% vim: tw=50
% 30/01/2023 10AM

\begin{proof}
    Since $\Log$ is inverse to $e^z$ then using
    the chain rule, $\Log z$ is holomorphic with
    $\dfrac{}{z} \Log z = \frac{1}{z}$. We have
    \[ \dfrac{}{z} = \frac{1}{z + 1} = 1 - z + z^2
    - z^3 + z^4 + \cdots ,\]
    which is the derivative of $\sum_{n =
    1}^\infty \frac{(-1)^{n - 1} z^n}{n}$. So
    $\Log(1 + z)$ agrees with this series up to a
    constant. Since $\Log(1) = 0$ the equality
    holds.
\end{proof}

\myskip
If $\alpha \in \CC$, define $z^\alpha \defeq
\exp(\alpha \Log z)$ gives a definition of
$z^\alpha$ on $\CC \setminus \RR_{\le 0}$. Can
compute $\dfrac{}{z} z^\alpha = \alpha z^{\alpha -
1}$.

\begin{warning*}
    Not necessarily true that $z^\alpha w^\alpha =
    (zw)^\alpha$.
\end{warning*}

\begin{note*}
    Note that if $f(z) = z^\alpha$, then the image
    of $f$ can be ``much smaller'' than $\CC$.
    For example, $\alpha = \half$,
    \begin{align*}
        z^{\half}
        &= \exp \left( \half \Log z \right) \\
        &= \exp \left( \half \ln |z| + \half
        i\theta \right)
        &&\theta \in (-\pi, \pi)
    \end{align*}
    So:
    \begin{center}
        \includegraphics[width=0.6\linewidth]
        {images/d02d573ea08711ed.png}
    \end{center}
\end{note*}

\subsection{Contour Integration}

\begin{flashcard}[complex-integral]
If $f : [a, b] \to \CC$ is continuous \cloze{(so
$\Re f$, $\Im f$ are integrable)} we define
\[ \int_a^b f(t) \dd t \defeq \cloze{\int_a^b \Re(f(t))
\dd t + i \int_a^b \Im(f(t)) \dd t} \]
\end{flashcard}

\begin{flashcard}[integral-length-bound]
\begin{proposition*}
    Let $f : [a, b] \to \CC$ be continuous.
    Then
    \[ \cloze{\left| \int_a^b f(t) \dd t \right|}
    \le \cloze{(b - a) \sup_{a \le t \le b}
    |f(t)|} ,\]
    with equality if and only if \cloze{$f$ is
    constant}.
\end{proposition*}

\begin{proof}
    Write $M = \sup_{a \le t \le b} |f(t)|$\cloze{,
    $\theta = \arg \left( \int_a^b f(t) \dd t
    \right)$.
    \begin{align*}
        \left| \int_a^b f(t) \dd t \right|
        &= e^{-i\theta} \int_a^b f(t) \dd t \\
        &= \int_a^b e^{-i\theta} f(t) \dd t \\
        &= \int_a^b \Re(e^{-i\theta} f(t)) \dd t
        \\
        &\le \int_a^b |f(t) \dd t \\
        &\le M(b - a)
    \end{align*}
    If we have equality, then $|f(t)| \equiv M$,
    and $\arg f(t) \equiv \theta$, so $f$ is
    constant.}
\end{proof}
\end{flashcard}

\begin{definition*}
    Let $\gamma : [a, b] \to \CC$ be a
    $\mathcal{C}^1$-smooth curve. Then we define
    the \emph{arc length of $\gamma$} to be
    \[ \length(\gamma) \defeq \int_a^b
    |\gamma'(t)| \dd t .\]
    We say $\gamma$ is simple if $\gamma(t_1) =
    \gamma(t_2) \implies t_1 = t_2$ or $\{t_1,
    t_2\} = \{a, b\}$. If $\gamma$ is simple, then
    $\length(\gamma) = \text{length of the image
    of $\gamma$}$.
\end{definition*}

\begin{flashcard}[integral-along-curve]
\begin{definition*}
    Let $f : \mathcal{U} \to \CC$ be continuous,
    $\mathcal{U}$ open, and $\gamma : [a, b] \to
    \mathcal{U}$\cloze{ be a $\mathcal{C}^1$-smooth curve.
    Then the integral of $f$ along $\gamma$ is}
    \[ \int_\gamma f(z) \dd z \defeq \cloze{\int_a^b
    f(\gamma(t)) \gamma'(t) \dd t} \]
\end{definition*}
\end{flashcard}

\subsubsection*{Basic properties}

\begin{enumerate}[(1)]
    \item linearity:
        \[ \int_\gamma c_1 f_1 + c_2 f_2 \dd z =
        c_1 \int_\gamma f_1 \dd z + c_2
        \int_\gamma f_2 \]
    \item additivity: if $a < a' < b$ then
        \[ \int_{\gamma |_{[a, a']}} f(z) \dd z +
        \int_{\gamma |_{[a', b]}} f(z) \dd z =
        \int_\gamma f(z) \dd z \]
    \item inverse path: if $(-\gamma)(t) \defeq
        \gamma(-t) : [-b, -a] \to \mathcal{U}$,
        then
        \[ \int_{-\gamma} f(z) \dd z =
        -\int_\gamma f(z) \dd z \]
    \item independence of paramterisation: if
        $\phi : [a', b'] \to [a, b]$ is
        $\mathcal{C}^1$-smooth, $\phi(a') = a$,
        $\phi(b') = b$, then for $\delta = \gamma
        \circ \phi$ we have
        \[ \int_\delta f(z) \dd z = \int_\gamma
        f(z) \dd z \]
\end{enumerate}

\begin{note*}
    We can usually assume without loss of
    generality that $\gamma : [0, 1] \to
    \mathcal{U}$.
\end{note*}

