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% 23/01/2023 10AM

\vspace{-1em}
\begin{itemize}
    \item [conformality] Let $f \colon \mathcal{U}
        \to \CC$ be holomorphic function on an
        open set $\mathcal{U}$, and $w \in
        \mathcal{U}$ with $f'(w) \neq = 0$.
        Geometric behaviour of $f$ at $w$?

        \noindent
        Claim: $f$ is conformal at $w$:
        \begin{center}
            \includegraphics[width=0.6\linewidth]
            {images/fe76f81e9b0711ed.png}
        \end{center}
        $\gamma_1, \gamma_2$
        $\mathcal{C}'$-paths through $w$,
        $\gamma_1, \gamma_2 \colon [-1, 1] \to
        \mathcal{U}$. $\gamma_1(0) =
        \gamma_2(0) = w$, $\gamma_1'(0) \neq
        0$. Write $\gamma_j(t) = w + r_j(t)
        e^{i\theta_j(t)}$, $j = 1, 2$. We have
        $\Arg(\gamma_j'(0)) = \theta_j(0)$ and
        \[ \Arg((f \circ  \gamma_j)'(0)) =
        \Arg(\gamma_j'(0) f'(\gamma_j(0)))
        = \Arg(\gamma_j'(0)) + \Arg(f'(w)) +
        2\pi n, n \in \ZZ \]
        so the direction of $\gamma_j$ at $w$
        under application of $f$ is rotated by
        $\Arg(f'(w))$, independent of $\gamma_j$.
        Since the angle between $\gamma_1,
        \gamma_2$ is a difference of arguments,
        the $f$ preserves this angle.
\end{itemize}

\begin{hiddenflashcard}[proof-holomorphic-is-conformal]
Proof that holomorphic \cloze{(with non-zero
derivative)} implies conformal? \\
\cloze{
Let $\gamma_1, \gamma_2$ be $C^1$ paths through
$w$, $\gamma_1, \gamma_2 : [-1, 1] \to
\mathcal{U}$, $\gamma_1(0) = \gamma_2(0) = w$,
$\gamma_i'(0) \neq 0$. After the setup, the proof
is trivial:
\[ \Arg((f \circ \gamma_j)'(0)) =
\Arg(\gamma_j'(0) f'(\gamma_j(0)))
= \Arg(\gamma_j'(0)) + \Arg(f'(w)) + 2\pi n \]
so the directions are rotated by the same amount.
}
\end{hiddenflashcard}

\begin{flashcard}[conformal-equivalence]
\begin{definition*}
    Let $\mathcal{U}, \mathcal{V}$ be domains in
    $\CC$. A map $f \colon \mathcal{U} \to
    \mathcal{V}$ is a \emph{conformal equivalence}
    of $\mathcal{U}$ and $\mathcal{V}$ \cloze{if $f$ is a
    \fcemph{bijective holomorphic} map with
    \prompt{$f'$ \fcemph{non-zero everywhere}}%
    \fcscrap{$f'(z) \neq 0
    ~\forall z \in \mathcal{U}$}.}
\end{definition*}
\end{flashcard}

\subsubsection*{Remarks}

\begin{enumerate}[(1)]
    \item On Example sheet 1, we will use the real
        inverse function theorem to show that if
        $f \colon \mathcal{U} \to \mathcal{V}$ is
        a holomorphic bijection of open sets with
        $f'(z) \neq 0 ~ \forall z \in
        \mathcal{U}$, then the inverse of $f$ is
        also holomorphic, so also conformal by the
        chain rule.

        \myskip
        So conformally equivalent domains are the
        same from the perspective of the
        holomorphic functions they admit.
    \item We will see later that injective and
        holomorphic on a domain implies that
        $f'(z) \neq 0 ~ \forall z \in
        \mathcal{U}$, so this requirement is
        redundant.
\end{enumerate}

\subsubsection*{Examples}

\begin{enumerate}[(1)]
    \item (Change of coordinates) On $\CC$, $f(z)
        = az + b$, $a \neq 0$ is a conformal
        equivalence $\CC \to \CC$. More generally
        a M\"obius map
        \[ f(z) = \frac{az + b}{cz + d}, \quad ad
        - bc \neq 0 \]
        is a conformal equivalence from the
        Riemann sphere to itself. Riemann sphere:
        add point $\infty$ to make a sphere
        $\CC_\infty$ (also sometimes written
        $\hat{\CC}$):
        \begin{center}
            \includegraphics[width=0.6\linewidth]
            {images/34a570409b0911ed.png}
        \end{center}
        or, imagine giving two copies of the unit
        disk with coordinates $z$, $\frac{1}{z}$
        (see Part II Riemann Surfaces). If $f
        \colon \CC_\infty \to \CC_\infty$ is
        continuous then:
        \begin{enumerate}[(1)]
            \item If $f(\infty ) = \infty$, $f$
                holomorphic at $\infty \iff g(z) =
                \frac{1}{f \left( \frac{1}{z}
                \right)}$ is holomorphic at
                $0$.
            \item If $f(\infty) \neq \infty$, $f$
                holomorphic at $\infty \iff f
                \left( \frac{1}{z} \right)$
                holomorphic at $0$.
            \item If $f(a) = \infty$, $a \in \CC$,
                then $f$ is holomorphic at $a \iff
                \frac{1}{f(z)}$ is holomorphic at
                $a$.
        \end{enumerate}
        M\"obius maps are change of coordinates
        for the sphere. Choosing $z_1 \mapsto 0$,
        $z_2 \mapsto \infty$, $z_3 \mapsto 1$
        defines a M\"obius map:
        \[ f(z) = \frac{(z - z_1)}{(z - z_2)}
        \cdot \frac{z_3 - z_2}{z_3 - z_1} \]
        for distinct $z_1, z_2, z_3 \in \CC$
        (recall Part IA Groups).
    \item asdf
\end{enumerate}