% vim: tw=50
% 23/02/2022 11AM

\begin{theorem}
    $f(z) = \sum_{n = 0}^\infty a_n z^n$ has
    radius of convergence $R$. Then $f$ is
    differentiable at all points with $|z| < R$
    with
    \[ f'(z) = \sum_{n = 1}^\infty na_n z^{n - 1} \]
\end{theorem}

\begin{proof}[(non-examinable)]
    We need two auxiliary lemmas:
    \begin{lemma}
        If $\sum_{n = 0}^\infty a_n z^n$ has
        radius of convergence $R$, so do
        \[ \sum_{n = 1}^\infty na_n z^{n - 1}
        \qquad \text{ and } \qquad \sum_{n =
        2}^\infty n(n - 1)a_n z^{n - 2} \]
    \end{lemma}
    \begin{lemma}
        \begin{enumerate}[(i)]
            \item ${n \choose r} \le n(n - 1) {n -
                2 \choose r - 2}$ for all $2 \le r
                \le n$
            \item $|(z + h)^n - z^n - nhz^{n - 1}|
                \le n(n - 1)(|z| + |h|)^{n -
                2}|h|^2$ for all $z \in \CC, h \in
                \CC$.
        \end{enumerate}
    \end{lemma}
    \begin{proof}[of 4.4]
        (after which we prove the lemmas) By Lemma
        4.5 we may define
        \[ f'(Z) := \sum_{n = 1}^\infty na_n z^{n
        - 1} \qquad |z| < R \]
        Then we are required to prove that
        \[ \lim_{h \to 0} \frac{f(z + h) - f(z) -
        hf'(z)}{h} = 0 \]
        \begin{align*}
            I
            &:= \frac{f(z + h) - f(z) - hf'(z)}{h}
            \\
            &= \frac{1}{h} \sum_{n = 0}^\infty a_n
            ((z + h)^n - z^n - hnz^{n - 1}) \\
            |I|
            &= \frac{1}{|h|} \left| \lim_{N \to
            \infty} \sum_{n = 0}^N a_n ((z + h)^n
            - z^n - nhz^{n - 1}) \right| \\
            &= \frac{1}{|h|} \lim_{N \to \infty}
            \left| \sum_{n = 0}^N a_n((z + h)^n -
            z^n - nhz^{n - 1}) \right| \\
            &\le \frac{1}{|h|} \sum_{n = 0}^N
            |a_n||(z + h)^n - z^n - nhz^{n - 1}| \\
            &\le \frac{1}{|h|} \sum_{n = 2}^\infty
            |a_n| n(n - 1)(|z| + |h|)^{n - 2}
            |h|^2 \\
            &= |h| \sum_{n = 2}^\infty |a_n| n(n -
            1)(|z| + |h|)^{n - 2}
        \end{align*}
        By Lemma 4.5, for $|h|$ small enough,
        \[ \sum_{n = 2}^\infty |a_n| n(n - 1)(|z|
        + |h|)^{n - 2} \]
        converges to $A(h)$, but $A(h) \le A(r)$
        for $|h| < r$ and $|z| + r < R$. Hence
        \[ |I| \le |h| A(h) \le |h|A(r) \to 0 \]
        as $h \to 0$.
    \end{proof}
    \begin{proof}[of Lemma 4.5]
        Take $z$ and $R_0$ such that $0 < |z| <
        R_0 < R$. Since $a_n R_0^n \to 0$,
        $\exists \,\,K$ such that $|a_n R_0^n| \le
        K$, $\forall\,\,n \ge 0$. Thus
        \begin{align*}
            |na_nz^{n - 1}|
            &= \frac{n}{|z|} |a_n R_0^n| \left|
            \frac{z}{R_0} \right|^n \\
            &\le \frac{Kn}{|z|} \left|
            \frac{z}{R_0} \right|^n
        \end{align*}
        But $\sum n \left| \frac{z}{R_0}
        \right|^n$ converges by the ratio test:
        \[ \frac{n + 1}{n} \left| \frac{z}{R_0}
        \right|^{n + 1} \left| \frac{R_0}{z}
        \right|^n = \frac{n + 1}{n} \left|
        \frac{z}{R_0} \right| \to \left|
        \frac{z}{R_0} \right| < 1 \]
        if $|z| > R$, the series diverges since
        $|a_n z^n|$ is unbounded hence so is
        $n|a_n z^n|$. The same proof applies to
        $\sum_{n = 2}^\infty n(n - 1) a_n z^{n -
        2}$.
    \end{proof}
    \begin{proof}[of Lemma 4.6]
        \begin{enumerate}[(i)]
            \item \eqnoskip
                \[ \frac{{n \choose r}}{{n - 2
                \choose r - 2}} =
                \frac{n!}{r!\cancel{(n - r)!}}
                \frac{(r - 2)!\cancel{(n - r)!}}{(n -
                2)!} = \frac{n(n - 1)}{r(r - 1)}
                \le n(n - 1) \]
            \item \eqnoskip
                \[ (z + h)^n - z^n - nhz^{n - 1} =
                \sum_{r = 2}^n {n \choose r} z^{n
                - r} h^r \]
                Thus
                \begin{align*}
                    |(z + h)^n - z^n - nhz^{n - 1}|
                    &\le \sum_{r = 2}^n {n \choose
                    r} |z|^{n - r} |h|^r \\
                    &\le n(n - 1) \left[ \sum_{r =
                    2}^n {n - 2 \choose r - 2}
                    |z|^{n - r} |h|^{r - 2}
                    \right] |h|^2 \\
                    &= n(n - 1) (|z| + |h|)^{n -
                    2} |h|^2
                \end{align*}
        \end{enumerate}
    \end{proof}
\end{proof}

\subsection{The Standard Functions}

(exponentials, logs, trigonometric, etc) \myskip
We have already seen that
\[ \sum_{n = 0}^\infty \frac{z^n}{n!} \]
has $R = \infty$. Define $e : \CC \to \CC$ by
\[ e(z) = \sum_{n = 0}^\infty \frac{z^n}{n!} \]
Straight from Theorem 4.4, $e$ is differentiable
and
\[ \boxed{e'(z) = e(z)} .\]

\begin{lemma*}
    If $F : \CC \to \CC$ has $F'(z) = 0$ for all
    $z \in \CC$, then $F$ is constant.
\end{lemma*}

\begin{proof}
    Consider $g(t) = F(tz)$. By chain rule:
    \[ g'(t) = F'(tz)z = 0 \]
    if $g(t) = u(t) + iv(t)$ then $g'(t) = u'(t) +
    iv'(t)$ so $u' = v' = 0$. Apply Corollary 3.5
    to get the claim.
\end{proof}

\myskip
Now let $a, b \in \CC$. Consider
\[ F(z) = e(a + b - z)e(z) \]
\[ F'(z) = -e(a + b - z)e(z) + e(a + b - z)e(z) =
0 \]
so $F$ is constant. Use $z = b$ and $z = 0$ to
deduce that
\[ \boxed{e(a)e(b) = e(a + b)} \]