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% 27/01/2023 10AM

Let's recall some facts about functions defined by
a power series or other sequences of functions.
\begin{enumerate}[(1)]
    \item A sequence $(f_n)_{n \in \NN}$ of
        functions \emph{converges uniformly} to a
        function $f$ on some set $S$ if $\forall
        \eps > 0$, $\exists N \in \NN$ such that
        $\forall n \ge N$, $\forall x \in S$,
        \[ |f_n(x) - f(x)| < \eps \]
    \item The uniform limit of continuous
        functions is continuous.
    \item
        \begin{flashcard}[Weierstrass-M-test]
            Weierstrass $M$-test:
            \prompt{\\}\cloze{if $(M_n)_{n \in
            \NN} \subseteq \RR_{> 0}$ and $0 \le
            |f_n(x)| \le M_n ~ \forall x \in S$ and
            all $n \in \NN$, then
            \[ \sum_{n = 1}^\infty M_n < \infty
            \implies \sum_{n = 1}^\infty f_n(x)
            \text{ converges uniformly on $S$ as $N \to
            \infty$} \]}
        \end{flashcard}
    \item Let $(c_n)_{n \in \NN} \subseteq \CC$,
        and fix $a \in \CC$. Then $\exists! R \in
        [0, 1\infty]$ such that the series
        \[ z \mapsto \sum_{n = 1}^\infty c_n(z -
        a)^n \]
        converges absolutely if $|z - a| < R$,
        diverges if $|z - a| > R$. If $0 < r < R$
        then the series converges uniformly on
        $D(a, r)$. $R$ is the \emph{radius of
        convergence} of the series. We can compute
        \[ R = \sup\{r \ge 0 \colon | c_n|r^n \to 0
        \text{ as } n \to \infty\} \]
        or
        \[ R = \frac{1}{\lambda},\qquad
        \lambda =
        \lim_{n \to \infty} \sup_{n' \ge n}
        |c_{n'}|^{1/n'} .\]
\end{enumerate}

\begin{theorem*}
    Let $f(z) = \sum_{n = 0}^\infty c_n(z - a)^n$
    be a complex power series with radius of
    convergence $R$. Then
    \begin{enumerate}[(i)]
        \item $f$ is holomorphic on $D(a, R)$
        \item $f$ has derivative
            \[ f'(z) = \sum_{n = 1}^\infty nc_n (z
            - a)^{n - 1} \]
            with radius of convergence $R$ about
            $a$.
        \item $f$ has derivatives of all orders on
            $D(a, R)$, and $f^{(n)}(a) = n! c_n$.
    \end{enumerate}
\end{theorem*}

\begin{hiddenflashcard}[power-series-derivative]
\begin{theorem*}
    Let $f(z) = \sum_{n = 0}^\infty c_n (z - a)^n$
    be a complex power series with radius of
    convergence $R$. Then
    \begin{enumerate}[(i)]
        \item \cloze{$f$ is holomorphic on $D(a,
            R)$.}
        \item \cloze{$f$ has the obvious term by
            term derivative, with radius of
            convergence $R$ about $a$.}
        \item \cloze{$f$ has derivatives of all
            orders on $D(a, R)$, with $f^{(n)}(a)
            = n! c_n$}
    \end{enumerate}
\end{theorem*}

\begin{proof}
\cloze{
    First prove (ii), which clearly implies (i)
    and also implies (iii) by simple induction.

    \myskip
    To prove (ii), first show that the term by term derivative
    series does indeed have radius of convergence
    $R$.
    Next, fix $z \in D(a, R)$. Pick $|z -
    a| < P < R$. Consider $g(z, w) = \frac{f(z)
    - f(w)}{z - w}$. Consider the partial sums
    \[ \sum_{n = 0}^N c_n \frac{z^n - w^n}{z - w}
    \]
    show that this converges uniformly to $g(z,
    w)$ on $D(a, P)$.
    Then show that $g(w, w)$ is equal to the
    desired series.
}
\end{proof}
\end{hiddenflashcard}

\begin{proof}
    Without loss of generality $a = 0$ by change
    of variables. Consider the series $\sum_{n =
    1}^\infty nc_n z^{n - 1}$. Since $|nc_n| \ge
    |c_n|$ the radius of convergence of this
    series is no larger than $R$. If $0 < R_1 <
    R$, then for $|z| < R$, we have
    \[ |nc_n z^{n - 1}| \le n|c_n|R_1^{n - 1} \cdot
    \frac{|z|^{n - 1}}{R_1^{n - 1}} \]
    $n \cdot \left( \frac{|z|}{R_1} \right)^{n -
    1} \to 0$ as $n \to \infty$. So applying the
    $M$-test with $M_n = c_n R_1^{n - 1}$ we have
    convergence of the series. So $\sum nc_n z^{n
    - 1}$ has radius of convergence $R$.

