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% 06/02/2023 10AM

\begin{flashcard}[CIF]
\begin{theorem*}[Cauchy's Integral Formula]
    Let \cloze{$\mathcal{U} \subseteq \CC$ be a domain,
    $f : \mathcal{U} \to \CC$ holomorphic, and
    $\ol{D(a, r)} \subseteq \mathcal{U}$.}
    \prompt{\\} Then \cloze{for
    all $z \in D(a, r)$,
    \[ f(z) = \frac{1}{2\pi i} \int_{\partial D(a,
    r)} \frac{f(w)}{w - z} \dd w \]}
\end{theorem*}
\end{flashcard}

\begin{hiddenflashcard}[CIF-proof]
Cauchy's Integral Formula proof? \\
\cloze{
Define $g(w) = \frac{f(w)}{w - z}$ for $w \neq z$
and $g(z) = f'(z)$, then $g$ is
\fcemph{continuous everywhere} since $f$
\fcemph{holomorphic}. Also, $g$
\fcemph{holomorphic} on $D(a,
r)$ except possibly at $z$. \\
Pick $r_1 > 0$ such that $D(a, r) \subseteq D(a,
r_1) \subseteq \mathcal{U}$. Apply \fcemph{Cauchy's
theorem in a disk} to $g$ on $D(a, r_1)$ with
$\gamma = \partial D(a, r)$ to deduce
\[ \int_{\partial D(a, r)} \frac{f(w)}{w - z} \dd
w = \int_{\partial D(a, r)} \frac{f(z)}{w - z} \dd
w \]
Now swap an integral and sum sign since the
partial sums are uniformly convergent (A\&T):
\begin{align*}
    \int_{\partial D(a, r)} \frac{f(z)}{w - z} \dd
    w
    &= \int_{\partial D(a, r)}
    \frac{f(z)}{w - a} \frac{1}{1 - \frac{z -
    a}{w - a}} \dd w \\
    &= \int_{\partial D(a, r)} \sum_{n = 0}^\infty
    f(z) \frac{(z - a)^n}{(w - a)^{n + 1}} \dd w \\
    &= \sum_{n = 0}^\infty f(z) (z - a)^n \int_{\partial D(a, r)}
    \frac{1}{(w - a)^{n + 1}} \dd w \\
    &= \sum_{n = 0}^\infty f(z) (z - a)^n \cdot
    \begin{cases}
        2\pi i & n = 0 \\
        0 & n \neq 0
    \end{cases} \\
    &= 2\pi i f(z)
\end{align*}
}
\end{hiddenflashcard}

\begin{proof}
    Define
    \[ g(w) = \begin{cases}
        \frac{f(w) - f(z)}{w - z} - f'(z) &
        w \neq z \\
        0 & w = z
    \end{cases} \]
    Then $g$ is continuous at $z$, holomorphic on
    $D(a, r)$ except possible at $z$. Find $r_1 >
    0$ such that $D(a, r) \subseteq D(a, r_1)
    \subseteq \mathcal{U}$. Apply Cauchy's theorem
    to $g$ on $D(a, r_1$ with curve $\gamma =
    \partial D(a, r)$, then
    \[ \int_{\partial D(a, r)} g(w) \dd w = 0 \]
    i.e.
    \[ \int_{\partial D(a, r)} \frac{f(w)}{w - z}
    \dd w =  \int_{\partial D(a, r)} \frac{f(z)}{w
    - z} \dd w \]
    Useful expansion: since $|w - a| = r > |z -
    a|$
    \[ \frac{1}{w - z} = \frac{1}{(w - a) \left[ 1
    - \frac{z - a}{w - a} \right]} = \sum_{n =
    0}^\infty \frac{(z - a)^n}{(w - a)^{n + 1}} ,\]
    by geometric expansion. So
    \[ \int_{\partial D(a, r)} \frac{f(z)}{w - z}
    \dd w = \sum_{n = 0}^\infty \left[ f(z) (z -
    a)^n \int_{\partial D(a, r)} \frac{1}{(w - a)^{n +
    1}} \dd w \right] \]
    We have computed that the integral in the
    brackets on the right is $0$ unless $n = 0$,
    in which case it is $2\pi i$. So
    \[ \int_{\partial D(a, r)} \frac{f(w)}{w - z}
    \dd w = 2\pi i f(z) \]
    as claimed.
\end{proof}

