Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

The hypothesis is similar to that of a previous question of mine, namely that we have a complex polynomial $f: \mathbb{C} \to \mathbb{C}$ with $\dfrac{\partial}{\partial \bar z} f^2 = 0$ . (In the previous question, $f$ was not squared, and the partial w.r.t. $z$ was also $0$)

What can be said about f?

At first glance, it seems that the square of $f$ doesn't depend on $\bar z$ (i.e. can be written in terms of $z$ alone). Is that all there is to this question? Are there some analytic/computational equivalent statements that I should be considering instead?


Added: I have tried to explicitly calculate $f^2$ and take its partial with respect to $\bar z$, but that doesn't seem to get me anywhere. Is there a less computational approach?

share|improve this question
    
Dear The Chaz: Have you tried anything? –  Pierre-Yves Gaillard Feb 8 '12 at 16:09
    
@Pierre: Well, I didn't know if making such a claim (as in the body of the question) would suffice. I'll add to the question. –  The Chaz 2.0 Feb 8 '12 at 16:12
    
Dear The Chaz: I see that you have some answers now. I see two interpretations of the question. It depends what you mean by "complex polynomial". Is it $\mathbb C[z]$ or $\mathbb C[z,\overline z]$? (Answerers have chosen the first.) (In any event, I'd have answered completely differently.) –  Pierre-Yves Gaillard Feb 8 '12 at 16:20
    
@Pierre: The proposition that I reference in my previous question states that a polynomial in $z, \bar z$ can be written in terms of $z$ alone iff $\dfrac{\partial f}{\partial \bar z} \equiv 0$ –  The Chaz 2.0 Feb 8 '12 at 16:42
    
Okay, editing accordingly. –  JakeR Feb 8 '12 at 16:51
add comment

2 Answers 2

up vote 2 down vote accepted

Well, if a complex-valued function $g$ satisfies $\dfrac{\partial}{\partial \bar z} g = 0$, we know that $\dfrac{\partial g}{\partial x} + i\dfrac{\partial g}{\partial y} = 0$ (nearly exactly from the definition of $\dfrac{\partial}{\partial \bar z}$) and so with a small bit of manipulation, we see that $i(\dfrac{\partial g}{\partial x} + i\dfrac{\partial g}{\partial y}) = i\dfrac{\partial g}{\partial x} - \dfrac{\partial g}{\partial y} = 0$ and $\dfrac{\partial g}{\partial y} = i\dfrac{\partial g}{\partial x}$.

This is an equivalent formulation of the Cauchy-Riemann equations, which tells us our function $g$ is holomorphic and equivalently analytic. In fact, a convenient test for holomorphicity is to check if $\dfrac{\partial}{\partial \bar z} = 0$. So we know that our function $f^2$ is holomorphic, but all polynomials (and their squares, of course) are naturally holomorphic! Hence we could have concluded $\dfrac{\partial}{\partial \bar z}f = 0$ from our original hypothesis. We have not actually constrained our function any further.

EDIT: Given your previous approach, a helpful fact to establish is that the product rule is valid for $\dfrac{\partial}{\partial \bar z}$. Then $\dfrac{\partial f^2}{\partial \bar z} = 2f\dfrac{\partial f}{\partial \bar z}$, and so if $\dfrac{\partial f^2}{\partial \bar z} = 0$ and our polynomial isn't $0$ everywhere, then $\dfrac{\partial f}{\partial \bar z} = 0$ and we can write $f$ solely in terms of $z$, hence we can do the same for $f^2$.

share|improve this answer
    
Jake, f holomorphic was defined to be $\dfrac{\partial f}{\partial \bar z} = 0$ –  The Chaz 2.0 Feb 8 '12 at 16:39
    
Well, the conclusion is then even quicker! Unless you mean a polynomial with powers in both $z$ and $\bar z$? Then there is a bit of computation that can be done. I'm not quite sure why I understand why there's anything to the question in this case where the polynomial is in $z$, if computational approaches are discouraged. –  JakeR Feb 8 '12 at 16:45
    
Jake, please see my comment to Pierre, above. I have no preference about the computational approach vs non-computational approach; I just seemed to be getting a "non conclusion" in either case! –  The Chaz 2.0 Feb 8 '12 at 16:51
    
@TheChaz: Hopefully the above is more helpful to you! –  JakeR Feb 8 '12 at 16:56
    
Yes, thank you. That seems like a reasonable solution (based on what the reader is "supposed to know" at this point in the book!) –  The Chaz 2.0 Feb 8 '12 at 17:11
add comment

Nothing can be said.

All complex polynomials, and their squares, are complex differentiable and therefore have zero $\frac{\partial}{\partial\overline z}$.

share|improve this answer
    
Henning, the next section of the book discussed complex differentiability (which is equivalent to being holomorphic, right?). Would you care to speculate as to what the authors' intent was in asking this question before showing such results? –  The Chaz 2.0 Feb 8 '12 at 16:38
1  
@TheChaz: Nothing immediately comes to mind, I'm afraid. –  Henning Makholm Feb 8 '12 at 16:48
    
Thanks for the answers. –  The Chaz 2.0 Feb 8 '12 at 16:50
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.