    \begin{align*}
        \int_\gamma g(z) \dd z
        &= \int_0^1 \left[ \int_a^b
        \phi(\gamma(t), s) \dd s \right]
        \gamma'(t) \dd t \\
        {}^*&= \int_a^b  \left[ \int_0^1
        \phi(\gamma(t), s) \gamma'(t) \dd t
    \right] \dd s
    \end{align*}
    $*$ is Fubini's theorem: Suppose $f : [a, b]
    \times [c, d] \to \CC$ is a continuous
    function. Then we have
    \[ \int_a^b \left( \int_c^d f(x,y) \dd y
    \right) \dd x = \int_c^d \left( \int_a^b f(x,
    y) \dd x \right) \dd y \]
    and $x \mapsto \int_c^d f(x, y) \dd y$ and $y
    \mapsto \int_a^b f(x, y) \dd x$ are
    continuous. This clearly holds if $f$ is
    constant, so also when $f$ is a step function.
    Since $[a, b] \times [c, d]$ is closed and
    bounded, $f$ is uniformly continuous. So $f$
    is a uniform limit of step functions, and so
    we exchange the order as claimed. End of proof
    of $*$.
    \[ \int_\gamma g(z) \dd z = \int_a^b \left(
    \int_\gamma \phi(z, s) \dd z \right) \dd s \]
    Since $z \mapsto \phi(z, s)$ is holomorphic,
    this is $0$ by Cauchy in a disk. So
    \[ \int_\gamma g = 0 \]
    and by Morera, $g$ is holomorphic as claimed.
\end{proof}
}
    So $h$ is holomorphic as claimed and the
    generalised CIF follows.
}
\end{proof}
\end{flashcard}

\begin{hiddenflashcard}[general-CIF-sublemma]
\begin{lemma*}[General CIF sublemma]
    \cloze{
    Let $\mathcal{U} \subseteq \CC$ be open and
    $\phi : \mathcal{U} \times [a, b] \to \CC$
    continuous with $z \mapsto \phi(z, s)$
    holomorphic on $\mathcal{U}$ for every $s \in
    [a, b]$. Then
    \[ g(z) \defeq \int_a^b \phi(z, s) \dd s \]
    is holomorphic on $\mathcal{U}$.
}
\end{lemma*}

\begin{proof}
    \cloze{
    Idea: Morera. WLOG, $\mathcal{U}$ is a disk.
    For any closed curve $\gamma : [0, 1] \to
    \mathcal{U}$ we have
    \begin{align*}
        \int_\gamma g(z) \dd z
        &= \int_0^1 \left[ \int_a^b
        \phi(\gamma(t), s) \dd s \right]
        \gamma'(t) \dd t \\
        {}^*&= \int_a^b  \left[ \int_0^1
        \phi(\gamma(t), s) \gamma'(t) \dd t
    \right] \dd s
    \end{align*}
    $*$ is Fubini's theorem: Suppose $f : [a, b]
    \times [c, d] \to \CC$ is a continuous
    function. Then we have
    \[ \int_a^b \left( \int_c^d f(x,y) \dd y
    \right) \dd x = \int_c^d \left( \int_a^b f(x,
    y) \dd x \right) \dd y \]
    and $x \mapsto \int_c^d f(x, y) \dd y$ and $y
    \mapsto \int_a^b f(x, y) \dd x$ are
    continuous. This clearly holds if $f$ is
    constant, so also when $f$ is a step function.
    Since $[a, b] \times [c, d]$ is closed and
    bounded, $f$ is uniformly continuous. So $f$
    is a uniform limit of step functions, and so
    we exchange the order as claimed. End of proof
    of $*$.
    \[ \int_\gamma g(z) \dd z = \int_a^b \left(
    \int_\gamma \phi(z, s) \dd z \right) \dd s \]
    Since $z \mapsto \phi(z, s)$ is holomorphic,
    this is $0$ by Cauchy in a disk. So
    \[ \int_\gamma g = 0 \]
    and by Morera, $g$ is holomorphic as claimed.
}
\end{proof}
\end{hiddenflashcard}

\begin{corollary*}[Cauchy's theorem for simply connected domains]
    Let $f : \mathcal{U} \to \CC$ be holomorphic
    on simply connected domain $\mathcal{U}$. Then
    $\forall$ closed curves $\gamma$ in
    $\mathcal{U}$,
    \[ \int_\gamma f = 0 .\]
\end{corollary*}

\noindent
Fact: If $\mathcal{U} \subseteq \CC$ is open, then
$\mathcal{U}$ is simply connected if and only if
the complement of $\mathcal{U}$ in $\CC_\infty$ Is
connected.

\subsubsection*{Examples}

\begin{enumerate}[(i)]
    \item $D(a, R) \subseteq \CC$, has disk
        complement in $\CC_\infty$ so simply
        connected.
        \begin{center}
            \includegraphics[width=0.2\linewidth]
            {images/c682511aaeb211ed.png}
        \end{center}
    \item Convex and starlike sets are simply
        connected.
    \item Annulus \emph{not} simply connected.
\end{enumerate}

\subsubsection*{Isolated singularities of holomorphic maps}

\begin{flashcard}[isolated-singularity]
\begin{definition*}[Isolated singularity]
    A point $a \in \CC$ is an \emph{isolated
    singularity} of $f : \mathcal{U} \to \CC$
    holomorphic if \cloze{$\exists r > 0$ such that $f$
    is holomorphic on $D(a, r) \setminus \{a\}$,
    denoted $D(a, r)^\times$.}
\end{definition*}
\end{flashcard}

\subsubsection*{Examples}

\begin{enumerate}[(i)]
    \item $a = 0$, $f(z) = \frac{\sin z}{z}$. Use
        the identity theorem or expansion of $e^z$
        to show that
        \[ \sin z = z - \frac{z^3}{3!} +
        \frac{z^5}{5!} - \frac{z^7}{7!} + \cdots \]
        about $0$. So
        \[ f(z) = 1 - \frac{z^2}{3!} +
        \frac{z^4}{5!} - \frac{z^6}{7!} + \cdots \]
        about $0$. So $f$ is restriction of a
        holomorphic function on $\CC$, call it
        $f$, and $f(0) = 1$.
    \item $a = 0$, $g(z) = \frac{1}{z^6}$, $g$
        holomorphic on $\CC^\times$, and $|g(z)|
        \to \infty$ as $|z| \to 0$, so there
        doesn't exist continuous extension at $0$.
    \item Recall the action $w \mapsto e^w =
        e^{\Re w} e^{i\Im w}$
        \begin{center}
            \includegraphics[width=0.6\linewidth]
            {images/f46edad0aeb211ed.png}
        \end{center}
        So $h(z) = e^{\frac{1}{z}}$ maps any
        $D(0,\eps)^\times$ to all of $\CC^\times$.
\end{enumerate}