\begin{flashcard}[mean-value-property]
\begin{corollary*}[Mean Value Property]
    \cloze{
    If $f : \mathcal{U} \to \CC$ is holomorphic on
    domain $\mathcal{U}$, and $\ol{D(a, r)}
    \subseteq \mathcal{U}$, then
    \[ f(a) = \int_0^1 f(a + re^{2\pi i t}) \dd t \]
    }
    \fcscrap{i.e. $f$ takes the average value on a circle
    about a point.}
\end{corollary*}
\end{flashcard}

\begin{proof}
    Applying Cauchy's Integral Formula, with $t
    \mapsto a + re^{2\pi i t}$ on $[0, 1]$ for
    $\partial D(a, r)$.
\end{proof}

\subsubsection*{Applications of CIF}

\begin{flashcard}[local-maximum-principle]
\begin{corollary*}[Local Maximum Principle]
    \cloze{Let $f : D(a, r) \to \CC$ be holomorphic. If
    $|f(z)| \le |f(a)|$ for all $z \in D(a, r)$,
    then $f$ is constant.}\fcscrap{ ``non-constant
    holomorphic maps cannot achieve maximum on an
    open set''.}
\end{corollary*}
\end{flashcard}

\begin{proof}
    By mean value property, $\forall 0 < \rho < r$ we
    have
    \begin{align*}
        |f(a)|
        &= \left| \int_0^1 f(a + \rho e^{2\pi it} \dd
        t \right| \\&\le \sup_{|z - a| = \rho}
        |f(z)| \\
        &\le |f(a)|
    \end{align*}
    Since we have equality at each step, we have
    $|f(z)| = |f(a)|$ for all $|z - a| = \rho$. So
    $|f|$ is a constant function on $D(a, r)$.
    Hence $f$ is constant on $D(a, r)$.
\end{proof}

\begin{flashcard}[liouville]
\begin{theorem*}[Liouville's Theorem]
    \cloze{Every bounded entire function is
    constant.}
\end{theorem*}
\end{flashcard}

\begin{flashcard}[liouville-proof]
\prompt{Proof of Liouville's theorem? \\}
\begin{proof}
\cloze{
    With $|f(z)| \le M$ for $f$ entire, and $R \gg
    1$ we have for any $0 < |z| < \frac{R}{2}$
    that
    .image
    \begin{align*}
        |f(z) - f(0)|
        &= \frac{1}{2\pi} \left| \int_{\partial
        D(0, R)} f(w) \left[ \frac{1}{w - z} -
        \frac{1}{w} \right] \dd w \right| \\
        &= \frac{1}{2\pi} \left| \int_{\partial
        D(0, R)} f(w) \frac{z}{(w - z)w} \dd w \right|
    \end{align*}
    Since $|z - w| > \frac{R}{2}$ and $|w| = R$,
    we have
    \begin{align*}
        |f(z) - f(0)|
        &\le \frac{1}{2\pi} \cdot
        2\pi R \cdot \sup_{w \in \partial D(0, R)}
        |f(w)| \cdot |z| \cdot \frac{1}{R \cdot
        \frac{R}{2}} \\
        &\le M \cdot |z| \cdot \frac{1}{R/2} \\
        &\to 0
    \end{align*}
    as $R \to \infty$. So $f(z) = f(0)$, so $f
    \equiv f(0)$ is constant.
}
\end{proof}
\end{flashcard}

