% vim: tw=50
% 27/02/2023 10AM

\begin{remark*}[Jordan Curve Theorem]
    Every simple closed continuous curve in the
    plane separates $\CC$ into two connected
    components, one bounded, one unbounded.
\end{remark*}

\subsubsection*{Computing residues}

\begin{enumerate}[(i)]
    \item If $f$ has a simple ($=$ order $1$) pole
        at $z = a$, then the Laurent expansion at
        $a$ is
        \[ f(z) = \frac{c_{-1}}{z - a} + c_0 +
        c_1(z - a) + c_2(z - a)^2 + \cdots \]
        so
        \[ \Res_{z = a} (f(z)) = \lim_{z \to a} (z
        - a) f(z) \]
        \begin{example*}
            $f(z) = \frac{1}{1 + z^2}$ at $z = i$:
            $(z - i) f(z) = \frac{1}{z + i}$, so
            $\Res_{z = i} f(z) = \frac{1}{2i}$.
        \end{example*}
    \item If $f = \frac{g(z)}{h(z)}$, where $g$ is
        holomorphic and non-zero at $z = a$, and
        $h$ is holomorphic and has a simple zero
        at $z = a$:
        \[ g(z) = g(a) + (z - a) \tilde{g}(z) \]
        $\tilde{g}$ holomorphic at $z = a$,
        \[ h(z) = h'(a)(z - a) \tilde{h}(z) \]
        $\tilde{h}(a) = 1$ at $z = a$, and is
        holomorphic at $a$. So
        \[ \frac{g(z)}{h(z)} = \frac{g(a)}{h'(a)
        (z - a) \tilde{h}(z)} +
        \boxed{\frac{\tilde{g}(z)}{h'(z) \tilde{h}(z)}} \]
        (the boxed expression is holomorphic at
        $a$). Applying (i) to $\frac{g(a)}{h'(a)
        (z - a)\tilde{h}(z)}$, we see that
        \[ \Res_{z = a} f(z) = \frac{g(a)}{h'(a)} \]
        \begin{example*}
            $f(z) = \frac{e^z}{z^2 + 1}$ at $z =
            i$. (ii) implies
            \[ \Res_{z = i} f(z) \frac{e^i}{2i} \]
        \end{example*}
    \item If $f(z) = \frac{g(z)}{(z - a)^k}$, $g$
        holomorphic at $a$. Then $\Res_{z = a}
        f(z)$ is the coefficient of $(z - a)^{k -
        1}$ in the expansion of $g$, which is
        \[ \frac{f^{(k - 1)}(a)}{(k - 1)!} \]
\end{enumerate}
Let's explore applications to real integrals.

