178 lines
6.5 KiB
TeX
178 lines
6.5 KiB
TeX
% !TeX root = ../../script.tex
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\documentclass[../../script.tex]{subfiles}
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\begin{document}
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\section{Complex Differentiability}
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\begin{defi}
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Let $f: U \rightarrow \cmpln$, with $U \subset \cmpln$ open. $f$ is said to be complex differentiable in $z_0 \in U$ if
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\[
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\limes{z}{z_0} \frac{f(z) - f(z_0)}{z - z_0} =: f'(z_0)
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\]
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exists. If $f$ is complex differentiable on all of $U$, $f$ is said to be holomorphic.
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A funciton that is holomorphic on all of $\cmpln$ is entire.
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An equivalent formulation wiould be
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\[
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\forall \epsilon > 0 ~\exists \delta > 0: ~~\abs{z - z_0} < \delta \implies \abs{f(z) - f(z_0) - a(z - z_0)} < \epsilon
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\]
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In this case $a = f'(z_0)$.
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\end{defi}
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\begin{thm}
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\begin{enumerate}[(i)]
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\item $f$ complex differentiable in $z_0 \in \cmpln \implies f$ continuous in $z_0$
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\item $f, g$ complex differentiable in $z_0$, then $f + g$ and $f \cdot g$ are complex differentiable in $z_0$, and
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\begin{align*}
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(f + g)'(z_0) &= f'(z_0) + g'(z_0) \\
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(fg)'(z_0) &= f'(z_0)g(z_0) + f(z_0)g'(z_0)
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\end{align*}
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If $g(z_0) \ne 0$, then $\frac{f}{g}$ is complex differentiable and
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\[
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\left(\frac{f}{g}\right)'(z_0) = \frac{g(z_0)f'(z_0) - g'(z_0)f(z_0)}{g(z_0)^2}
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\]
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\item Let $f: U \rightarrow \cmpln$, $U \subset \cmpln$ open and $C \subset \cmpln$ open with $f(U) \subset V$, and let $g: V \rightarrow \cmpln$.
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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
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$g \circ f$ is complex differentiable in $z_0$ with
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\[
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(g \circ f)'(z_0) = g'(f(z_0)) f'(z_0)
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\]
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\item If $f$ is complex differentiable in $z_0$, $f'(z_0) \ne 0$
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and if $\exists \delta > 0$ such that $f: \oball[\delta](z_0) \rightarrow U \subset \cmpln$ is bijective,
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then the inverse function $g$ is complex differentiable in $f(z_0)$, with
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\[
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g'(f(z_0)) = \rec{f'(z_0)}
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\]
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\end{enumerate}
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\end{thm}
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\begin{proof}
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\reader
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\end{proof}
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\begin{rem}[Complex vs. Real Differentiability]
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Consider $f: U \rightarrow \cmpln$, $U \subset \cmpln$ open. Let
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\begin{align*}
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x = \Re z && y = \Im z
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\end{align*}
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and define
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\[
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\tilde{U} _= \set[x + iy \in U]{(x, y) \in \realn^2}
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\]
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and
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\begin{align*}
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\tilde{f}: \tilde{U} &\longrightarrow \realn^2 \\
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(x, y) &\longmapsto (\Re(f(x + iy)), \Im(f(x + iy))) =: (u(x, y), v(x, y))
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\end{align*}
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Then $f$ is complex differentiable in $z = x + iy$.
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\begin{enumerate}[(i)]
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\item We have
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\begin{align*}
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f'(z) &= \limes{h}{0} \frac{f(z + h) - f(z)}{h} \\
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&= \limes{h}{0} \frac{u(x + h, y) + iv(x, y + h) - u(x, y) - iv(x, y)}{h} \\
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&= \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} \\
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&= \pdv{x} u(x, y) + i \pdv{x} v(x, y)
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\end{align*}
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\item And also
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\begin{align*}
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f'(z) &= \limes{h}{0} \frac{f(z + ih) - f(z)}{ih} \\
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&= -i \limes{h}{0} \frac{u(x, y + h) + iv(x, y + h) - u(x, y) - iv(x, y)}{h} \\
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&= -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} \\
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&= - \pdv{y} u(x, y) + \pdv{y}(x, y)
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\end{align*}
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\end{enumerate}
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This results in the Cauchy-Riemann equations:
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\begin{align*}
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\pdv{x} u(x, y) &= \pdv{y} v(x, y) \\
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\pdv{y} u(x, y) &= -\pdv{x} v(x, y)
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\end{align*}
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if $f$ is complex differentiable in $z = x + iy$.
