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<record version="3" id="11476">
 <title>antiholomorphic</title>
 <name>Antiholomorphic2</name>
 <created>2009-01-07 13:42:28</created>
 <modified>2009-01-07 15:57:31</modified>
 <type>Definition</type>
<parent id="6226">complex function</parent>
 <creator id="2872" name="pahio"/>
 <author id="2872" name="pahio"/>
 <classification>
	<category scheme="msc" code="30A99"/>
 </classification>
 <synonyms>
	<synonym concept="antiholomorphic" alias="antiholomorphic function"/>
 </synonyms>
 <related>
	<object name="ComplexConjugate"/>
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 <content>\PMlinkescapeword{iff}

A complex function \,$f\!: D \to \mathbb{C}$,\, where $D$ is a domain of the complex plane, having the derivative 
$$\frac{df}{d \overline{z}}$$
in each point $z$ of $D$, is said to be {\em antiholomorphic} in $D$.\\

The following conditions are \PMlinkname{equivalent}{Equivalent3}:
\begin{itemize}

\item $f(z)$ is antiholomorphic in $D$.

\item \, $\overline{f(z)}$\, is holomorphic in $D$. 

\item $f(\overline{z})$ is holomorphic in\, $\overline{D} \,:=\, \{\overline{z}\;\vdots\;\, z \in D\}$.

\item $f(z)$ may be \PMlinkescapetext{expanded} to a power series $\sum_{n=0}^\infty a_n(\overline{z}-u)^n$ at each\, $u \in D$.

\item The real part \,$u(x,\,y)$\, and the imaginary part \,$v(x,\,y)$\, of the function $f$ satisfy the equations
$$\frac{\partial u}{\partial x} \;=\; -\frac{\partial v}{\partial y}, \qquad 
  \frac{\partial u}{\partial y} \;=\; \frac{\partial v}{\partial x}.$$
N.B. the \PMlinkescapetext{place} of minus; cf. the \PMlinkname{Cauchy--Riemann equations}{CauchyRiemannEquations}.\\

\end{itemize}

\textbf{Example.}\, The function\, $\displaystyle z \mapsto \frac{1}{\overline{z}}$ is antiholomorphic in\, 
$\mathbb{C}\!\smallsetminus\!\{0\}$.\, One has 
$$f(z) \,=\, \frac{z}{|z|^2} \,=\, \underbrace{\frac{x}{x^2+y^2}}_{u}+i\underbrace{\frac{y}{x^2+y^2}}_{v},$$
$$\frac{\partial u}{\partial x} \;=\; \frac{y^2\!-\!x^2}{(x^2\!+\!y^2)^2}, \quad
\frac{\partial v}{\partial y} \;=\; \frac{x^2\!-\!y^2}{(x^2\!+\!y^2)^2}, \quad 
\frac{\partial u}{\partial y} \;=\; -\frac{2xy}{(x^2\!+\!y^2)^2}, \quad
\frac{\partial v}{\partial x} \;=\; -\frac{2xy}{(x^2\!+\!y^2)^2}.$$

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