\begin{example*}
    Evaluate $\int_0^\infty \frac{1}{1 + x^4} \dd
    x$. Note:
    \begin{enumerate}[(1)]
        \item $\frac{1}{1 + x^4} = \frac{1}{1 +
            (-x)^4}$
        \item $|x| \gg 1 \implies \left|
            \frac{1}{1 + x^4} \ll 1 \right|$.
    \end{enumerate}
    Consider:
    \begin{center}
        \includegraphics[width=0.3\linewidth]
        {images/a6477072b68911ed.png}
    \end{center}
    $1 + x^4$ has 4 simple zeroes: $e^{\pi i/4}$,
    $e^{3\pi i/4}$, $e^{5\pi i/4}$ and $e^{7\pi
    i/4}$. $\gamma_R$ has winding number $1$
    around $e^{\pi i/4}$, $e^{3\pi i/4}$, and $0$
    around the others. $\Res_{z = e^{\pi i/4}}
    \frac{1}{1 + z^4} = \frac{1}{4e^{3\pi i/4}}$,
    and $\Res_{z = e^{3\pi i/4}} \frac{1}{1 + z^4}
    = \frac{1}{4e^{9\pi i/4}} = \frac{1}{4e^{\pi
    i/4}}$. (Computed using (ii), with $g(z)
    \equiv 1$, $h(z) = 1 + z^4$). We have:
    \[ \int_{\gamma_R} \frac{1}{z^4 + 4} \dd z =
    \int_{C_R'} \ub{\frac{1}{1 + z^4} \dd z}_{I_1}
    + \ub{\int_{-R}^R \frac{1}{1 + z^4} \dd z}_{I_2} \]
    For $I_1$, parametrise $z = Re^{i\theta}$,
    $\theta \in [0, \pi]$. Then
    \[ I_1 = \int_0^\pi \frac{1}{1 + R^4
    e^{4i\theta}} iRe^{i\theta} \dd \theta \]
    $|I_1| \le \frac{\pi R}{R^4 + 1} \to 0$ as $R
    \to \infty$. So
    \begin{align*}
        I_2
        &= \int_{\gamma_R} \frac{1}{1 + z^4} \dd z
        - \int_{C_R'} \frac{1}{1 + z^4} \dd z \\
        &= 2\pi i \left[ \frac{1}{4e^{3\pi i/4}} +
        \frac{1}{4e^{\pi i/4}} \right] -
        \int_{C_R'} \frac{1}{1 + z^4} \dd z \\
        &\to 2\pi i \left[ \frac{1}{4e^{3\pi i/4}} +
        \frac{1}{4e^{\pi i/4}} \right] - 0
    \end{align*}
    So
    \[ I_2 \to \half \pi i \left( e^{-3\pi i/4} +
    e^{-\pi i/4} \right) = \half \pi i \left(
    -\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} i +
    \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} i
    \right) = \frac{\pi}{\sqrt{2}} \]
    So $\int_0^\infty \frac{1}{1 + z^4} \dd z =
    \half \int_{-\infty}^\infty \frac{1}{1 + z^4}
    \dd z = \frac{\pi}{2\sqrt{2}}$.
\end{example*}

\begin{example*}
    Compute $\int_{-\infty}^\infty
    \frac{\cos(x)}{1 + x + x^2} \dd x$. Note
    \[ \cos z = \frac{e^{iz} + e^{-iz}}{2} =
    \frac{e^{ix - y} + e^{-ix + y}}{2} \]
    However, $e^{ix} = \cos x + i\sin x$, so the
    function $\frac{e^{ix}}{1 + x + x^2}$ has real
    part $\frac{\cos x}{1 + x + x^2}$ on $\RR$.
    Notice then $e^{i(x + iy)} = e^{ix - y}$, so
    this function is bounded above by $1$ in
    modulus for $y \ge 0$.
    \begin{center}
        \includegraphics[width=0.3\linewidth]
        {images/0bcbab4ab68d11ed.png}
    \end{center}
    Roots of $1 + x + x^2$ are $e^{2\pi i/3}$,
    $e^{4\pi i/3}$, $\gamma_R$ winds around
    $e^{2\pi i/3}$ with winding number $1$.
    \[ \int_{\gamma_R} \frac{e^{iz}}{1 + z + z^2}
    \dd z = \ub{\int_{C_R'} \frac{e^{iz}}{1 + z +
    z^2}}_{I_1} + \ub{\int_{-R}^R \frac{e^{iz}}{1 + z
    + z^2}}_{I_2} \]
    $|I_1| \le \length(C_R') = \frac{1}{R^2 - R -
    1} = \frac{\pi R}{R^2 - R - 1} \to 0$ as $R
    \to \infty$. We have
    \[ \Res_{z = e^{2\pi i/3}} \frac{e^{iz}}{1 + z
    + z^2} = \frac{e^{ie^{2\pi i/3}}}{1 + 2e^{2\pi
    i/3}} \]
    so $I_2 \to 2\pi i \left[ \frac{e^{ie^{2\pi
    i/3}}}{1 +2e^{2\pi i/3}} \right] - 0$ (as $R
    \to \infty$). $e^{2\pi i/3} = -\half +
    \frac{\sqrt{3}}{2} i$, so $1 + 2e^{2\pi i/3} =
    \sqrt{3}i$.
    \[ e^{ie^{2\pi i/3}} = e^{i\left( -\half +
    \frac{\sqrt{3}}{2}i \right)} = e^{-i/2}
    e^{-\sqrt{3}/2} \]
    so $I_2 \to 2\pi i \left[ \frac{e^{-i/2}
    e^{-\sqrt{3}/2}}{\sqrt{3} i} \right] =
    \frac{2\pi}{\sqrt{3}}e^{-\sqrt{3}/2}
    e^{-i/2}$. So
    \begin{align*}
        \int_{-\infty}^\infty \frac{\cos x}{1 + x
        + x^2}
        &= \Re \left( \frac{2\pi}{\sqrt{3}}
        e^{-\sqrt{3}/2} e^{-i/2} \right) \\
        &= \frac{2\pi}{\sqrt{3}} e^{-\sqrt{3}/2}
        \cos \left( -\half \right)
    \end{align*}
\end{example*}

