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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).


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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 8 '11 at 18:36
get rid of firefox 4 and/or read my post meta.math.stackexchange.com/questions/1737/…. –  Fabian Mar 8 '11 at 18:39
@Fabian: All right, gotcha :) –  rubenvb Mar 8 '11 at 18:46
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 8 '11 at 18:49
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1 Answer

up vote 10 down vote accepted

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. Thanks –  rubenvb Mar 8 '11 at 18:34
@Moron: Thanks for the adding the link. –  user17762 Mar 8 '11 at 18:41
@Siva: You are welcome :-) –  Aryabhata Mar 8 '11 at 18:49
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 8 '11 at 23:05
@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 8 '11 at 23:31
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