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I'm currently learning about complex analysis, and I keep coming across expressions involving $\frac{\partial f}{\partial \bar{z}}$. But I don't understand what this means. For example people might write $$\frac{\partial }{\partial \bar{z}}z\bar{z}=z$$ But how does that make any sense? Why can we regard $z$ and $\bar{z}$ as independent variable while they clearly aren't?

In some material I've also read that $\frac{\partial f}{\partial \bar{z}}=0$ can be interpreted as meaning that $f$ is 'independent' of $\bar{z}$. But $f$ depends on $z$, and $z$ can be thought of as 'depending' on $\bar{z}$, so clearly $f$ depends as much on $\bar{z}$ as it does on $z$.

This $\frac{\partial f}{\partial \bar{z}}=0$ condition, whatever it even means, seems to also be equivalent to $f$ being analytic, which then leads to people saying things like 'If $f$ can be expressed without involving $\bar{z}$ then it is analytic'. But this again seems very strange to me. For example $$f(z)=\bar{z}=z-2\Im(z)$$ Can certainly 'be expressed' without explicitly involving $\bar{z}$, but is not analytic, so this also doesn't make any sense to me.

Basically I'm asking for clarification on why in complex analysis people seem to be doing these dodgy things, and what they mean by them.

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    $\begingroup$ The imaginary part is $\dfrac{z-\overline{z}}{2i}$. So you have used $\overline{z}$. It's just hiding behind another name. $\endgroup$ – Matt Samuel May 5 '15 at 19:34
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    $\begingroup$ Dodgy, lol, indeed it is a bit weird at the beginning. But soon you'll be drinking the kool-aid. $\endgroup$ – zhw. May 5 '15 at 19:34
  • $\begingroup$ @MattSamuel Oke yes of course. But maybe something similar is going on with every function that 'seems' to only be in $z$. Maybe we can write $z^2$ as a function not involving $z$ at all and only involving $\bar{z}$. Just like we might also through some complicated method be able to write $\bar{z}$ as a function in $z$ alone. I'm just not seeing how this criterion works. $\endgroup$ – user2520938 May 5 '15 at 19:39
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    $\begingroup$ It's not complicated, to write $\overline{z}$ in terms of $z$ just use the conjugation function. But there are no algebraic functions (i.e. quotients of polynomials) that can express one in terms of the other. This doesn't really answer your question but it gives some evidence that they can be treated as independent. $\endgroup$ – Matt Samuel May 5 '15 at 19:42
  • $\begingroup$ write the differentiation operators as linear combinations of $d_x$ and $d_y$ and apply them to some functions like $z, z^*,|z|^2$ you will see that everything is consistent $\endgroup$ – tired May 5 '15 at 20:11
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$\newcommand{\dd}{\partial}$The complex differential operators may be defined as complex linear combinations of ordinary partial derivatives: $$ \frac{\dd}{\dd z} = \frac{1}{2}\left(\frac{\dd}{\dd x} - i\frac{\dd}{\dd y}\right),\qquad \frac{\dd}{\dd \bar{z}} = \frac{1}{2}\left(\frac{\dd}{\dd x} + i\frac{\dd}{\dd y}\right). \tag{1} $$ This is what the chain rule would give if you treated $z = x + iy$ and $\bar{z} = x - iy$ as independent (so that $x = \frac{1}{2}(z + \bar{z})$ and $y = \frac{1}{2i}(z - \bar{z})$).

Then it's just a matter of formalities to check that $$ \frac{\dd z}{\dd z} = \frac{\dd \bar{z}}{\dd \bar{z}} = 1,\qquad \frac{\dd \bar{z}}{\dd z} = \frac{\dd z}{\dd \bar{z}} = 0 $$ and $$ \frac{\dd}{\dd z} f(z, \bar{z}) = D_{1}f(z, \bar{z}),\qquad \frac{\dd}{\dd \bar{z}} f(z, \bar{z}) = D_{2}f(z, \bar{z}), $$ i.e., that a function expressed in terms of $z$ and $\bar{z}$ can be (formally) differentiated using the standard manipulations as if $z$ and $\bar{z}$ were independent variables.


If you must know, there's a rigorous construction (well-known to all students of complex geometry) that involves complexifying the tangent bundle of $\mathbf{R}^{2}$—equipped with the "complex structure" $J$ that rotates each tangent plane a quarter turn—and splitting the complexified tangent bundle into eigenbundles of (the natural extension of) $J$. The formulas (1) then define a complex-linear isomorphism between $(T\mathbf{R}^{2}, J)$ and the $i$-eigenbundle of $J$, and an anti-linear isomorphism with the $(-i)$-eigenbundle.

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  • $\begingroup$ Thanks for this answer. After reading this I have another question though. Why is it that we can talk about $f$ as a function in $z$ and $\bar{z}$? Why is it that we cannot 'reduce' a function in $\bar{z}$ to a function in $z$ alone? Because for $D_1 f(z,\bar{z})$ to 'make sense' (ie be uniquely determined) would require it to be impossible to rewrite any expression involving $\bar{z}$ into one in $z$ alone right? $\endgroup$ – user2520938 May 5 '15 at 20:34
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    $\begingroup$ I'm guessing the root cause of your unease is that "$z$ determines both $x$ and $y$, so isn't $\bar{z}$ redundant?" Here are a couple of observations that may help make the $\bar{z}$ formalism more acceptable: 1. The fact that "the partial of $\bar{z}$ with respect to $z$" vanishes identically means "$\bar{z}$ is independent of (or constant with respect to) $z$". 2. The (linearly independent) sets $\{x, y\}$ and $\{z, \bar{z}\}$, viewed as complex-valued functions on $\mathbf{C}$, span the same two-dimensional complex vector space. $\endgroup$ – Andrew D. Hwang May 5 '15 at 22:37

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