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\section{Complex Differentiability}

\begin{defi}
    Let $f: U \rightarrow \cmpln$, with $U \subset \cmpln$ open. $f$ is said to be complex differentiable in $z_0 \in U$ if 
    \[
        \limes{z}{z_0} \frac{f(z) - f(z_0)}{z - z_0} =: f'(z_0)
    \]
    exists. If $f$ is complex differentiable on all of $U$, $f$ is said to be holomorphic. 
    A funciton that is holomorphic on all of $\cmpln$ is entire.

    An equivalent formulation wiould be 
    \[
        \forall \epsilon > 0 ~\exists \delta > 0: ~~\abs{z - z_0} < \delta \implies \abs{f(z) - f(z_0) - a(z - z_0)} < \epsilon
    \]
    In this case $a = f'(z_0)$.
\end{defi}

\begin{thm}

    \begin{enumerate}[(i)]
        \item $f$ complex differentiable in $z_0 \in \cmpln \implies f$ continuous in $z_0$ 
        \item $f, g$ complex differentiable in $z_0$, then $f + g$ and $f \cdot g$ are complex differentiable in $z_0$, and 
        \begin{align*}
            (f + g)'(z_0) &= f'(z_0) + g'(z_0) \\
            (fg)'(z_0) &= f'(z_0)g(z_0) + f(z_0)g'(z_0) 
        \end{align*}
        If $g(z_0) \ne 0$, then $\frac{f}{g}$ is complex differentiable and 
        \[
            \left(\frac{f}{g}\right)'(z_0) = \frac{g(z_0)f'(z_0) - g'(z_0)f(z_0)}{g(z_0)^2}
        \]
        \item Let $f: U \rightarrow \cmpln$, $U \subset \cmpln$ open and $C \subset \cmpln$ open with $f(U) \subset V$, and let $g: V \rightarrow \cmpln$.
        Then $g \circ f: U \rightarrow \cmpln$. If $f$ is complex differentiable in $z_0$, and $g$ is complex differentiable in $f(z_0)$, then 
        $g \circ f$ is complex differentiable in $z_0$ with 
        \[
            (g \circ f)'(z_0) = g'(f(z_0)) f'(z_0)
        \]
        \item If $f$ is complex differentiable in $z_0$, $f'(z_0) \ne 0$
        and if $\exists \delta > 0$ such that $f: \oball[\delta](z_0) \rightarrow U \subset \cmpln$ is bijective, 
        then the inverse function $g$ is complex differentiable in $f(z_0)$, with 
        \[
            g'(f(z_0)) = \rec{f'(z_0)}
        \]
    \end{enumerate}
\end{thm}
\begin{proof}
    \reader
\end{proof}

\begin{rem}[Complex vs. Real Differentiability]
    Consider $f: U \rightarrow \cmpln$, $U \subset \cmpln$ open. Let 
    \begin{align*}
        x = \Re z && y = \Im z
    \end{align*}
    and define 
    \[
        \tilde{U} _= \set[x + iy \in U]{(x, y) \in \realn^2}
    \]
    and 
    \begin{align*}
        \tilde{f}: \tilde{U} &\longrightarrow \realn^2 \\
        (x, y) &\longmapsto (\Re(f(x + iy)), \Im(f(x + iy))) =: (u(x, y), v(x, y))
    \end{align*}
    Then $f$ is complex differentiable in $z = x + iy$.
    \begin{enumerate}[(i)]
        \item We have 
        \begin{align*}
            f'(z) &= \limes{h}{0} \frac{f(z + h) - f(z)}{h} \\
            &= \limes{h}{0} \frac{u(x + h, y) + iv(x, y + h) - u(x, y) - iv(x, y)}{h} \\
            &= \limes{h}{0} \frac{u(x + h, y) - u(x, y)}{h} + i \limes{h}{0} \frac{v(x + h, y) - v(x, y)}{h} \\
            &= \pdv{x} u(x, y) + i \pdv{x} v(x, y)
        \end{align*}

        \item And also 
        \begin{align*}
            f'(z) &= \limes{h}{0} \frac{f(z + ih) - f(z)}{ih} \\
            &= -i \limes{h}{0} \frac{u(x, y + h) + iv(x, y + h) - u(x, y) - iv(x, y)}{h} \\ 
            &= -i \limes{h}{0} \frac{u(x, y + h) - u(x, y)}{h} + \limes{h}{0} \frac{v(x, y + h) - v(x, y)}{h} \\
            &= - \pdv{y} u(x, y) + \pdv{y}(x, y)
        \end{align*}
    \end{enumerate}
    This results in the Cauchy-Riemann equations:
    \begin{align*}
        \pdv{x} u(x, y) &= \pdv{y} v(x, y) \\
        \pdv{y} u(x, y) &= -\pdv{x} v(x, y)
    \end{align*}
    if $f$ is complex differentiable in $z = x + iy$.

