6
$\begingroup$

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!

$\endgroup$
4
  • $\begingroup$ 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. $\endgroup$
    – rubenvb
    Mar 8, 2011 at 18:36
  • $\begingroup$ get rid of firefox 4 and/or read my post meta.math.stackexchange.com/questions/1737/…. $\endgroup$
    – Fabian
    Mar 8, 2011 at 18:39
  • $\begingroup$ @Fabian: All right, gotcha :) $\endgroup$
    – rubenvb
    Mar 8, 2011 at 18:46
  • 1
    $\begingroup$ 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. $\endgroup$ Mar 8, 2011 at 18:49

1 Answer 1

10
$\begingroup$

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

$\endgroup$
6
  • $\begingroup$ That's what I thought to have remembered from my course Complex Analysis. Thanks $\endgroup$
    – rubenvb
    Mar 8, 2011 at 18:34
  • $\begingroup$ @Moron: Thanks for the adding the link. $\endgroup$
    – user17762
    Mar 8, 2011 at 18:41
  • $\begingroup$ @Siva: You are welcome :-) $\endgroup$
    – Aryabhata
    Mar 8, 2011 at 18:49
  • 2
    $\begingroup$ 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. $\endgroup$
    – Zhen Lin
    Mar 8, 2011 at 23:05
  • 1
    $\begingroup$ @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 $\mathbb{R}^2$? $\endgroup$
    – Zhen Lin
    Mar 8, 2011 at 23:31

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .