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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?

Martin Sleziak
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The Chaz 2.0
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2 Answers2

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Nothing can be said.

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

hmakholm left over Monica
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  • 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 08 '12 at 16:38
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    @TheChaz: Nothing immediately comes to mind, I'm afraid. – hmakholm left over Monica Feb 08 '12 at 16:48
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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 $f^2$.

JakeR
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  • Jake, f holomorphic was defined to be $\dfrac{\partial f}{\partial \bar z} = 0$ – The Chaz 2.0 Feb 08 '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 08 '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 08 '12 at 16:51
  • @TheChaz: Hopefully the above is more helpful to you! – JakeR Feb 08 '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 08 '12 at 17:11