    \myskip
    For $|z|, |w| < R$, we want to consider
    $\frac{f(z) - f(w)}{z - w}$. Taking partial
    sums:
    \[ \sum_{n = 0}^N c_N \cdot \frac{z^n - w^n}{z
    - w} = \sum_{n = 0}^N c_n \left( \sum_{j =
    0}^{n - 1} z^j w^{n - 1 - j} \right) \tag{$*$} \]
    For $|z|, |w| < P < R$, we have
    \[ \left| c_n \left( \sum_{j = 0}^{n - 1} z^j
    w^{n - 1 - j} \right) \right| \le |c_n| \cdot
    n \cdot P^{n - 1} \]
    so ($*$) converges uniformly on $\{(z, w)
    \colon |z|, |w| < P\}$. So the series
    converges to a continuous limit on $\{|z|, |w|
    < R\}$, call it $g(z, w)$. When $z \neq w$,
    $g(z, w) = \frac{f(z) - f(w)}{z - w}$. When $z
    = w$, $g(w, w) = \sum_{n = 0}^\infty nc_n w^{n
    - 1}$, so by continuity of $g$, (i) and (ii)
    are proved. (iii) is a simple induction.
\end{proof}

\begin{corollary*}
    Suppose $0 < \eps < R$, where $R$ is the radius
    of convergence of the complex power series
    \[ f(z) = \sum_{n = 0}^\infty c_n(z - a)^n ,\]
    and $f(z) = 0 ~\forall z \in F(a, \eps)$. Then
    $f \equiv 0$ on $D(a, R)$.
\end{corollary*}

\begin{proof}
    Since $f \equiv 0$ on $D(a, \eps)$, we have
    $f^{(n)}(a) = 0 ~\forall n$. So by part (iii)
    of the previous theorem, we have $c_n = 0
    ~\forall n$, and $f \equiv 0$ on $D(a, R)$.
\end{proof}

\subsubsection*{The Exponential and The Logarithm}

We define the complex exponential
\[ e^z = \exp(z) \defeq \sum_{n = 0}^\infty
\frac{z^n}{n!} .\]
Properties:
\begin{enumerate}[(1)]
    \item Radius of convergence is $\infty$, so
        this function is entire, and we have
        $\dfrac{}{z} e^z = e^z$.
    \item For all $z, w \in \CC$, $e^{z + w} = e^z
        e^w$, and $e^z \neq 0$. Proof: fix $w \in
        \CC$, and consider $F(z) \defeq e^{z + w}
        \cdot e^{-z}$. We have
        \[ F'(z) = e^{z + w} e^{-z} - e^{z + w}
        e^{-z} = 0 ,\]
        so $F$ is constant. Since $e^0 = 1$, $F(z)
        = e^w$, so $e^{z + w} = e^z e^w$. Since
        $e^z \cdot e^{-z} = e^0 = 1 ~\forall z \in
        \CC$, $e^z \neq 0$.
    \item $z = x + iy$. Then $e^z e^{x + iy} = e^x
        e^{iy}$, $x, y \in \RR$.
        \[ e^{iy} = \cos y + i\sin y ;\]
        note then $|e^{iy}| = 1$. So
        \[ e^z = e^x(\cos y + i\sin y), \]
        and $|e^z| = e^x = e^{\Re(z)}$. $e^z = 1$
        if and only if $x = 0$ and $y = 2\pi k$
        for some $k \in \ZZ$. In fact, $\forall w
        \in \CC^\times$, $\exists$ infinitely many
        $z \in \CC$ such that $e^z = w$,
        differing by integer multiples of $2\pi
        i$.
        \begin{center}
            \includegraphics[width=0.6\linewidth]
            {images/32cf03f09e3111ed.png}
        \end{center}
\end{enumerate}

\begin{definition*}
    Let $\mathcal{U} \subseteq \CC^\times$ be an
    open set. We say a continuous function
    $\lambda \colon \mathcal{U} \to \CC$ is a
    \emph{branch of the logarithm} if $\forall z
    \in \mathcal{U}$, $\exp(\lambda(z)) = z$.
    Useful example: $\mathcal{U} = \CC \setminus
    \RR_{\le 0}$.
    \begin{center}
        \includegraphics[width=0.3\linewidth]
        {images/c6c426089e3111ed.png}
    \end{center}
    Define $\Log
    \colon \mathcal{U} \to \CC$ by
    \[ \Log(Z) \defeq \ln|z| +
    i\theta \]
    $\theta = \arg z$, $\theta \in (-\pi, \pi)$.
    This is the \emph{principal branch of the
    logarithm}.
\end{definition*}

\begin{hiddenflashcard}[principle-branch-of-log]
Principal branch of the logarithm? \\
\begin{align*}
    \Log(1 + z)
    &= \cloze{z - \frac{z^2}{2} + \frac{z^3}{3} -
    \cdots} \\
    &= \cloze{\sum_{n = 1}^\infty \frac{(-1)^{n -
    1} z^n}{n}}
\end{align*}
\end{hiddenflashcard}

\begin{proposition*}
    $\Log(z)$ is holomorphic on $\CC \setminus
    \RR_{\le 0}$ with derivative $\frac{1}{z}$.
    Moreover, if $|z| < 1$, then
    \[ \Log(1 + z) = \sum_{n = 1}^\infty
    \frac{(-1)^{n - 1} z^n}{n} \]
\end{proposition*}

\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/7afe85149e3211ed.png}
\end{center}