\noindent
Common types of curves we work with:
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/db305426a08a11ed.png}
\end{center}
We can loosen the $\mathcal{C}^1$-smooth
restriction and allow $\gamma$ to be
\emph{piecewise-$\mathcal{C}^1$-smooth}: i.e. $a =
a_0 < a_1 < a_2 < \cdots < a_n = b$ such that
$\gamma_i \defeq \gamma |_{[a_{i - 1}, a_i]}$ is
$\mathcal{C}^1$-smooth. Define then
\[ \int_\gamma f(z) \dd z = \sum_{i = 1}^n
\int_{\gamma_i} f(z) \dd z \]
(which is well-defined by additivity).

\begin{remark*}
    Any piecewise-$\mathcal{C}^1$-smooth curve can
    be re parametrized to by $\mathcal{C}^1$: for
    such a $\gamma$ as above, replace $\gamma_i$
    by $\gamma_i \circ h_i$ where $h_i$ is
    monotonic $\mathcal{C}^1$-smooth bijection
    with endpoint derivatives $0$. So
    $\mathcal{C}^1$-smooth paths can have corners.
    For example,
    \[ \gamma(t) \defeq \begin{cases}
        1 + i\sin(\pi t) & t \in \left[ 0, \half
        \right] \\
        \sin(\pi t) + i & t \in \left[ \half,
        1 \right] 
    \end{cases} \]
\end{remark*}

\subsubsection*{Terminology}

\begin{itemize}
    \item ``curve'':
        piecewise-$\mathcal{C}^1$-smooth path.
    \item ``contour'': simple \emph{closed}
        (endpoints are equal)
        piecewise-$\mathcal{C}^1$-smooth path.
\end{itemize}

\begin{flashcard}[integral-bound-by-length]
\begin{proposition*}
    For any continuous $f : \mathcal{U} \to \CC$,
    $\mathcal{U}$ open, and any curve $\gamma :
    [a, b] \to \mathcal{U}$,
    \[ \cloze{\left| \int_\gamma f(z) \dd z \right| \le
    \length(\gamma) \sup_{z \in \gamma} |f(z)|} \]
\end{proposition*}
\end{flashcard}

\begin{proof}
    \begin{align*}
        \left| \int_\gamma f(z) \dd z \right|
        &= \left| \int_a^b f(\gamma(t)) \gamma'(t)
        \dd t\right| \\
        &\le \int_a^b |f(\gamma(t)) \gamma'(t))|
        \dd t
        &&(\text{by similar trick to previous
        proof}) \\
        &\le \sup_{z \in \gamma}|f(z)|
        \length(\gamma)
        && \qedhere
    \end{align*}
\end{proof}

\begin{flashcard}[function-integral-converging-on-curve]
\begin{proposition*}
    If $f_n : \mathcal{U} \to \CC$ for $n \in \NN$
    and $f : \mathcal{U} \to \CC$ are continuous,
    and $\gamma : [a, b] \to \mathcal{U}$ is a
    curve in $\mathcal{U}$ with $f_n \to f$
    uniformly on $\gamma$, then
    \[ \cloze{\int_\gamma f_n(z) \dd z \to \int_\gamma
    f(z) \dd z} \]
    as $n \to \infty$.
\end{proposition*}

\begin{proof}
\cloze{
    By uniform convergence $\sup_{z \in \gamma}
    |f(z) - f_n(z)| \to 0$ as $n \to \infty$.
    By previous proposition,
    \begin{align*}
        \left|\int_\gamma f(z) \dd z - \int_\gamma
        f_n(z) \dd z\right|
        &\le \length(\gamma)
        \sup_\gamma |f - f_n| \\
        &\to 0
    \end{align*}
    as $n \to \infty$.
}
\end{proof}
\end{flashcard}

\begin{example*}
    $f_n(z) = z^n$, $n \in \ZZ$, on $\CC^* \eqdef
    \mathcal{U}$, and $\gamma : [0, 2\pi] \to
    \mathcal{U}$, $\gamma(t) = e^{it}$.
    \begin{center}
        \includegraphics[width=0.2\linewidth]
        {images/3f64f008a08d11ed.png}
    \end{center}
    \begin{align*}
        \int_\gamma f_n(z) \dd z
        &= \int_0^{2\pi} e^{nit} ie^{it} \dd t \\
        &= i\int_0^{2\pi} e^{(n + 1)it} \dd t \\
        &= \begin{cases}
            2\pi i & n = -1 \\
            0 & n \neq -1
        \end{cases}
    \end{align*}
\end{example*}