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From the Cauchy-Riemann equations and the real differentiability of the function $\tilde{f}: \tilde{U} \rightarrow \realn^2$ follows
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\begin{align*}
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D\tilde{f}(x, y) = \begin{pmatrix}
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\partial_x u(x, y) & \partial_y u(x, y) \\
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\partial_x v(x, y) & \partial_y v(x, y)
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\end{pmatrix} &= \begin{pmatrix}
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\partial_x u(x, y) & -\partial_x v(x, y) \\
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\partial_x v(x, y) & \partial_x u(x, y)
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\end{pmatrix} \\
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&=: \begin{pmatrix}
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a & -b \\
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b & a
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\end{pmatrix}
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\end{align*}
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and thus for $h = (h_1, h_2) \in \realn^2$
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\begin{align*}
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\tilde{f}(x + h_1, y + h_2) - \tilde{f}(x, y) &= D\tilde{f}(x, y) h + \bigo(\abs{h}) \\
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&= \begin{pmatrix}
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ah_1 - bh_2 \\
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bh_1 + ah_2
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\end{pmatrix} + \bigo(\abs{h})
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\end{align*}
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A side calculation:
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\[
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(a +ib)(h_1 + ih_2) = ah_1 - bh_2 + i(bh_1 + ah_2)
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\]
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\[
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\implies \begin{pmatrix}
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ah_1 - bh_2 \\
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bh_1 + ah_2
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\end{pmatrix} + \bigo(\abs{h}) =
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\begin{pmatrix}
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\Re(a + ib)(h1 + ih_2) \\
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\Im(a + ib)(h_1 + ih_2)
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\end{pmatrix} + \bigo(\abs{h})
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\]
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So for $h = h_1 + ih_2$ we get
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\[
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f(z + h) - f(z) = (a + ib)h + \bigo(\abs{h})
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\]
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So $f$ is complex differentiable in $z$ with $f'(z) = a + ib$. In short, we have shown the following theorem.
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\end{rem}
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\begin{thm}
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Let $f: U \rightarrow \cmpln$ with $U \subset \cmpln$ open. $f$ is complex differentiable in $z \in U$ if and only if
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$\tilde{f}: \tilde{U} \rightarrow \realn^2$ is real differentiable in $(x, y) \in \tilde{U}$, and if the Cauchy-Riemann equations are satisfied.
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\end{thm}
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\begin{proof}
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Proof is in the previous remark.
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\end{proof}
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\begin{eg}
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\begin{enumerate}[(i)]
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\item Power series like
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\[
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f(z) = \sum_{n=0}^{\infty} a_n z^n, \quad (a_n) \subset \cmpln
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\]
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with convergence radius $\rho \in [0, \infty]$ are holomorphic on $\oball[\rho](0)$. The following holds
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\[
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f'(z) = \sum_{n=0}^{\infty} n a_n z^{n-1}
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\]
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Especially, the funciton
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\[
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f(z) = e^{\alpha z}, \quad \alpha \in \cmpln
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\]
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is holomorphic on all of $\cmpln$ with
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\[
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f'(z) = \alpha e^{\alpha z}
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\]
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\item The function
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\[
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f(z) = \frac{1}{z^n}
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\]
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is holomorphic $\cmpln \setminus \set{0}$ with
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\[
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f'(z) = -n \rec{z^{n+1}}
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\]
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\item Functions that are not complex differentiable include
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\begin{align*}
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f(z) = \conj{z} && f(z) = z\conj{z} \\
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(\partial_x u = 1 \ne \partial_y v = -1) && (\partial_x u = 2x^2 \ne \partial_y v = 0)
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\end{align*}
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\end{enumerate}
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\end{eg}
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\end{document} |