antiholomorphic

A complex function  $f:D\to ℂ$,  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 antiholomorphic in $D$.

The following conditions are equivalent (http://planetmath.org/Equivalent3):

• $f(z)$ is antiholomorphic in $D$.

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

• $f(\overline{z})$ is holomorphic in  $\overline{D}:=\{\overline{z}\mathrm{⋮}z\in D\}$.

• $f(z)$ may be to a power series ${\sum }_{n=0}^{\mathrm{\infty }}{a}_n{(\overline{z}-u)}^n$ at each  $u\in D$.

• The real part  $u(x,y)$  and the imaginary part $v(x,y)$  of the function $f$ satisfy the equations

  $$\frac{\partial u}{\partial x}=-\frac{\partial v}{\partial y},\frac{\partial u}{\partial y}=\frac{\partial v}{\partial x}.$$

  N.B. the of minus; cf. the Cauchy–Riemann equations (http://planetmath.org/CauchyRiemannEquations).
  

Example.  The function  $z\mapsto {\displaystyle \frac{1}{\overline{z}}}$ is antiholomorphic in $ℂ\setminus \{0\}$.  One has

$$f(z)=\frac{z}{{|z|}^2}=\underset{u}{\underbrace{\frac{x}{x^2+y^2}}}+i\underset{v}{\underbrace{\frac{y}{x^2+y^2}}}$$

and thus

$$\frac{\partial u}{\partial x}=\frac{y^2-x^2}{{(x^2+y^2)}^2},\frac{\partial v}{\partial y}=\frac{x^2-y^2}{{(x^2+y^2)}^2},\frac{\partial u}{\partial y}=-\frac{2xy}{{(x^2+y^2)}^2},\frac{\partial v}{\partial x}=-\frac{2xy}{{(x^2+y^2)}^2}.$$