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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!

rubenvb
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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. – rubenvb Mar 08 '11 at 18:36
  • get rid of firefox 4 and/or read my post http://meta.math.stackexchange.com/questions/1737/firefox-4-mathml-vs-mathjax. – Fabian Mar 08 '11 at 18:39
  • @Fabian: All right, gotcha :) – rubenvb Mar 08 '11 at 18:46
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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. – Jonas Meyer Mar 08 '11 at 18:49

1 Answers1

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

Community
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  • That's what I thought to have remembered from my course Complex Analysis. Thanks – rubenvb Mar 08 '11 at 18:34
  • @Moron: Thanks for the adding the link. –  Mar 08 '11 at 18:41
  • @Siva: You are welcome :-) – Aryabhata Mar 08 '11 at 18:49
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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. – Zhen Lin Mar 08 '11 at 23:05
  • @Zhen Lin: $z$ and $\bar{z}$ are linearly independent. –  Mar 08 '11 at 23:09
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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 $\mathbb{R}^2$? – Zhen Lin Mar 08 '11 at 23:31