% vim: tw=50
% 06/03/2023 10AM

\begin{proposition*}
    Let $f$ have a zero (respectively pole) of
    order $k > 0$ at $z = a$. Then
    $\frac{f'(z)}{f(z)}$ has a simple pole at $z =
    a$, of residue $k$ (respectively $-k$).
\end{proposition*}

\begin{remark*}
    By Example Sheet 2, if $f : \mathcal{U} \to
    \CC$ with $f(\mathcal{U})$ contained in a
    simply connected set which omits $0$, then
    there exists holomorphic function $g(z) = \log
    f(z)$ on $\mathcal{U}$, so
    $\frac{f'(z)}{f(z)}$ has holomorphic
    antiderivative $\log f$ on $\mathcal{U}$. We
    call $\frac{f'(z)}{f(z)}$ the ``logarithmic
    derivative'' of $f$.
\end{remark*}

\begin{proof}
    Suppose $f(z) = (z - a)^k g(z)$ near $a$, with
    $g(a) \neq 0$, then $f'(z) = k(z - a)^{k - 1}
    g(z) + (z - a)^k g'(z)$, so
    \[ \frac{f'(z)}{f(z)} = \frac{k}{z - a} +
    \frac{g'(z)}{g(z)} \]
    Since $g(a) \neq 0$, $\frac{g'}{g}$ is
    holomorphic at $a$. So $\Res_{z = a}
    \frac{f'}{f} = k$. (Similarly for the pole
    case).
\end{proof}

\begin{theorem*}[Argument Principle]
    Let $\gamma$ be a closed curve bounding a
    domain $D$, and $f$ a function meromorphic on
    an open neighbourhood of $D \cup \gamma$. If
    $f$ has no zeroes or poles on $\gamma$, then
    \[ I(f \circ \gamma; 0) = \frac{1}{2\pi i}
    \int_\gamma \frac{f'(z)}{f(z)} \dd z =
    \text{\# of zeroes of $f$ in $D$ } -
    \text{ \# of poles of $f$ in $D$} \]
    where zeroes and poles are counted with
    multiplicity.
\end{theorem*}

\begin{hiddenflashcard}[argument-principle]
\begin{theorem*}[Argument Principle]
    \begin{itemize}
        \item \cloze{$\gamma$ closed curve bounding
            a domain $D$}
        \item \cloze{$f$ meromorphic on open
            neighbourhood of $D \cup \gamma$}
        \item \cloze{$f$ has no zeroes or poles on
            $\gamma$}
    \end{itemize}
    Then:
    \[
        \cloze{
        I(f \circ \gamma; 0)
        =
        \frac{1}{2\pi i}
        \int_\gamma \frac{f'(z)}{f(z)} \dd z
        =
        \text{\# of zeroes of $f$ in $D$ }
        - \text{ \# of poles of $f$ in $D$}
        }
    \]
\end{theorem*}
\end{hiddenflashcard}

\begin{flashcard}[argument-principle-proof]
\prompt{Argument principle proof? \\}
\begin{proof}
    \cloze{
    We have
    \begin{align*}
        I(f \circ \gamma; 0)
        &= \frac{1}{2\pi i}
        \int_{f \circ \gamma} \frac{\dd w}{w} \\
        &= \frac{1}{2\pi i} \int_\gamma
        \frac{f'(z)}{f(z)} \dd z
    \end{align*}
    By residue theorem, this is
    \[ \sum_{\substack{\text{poles $\alpha$ in
    $D$} \\ \text{of $f' / f$}}} \Res_{z = \alpha}
    \frac{f'}{f} \]
    but by previous proposition this equals
    \[ \text{number of zeroes of $f$ in $D$} -
    \text{number of poles of $f$ in $D$} \]
    counting multiplicity.
}
\end{proof}
\end{flashcard}

\subsubsection*{Remarks}

\begin{enumerate}[(1)]
    \item Recall $\gamma$ is compact, $f \circ \gamma$
        is also a closed curve (and compact).
    \item Morally: this says
        \[ 2\pi (\text{\# of zeroes of $f$ in
        $D$ } - \text{ \# of poles of $f$ in
        $D$}) \]
        is the change in $\arg f(z)$ as $z$
        travels $\gamma$.
\end{enumerate}
The argument principle has important consequences
for local behaviour of $f$.

\begin{flashcard}[local-degree-defn]
\begin{definition*}
    If $f$ is holomorphic and non-constant near $z
    = a$, then the \emph{local degree} of $f(z)$
    at $z = a$ is \cloze{$\deg_{z = a} f(z)$, the order
    of the zero of $f(z) - f(a)$ at $z = a$.}
\end{definition*}
\end{flashcard}

\noindent
If $f$ is non-constant, we can write $f(z) - f(a) =
(z - a)^k g(z)$, $g$ holomorphic at $a$, and the
zero at $z = a$ of $f(z) - f(a)$ is isolated. So
$0 < |z - a|$ sufficiently small implies $f(z) -
f(a) \neq 0$. SO for small $\eps > 0$, the circle
$\gamma(t) = a + \eps e^{it}$, $t \in [0, 2\pi]$,
about $a$ gives
\begin{align*}
    I(f \circ \gamma; f(a))
    &= I(f(\gamma(t)) - f(a); 0) \\
    &= \text{\# of zeroes in $D(a, \eps)$ of $f(z)
    - f(a)$ } - \text{ \# of poles of $f(z) -
    f(a)$ in $D(a, \eps)$} \\
    &= \deg_{z = a} f(z)
\end{align*}
Consider the local behaviour of $f(z) = z^k$ at $z
= 0$ for $k > 0$. We have $\deg_{z = 0} f(z) = k$.
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/24ef76c0bc0e11ed.png}
\end{center}
Note that $\forall w \in D(0, \eps)$, $w$ has $k$
preimages under $f$ in $D(0, \eps^{1/k})$.

\begin{flashcard}[local-mapping-thm]
\begin{theorem*}[Local mapping degree theorem]
    \cloze{
    Let $f : D(a, R) \to \CC$ be holomorphic and
    non-constant, with local degree $k > 0$. Then
    for $r > 0$ sufficiently small, there exists
    $\eps > 0$ such that if $0 < |w - f(a)| <
    \eps$, then $f(z) = w$ has exactly $k$ (simple)
    roots in $D(a, r)$.
}
\end{theorem*}

\begin{proof}
    \cloze{
    Choose $r > 0$ such that $f(z) - f(a)$ has no
    zeroes for $0 < |z - a| \le r$ and $f'(z) \neq
    0$ for $0 < |z - a| \le r$; $r$ exists by
    identity principle. Let $\gamma$ be the circle
    of radius $r$ about $a$. Then $f \circ \gamma$
    doesn't contain $f(a)$, so there exists $\eps
    > 0$ such that $D(f(a), \eps) \cap f \circ
    \gamma = \emptyset$. For $w \in D(f(a),
    \eps)$, the number of zeroes of $f(z) = w$ in
    $D(a, r)$ is $I(f \circ \gamma; w)$. But $I(f
    \circ \gamma; w) = I(f \circ \gamma, f(a)) =
    k$\prompt{ (since the winding number is
    \fcemph{constant} on
    \fcemph{connected components} of $\CC \setminus f \circ
    \gamma$)}. Since $f(z) - w$ has nonzero derivative in
    $D(a, r)^\times$, so the preimages of $w$ are
    simple.
}
\end{proof}
\end{flashcard}

\myskip
Note $I(f \circ \gamma; w) = I(f \circ \gamma;
f(a))$ because the winding number is constant on
connected components of $\CC \setminus f \circ
\gamma$.