\begin{flashcard}[jordans-lemma]
\begin{lemma*}[Jordan's Lemma]
    Suppose $f(z)$ is holomorphic on $\{|z| > r\}$
    for some $r > 0$, and \cloze{$zf(z)$ is bounded. Then
    for all $\alpha > 0$, we have
    \[ \int_{C_R'} f(z) e^{i\alpha z} \dd z \to 0 \]
    }
    as $R \to \infty$, where $C_R' : [0, \pi] \to
    \CC$, $C_R'(t) = Re^{it}$.
\end{lemma*}
\end{flashcard}

\begin{proof}
    We have for $z = Re^{it}$, that $|e^{i\alpha
    z}| = e^{-\alpha R \sin t}$, and so using the
    basic estimate $\frac{\sin t}{t} \ge
    \frac{2}{\pi}$ on $[0, \pi/2]$ (since
    $\frac{\sin t}{t}$ decreases on $[0, \pi/2]$),
    we have
    \[ |e^{i\alpha z}| \le \begin{cases}
        e^{-\alpha R \frac{2t}{\pi}} & t \in [0,
        \pi/2] \\
        e^{-\alpha R \frac{2t'}{\pi}} & t' = \pi -
        t, t \in [\pi/2, \pi]
    \end{cases} \]
    BY hypothesis, there exists $M \in \RR$ such
    that $|zf(z)| \le M$. Putting these together,
    let $\tilde{C_R'}$ be $C_R'$ for $[0, \pi/2]$.
    Then
    \begin{align*}
        \left| \int_{\tilde{C_R'}} f(z) e^{i\alpha
        z} \dd z \right|
        &\le \int_0^{\pi/2} M e^{-\alpha R
        \frac{2t}{\pi}} \dd t \\
        &= \left[ M \left( \frac{1}{-\alpha R
        \frac{2}{\pi}} \right) e^{-\alpha R
        \frac{2t}{\pi}} \right]_{t = 0}^{t =
        \pi/2} \\
        &= \frac{(1 - e^{-\alpha R} \pi
        M}{2R\alpha} \\
        &\to 0
    \end{align*}
    as $R \to \infty$. Similarly for $t \in
    [\pi/2, \pi]$.
\end{proof}

\begin{hiddenflashcard}[jordans-lemma-proof]
\prompt{Jordan's Lemma proof}
\begin{itemize}
    \item \cloze{Estimate $\frac{\sin t}{t} \ge
        \frac{2}{\pi}$ on $[0, \pi/2]$.}
    \item \cloze{Since $|e^{i\alpha z}| = e^{-\alpha R
        \sin t}$,
        \[
            |e^{i\alpha z}|
            \le \begin{cases}
                e^{-\alpha R \frac{2t}{\pi}} & t \in
                [0, \pi/2] \\
                e^{-\alpha R \frac{2t'}{\pi}} & t' =
                \pi - t, t \in [\pi/2, \pi]
            \end{cases}
        \]}
    \item \cloze{Split the integral into $[0, \pi/2]$ and
        $[\pi/2, \pi]$ and show that each goes to
        0.}
\end{itemize}
\end{hiddenflashcard}