\begin{flashcard}[fundamental-theorem-of-algebra]
\begin{corollary*}[Fundamental Theorem of Algebra]
    \cloze{Every \fcemph{non-constant} polynomial with complex
    coefficients has a root in $\CC$.}
\end{corollary*}

\begin{proof}
\cloze{
    If $p(z)$ has no root in $\CC$, then $f(z)
    \defeq \frac{1}{p(z)}$ is entire. $p(z)$
    non-constant implies $a_d \neq 0$, $d \ge 1$.
    So $\frac{p(z)}{z^d} = a_d + a_{d - 1} +
    \frac{1}{z} + \cdots + a_0 \frac{1}{z^d}$
    shows that $|p(z)| \to \infty$ as $|z| \to
    \infty$. So $|f(z)| \to 0$ as $|z| \to
    \infty$; so there exists $R > 0$ such that
    $\forall z \not\in D(0, r)$, $|f(z)| \le 1$.
    Let $M \defeq \max_{z \in \ol{D(0, r)}}
    |f(z)|$
    Then $|f|$ is bounded by $\max\{1, M\}$, and
    so by Liouville is constant, contradicting the
    assumption that $p$ is non-constant.
}
\end{proof}
\end{flashcard}

\subsubsection*{Taylor-Expansion}

\begin{flashcard}[taylor-expansion]
\begin{theorem*}[Taylor Expansion]
\cloze{
    Let $f : D(a, r) \to \CC$ be holomorphic. Then
    $f$ is represented by convergent power series
    on $D(a, r)$:
    \[ f(z) = \sum_{n = 0}^\infty c_n(z - a)^n \]
    with
    \[ c_n = \frac{f^{(n)}(a)}{n!} = \frac{1}{2\pi
    i} \int_{\partial D(a, \rho)} \frac{f(w)}{(w -
    a)^{n + 1}} \dd w \]
    for $0 \le \rho < r$.
}
\end{theorem*}

\begin{proof}
\cloze{
    For $|z - a| < \rho < r$, CIF gives
    \begin{align*}
        f(z)
        &= \frac{1}{2\pi i} \int_{\partial D(a,
        \rho)} \frac{f(w)}{w - z} \dd w \\
        &= \frac{1}{2\pi i} \int_{\partial D(a,
        \rho)} f(w) \cdot \sum_{n = 0}^\infty
        \frac{(z - a)^n}{(w - a)^{n + 1}} \dd w \\
        &= \sum_{n = 0}^\infty \left[
        \frac{1}{2\pi i} \int_{\partial D(a,
        \rho)} f(w) \cdot \frac{1}{w - a^{n + 1}}
        \dd w \right] (z - a)^n
    \end{align*}
    proving the theorem. (We swap the sum and
    integral since the partial sums give rise to a
    sequence of functions that \fcemph{converge
    uniformly} on $\partial D(a, \rho)$).
}
\end{proof}
\end{flashcard}

\subsubsection*{Remarks}

\begin{enumerate}[(1)]
    \item ``analytic'' $=$ has power series
        representation on a disk in the domain. So
        holomorphic $\implies$ analytic.
    \item holomorhic functions have derivatives of
        all orders, which are holomorphic.
\end{enumerate}

\begin{flashcard}[morera-thm]
\begin{corollary*}[Morera's Theorem]
\cloze{
    Let $D$ be a disk and $f : D \to \CC$ such
    that $\int_\gamma f = 0$ for all closed curves
    $\gamma$ in $D$. Then $f$ is holomorphic.
}
\end{corollary*}

\begin{proof}
\cloze{
    By converse of Fundamental Theorem of
    Calculus, there exists holomorphic $F$ on
    $D$ with $F' = f$. So $f$ is holomorphic.
    (Because existence of Taylor expansion implies
    that the derivative of a holomorphic function
    is holomorphic).
}
\end{proof}
\end{flashcard}
