# Why are analytic functions functions of $z$ and not of $\bar{z}$?

I was reading a note on complex analysis and was stuck on one line:

Cauchy-Riemann equations

Alternatively, the C-R equations can be written as $$\frac{\partial f}{\partial\bar z}=0,$$ where $z=x+iy$ and $\bar z=x-iy$.

In some sense, analytic functions are truly functions of $z$, and not of $\bar z$. $$\begin{array}{c} \frac\partial{\partial z} = \frac 1 2 \left(\frac\partial{\partial x}-i\frac\partial{\partial y}\right) &\frac\partial{\partial \bar z}=\frac 1 2 \left(\frac\partial{\partial x}+i\frac\partial{\partial y}\right) \end{array}$$

I do not understand what it means that "Analytic functions are truly functions of $z$ and not of $\bar{z}$." Can someone explain? Thanks in advance for your time.

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In the future, please include a link to the actual PDF you're asking about, when possible. –  dfeuer Jul 28 '13 at 16:58
A related question. –  user26872 Jul 28 '13 at 18:19
Dear learner, I just want to remark that you can get a genuinely rigorous interpretation of this statement by understanding Ted Schifrin's answer below. Basically, for complex valued functions of $x$ and $y$ that are analytic (i.e. can be expressed as power series in $x$ and $y$), you can instead think of them as power series in $z$ and $\overline{z}$ (by making the obvious change of variables). The complex-analytic functions are then precisely the ones that don't involve $\overline{z}$. Regards, –  Matt E Jul 29 '13 at 15:09
@MattE thanks a lot sir for your interpretation. –  learner Jul 30 '13 at 8:22

This is to be taken "in some sense" or as a mnemonic for CR. As $\frac{\partial f}{\partial \bar z}=0$ suggests, the function value does not change when $\bar z$ changes. Of course this is nonsense insofar as if you know $\bar z$ the you know $z$ and then you can obtain the (changing) value of $f$, but then again a function such as $z\mapsto az+b\bar z+c$ is analytic if and only if $b=0$, that is, there is only a part depending on $z$ itself and the possible part depending on $\bar z$ must be constant.

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Editing slightly (thanks @Matt E): An alternative approach to give meaning to the sentence is this. A complex-valued real analytic function $f$ can be expanded in a power series in $x$ and $y$, hence, by substitution, in $z$ and $\bar z$. $f$ is complex analytic precisely when no $\bar z$'s appear in the latter expression. For example, $f(x,y)=(x^2-y^2)+2ixy = z^2$ is, of course, complex analytic, but $f(x,y)=(x^2+y^2)-2ixy = z\bar z +\frac12(\bar z^2-z^2)$ is not. Of course, this is equivalent to the calculation $\partial f/\partial\bar z \overset{?}{=} 0$.

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Dear Ted, I think that this is the best answer to this question. Perhaps you could amplify it by pointing out that $f$ should be taken to be a complex-valued real analytic function, just to make it clear why the substitution of $z$ and $\overline{z}$ in place of $x$ and $y$ is valid. Regards, –  Matt E Jul 29 '13 at 15:06

The question reflects a reasonable confusion, based on word-use, as partly reflected in @HagenvonEitzen's answer. That is, to say that $f$ "is a function of" $z$ is that knowing $z$ determines $f(z)$. But this colloquial (and often accurate!) sense is too broad. Indeed, if we know $z$, we know $\bar{z}$, and vice-versa, so the broader sense "$f$ is a function of $z$, not of $\bar{z}$" would be self-contradictory, indeed!

A more technical sense of "is not a function of" is that the derivative vanishes, the idea being that if a function does not change when "a variable" changes, it does not "depend" on that variable. (This is still vague!) Further, in the case at hand, the "definitions" of derivative-with-respect-to-$z$ and $\bar{z}$ are arguably partly artificial, so this interpretation is even more tenuous.

However, as in HagenvonEitzen's linear example, experiments with polynomials suggest that this derivative interpretation is correct.

Indeed, a fancier interpretation about "complexifying the tangent bundle" shows that there really is an underlying sense to this interpretation, although, admittedly, at the point it arises in one's education it's hard to get a grip on.

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Suppose that

$$f(z) = |z|^2 = z\cdot \bar{z}$$

$$\frac{df(z)}{d\bar{z}} = z$$

which isn't identically $0$ hence $f(z)$ is analytic nowhere because defining it to be analytic at $z=0$ is absurd . analytic functions are only defined on open sets .

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I think you mean $df/d\overline{z} = z$. –  user61527 Jul 28 '13 at 17:03
That's not quite right: rather, $d|z|^2/dz=\bar{z}$. –  paul garrett Jul 28 '13 at 17:04
Thanks I forgot the bar :/ –  Zaid Alyafeai Jul 28 '13 at 18:10