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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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    Offtopic: does the `\bar` command result in ugly output? In my Firefox it displays waaay to high above, mmucking with every line containing such a symbol.2011-03-08
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    get rid of firefox 4 and/or read my post http://meta.math.stackexchange.com/questions/1737/firefox-4-mathml-vs-mathjax.2011-03-08
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    @Fabian: All right, gotcha :)2011-03-08
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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

1 Answers 1

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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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    That's what I thought to have remembered from my course Complex Analysis. Thanks2011-03-08
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    @Moron: Thanks for the adding the link.2011-03-08
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    @Siva: You are welcome :-)2011-03-08
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    It just seems to be symbol-pushing though, and in "reality" $z$ and $\bar{z}$ aren't independent variables. Your equation for $\dfrac{\partial f}{\partial \bar{z}}$ is, I think, the only sensible _definition_ of the operator, but once you define it that way, it's no longer clear whether it obeys the laws of symbolic calculus you expect it to obey.2011-03-08
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    @Zhen Lin: $z$ and $\bar{z}$ are linearly independent.2011-03-08
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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