    From the Cauchy-Riemann equations and the real differentiability of the function $\tilde{f}: \tilde{U} \rightarrow \realn^2$ follows
    \begin{align*}
        D\tilde{f}(x, y) = \begin{pmatrix}
            \partial_x u(x, y) & \partial_y u(x, y) \\
            \partial_x v(x, y) & \partial_y v(x, y)
        \end{pmatrix} &= \begin{pmatrix}
            \partial_x u(x, y) & -\partial_x v(x, y) \\
            \partial_x v(x, y) & \partial_x u(x, y)
        \end{pmatrix} \\
        &=: \begin{pmatrix}
            a & -b \\ 
            b & a
        \end{pmatrix}
    \end{align*}
    and thus for $h = (h_1, h_2) \in \realn^2$ 
    \begin{align*}
        \tilde{f}(x + h_1, y + h_2) - \tilde{f}(x, y) &= D\tilde{f}(x, y) h + \bigo(\abs{h}) \\
        &= \begin{pmatrix}
            ah_1 - bh_2 \\ 
            bh_1 + ah_2
        \end{pmatrix} + \bigo(\abs{h})
    \end{align*}
    A side calculation:
    \[
        (a +ib)(h_1 + ih_2) = ah_1 - bh_2 + i(bh_1 + ah_2)
    \]
    \[
        \implies \begin{pmatrix}
            ah_1 - bh_2 \\
            bh_1 + ah_2
        \end{pmatrix} + \bigo(\abs{h}) = 
        \begin{pmatrix}
            \Re(a + ib)(h1 + ih_2) \\
            \Im(a + ib)(h_1 + ih_2)
        \end{pmatrix} + \bigo(\abs{h})
    \]
    So for $h = h_1 + ih_2$ we get 
    \[
        f(z + h) - f(z) = (a + ib)h + \bigo(\abs{h})
    \]
    So $f$ is complex differentiable in $z$ with $f'(z) = a + ib$. In short, we have shown the following theorem.
\end{rem}

\begin{thm}
    Let $f: U \rightarrow \cmpln$ with $U \subset \cmpln$ open. $f$ is complex differentiable in $z \in U$ if and only if
    $\tilde{f}: \tilde{U} \rightarrow \realn^2$ is real differentiable in $(x, y) \in \tilde{U}$, and if the Cauchy-Riemann equations are satisfied.
\end{thm}
\begin{proof}
    Proof is in the previous remark.
\end{proof}

\begin{eg}
    \begin{enumerate}[(i)]
        \item Power series like 
        \[
            f(z) = \sum_{n=0}^{\infty} a_n z^n, \quad (a_n) \subset \cmpln
        \]
        with convergence radius $\rho \in [0, \infty]$ are holomorphic on $\oball[\rho](0)$. The following holds 
        \[
            f'(z) = \sum_{n=0}^{\infty} n a_n z^{n-1}
        \]
        Especially, the funciton
        \[
            f(z) = e^{\alpha z}, \quad \alpha \in \cmpln
        \]
        is holomorphic on all of $\cmpln$ with 
        \[
            f'(z) = \alpha e^{\alpha z}
        \]

        \item The function 
        \[
            f(z) = \frac{1}{z^n}
        \]
        is holomorphic $\cmpln \setminus \set{0}$ with 
        \[
            f'(z) = -n \rec{z^{n+1}}
        \]

        \item Functions that are not complex differentiable include 
        \begin{align*}
            f(z) = \conj{z} && f(z) = z\conj{z} \\
            (\partial_x u = 1 \ne \partial_y v = -1) && (\partial_x u = 2x^2 \ne \partial_y v = 0)
        \end{align*}
    \end{enumerate}
\end{eg}
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