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Say $z \in \mathbb{C}$ and $\bar{z}$ the complex conjugate (i.e. with $\bar{z} z = \left|z \right|^2$).

Can a function of $z$ and $\bar{z}$ be analytical?

Example: $f(z,\bar{z}) = Az^3 + B \bar{z} z$

I thought no, because the partial derivatives will depend on the direction in the complex plane (i.e. the phase of the line along which you take the derivative limit).

Thanks!

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    If $f(z)=Az^3+B\overline{z}z$ were analytic, then $g(z)=\frac{1}{B}(f(z)-Az^3)=|z|^2$ would be. Or, away from $0$, $h(z)=\frac{1}{z}g(z)=\overline{z}$ would be.2011-03-08

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One of the many equivalent definitions for a function to be holomorphic is $\displaystyle \frac{\partial f}{\partial \bar{z}} = 0$

$\displaystyle \frac{\partial f}{\partial \bar{z}} = 0$ is equivalent to Cauchy Riemann equations as shown below.

$x = \frac{z+\bar{z}}{2} \text{ and } y = \frac{z-\bar{z}}{2i}$

$\frac{\partial f}{\partial \bar{z}} = \frac{\partial f}{\partial x} \frac{\partial x}{\partial \bar{z}} + \frac{\partial f}{\partial y} \frac{\partial y}{\partial \bar{z}} = \frac{1}{2} \left( \frac{\partial f}{\partial x} + i \frac{\partial f}{\partial y} \right)$

So if $f = u(x,y) + i v(x,y)$, where $u,v: \mathbb{R} \times \mathbb{R} \rightarrow \mathbb{R}$, then $\frac{\partial f}{\partial x} = \frac{\partial u}{\partial x} + i \frac{\partial v}{\partial x}$ $\frac{\partial f}{\partial y} = \frac{\partial u}{\partial y} + i \frac{\partial v}{\partial y}$

Hence,$\frac{\partial f}{\partial \bar{z}} = \frac{1}{2} \left( \frac{\partial u}{\partial x} - \frac{\partial v}{\partial y} \right) + \frac{i}{2} \left( \frac{\partial u}{\partial y} + \frac{\partial v}{\partial x} \right)$

Hence, you find that $\left( \frac{\partial f}{\partial \bar{z}} = 0 \right) \iff \left(\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} \text{ and } \frac{\partial u}{\partial y} = - \frac{\partial v}{\partial x} \right)$

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    @Sivaram: Yes, they are linearly independent over $\mathbb{R}$. But then how do you justify differentiating with respect to them, if we're treating them as vectors in $\mathbb{R}^2$?2011-03-08