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Note this is a problem 1.32 for Robert E Green, Steven G.Krantz, Function theory of one complex variable, 3rd edition: I guess their definition for a polynomial is not restricted on holomorphic function. Sorry to bother, but I found the same questions that already appeared at What can be said about a complex polynomial $f$ with $\frac{\partial}{\partial \bar z} f^2 = 0$?

For given polynomial $F : \mathbb{C} \rightarrow \mathbb{C}$, suppose \begin{align} \frac{\partial}{\partial \bar{z}} F^2=0 \end{align} Then what is the condition for $F$?

My approach :

Trivial case : $F=0$, $F=const$.

Nontrivial case.

\begin{align} 0=\frac{\partial}{\partial \bar{z}} F^2= 2 F\frac{\partial F}{\partial \bar{z}} \end{align} Thus $F$ is holomorphic function.

Is this all possible case?

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  • Since every polynomial is also holomorphic, that world imply that $F$ is constant. – Joel Cohen Sep 12 '16 at 13:39
  • @Joel: no, what you write is wrong, $F$ need not be constant at all. – Georges Elencwajg Sep 12 '16 at 13:56
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    It depends what you mean by polynomial. If $F\in \mathbb C[z]$ the equality $\frac{\partial}{\partial \bar{z}} F^2=0$ is always true and the condition is empty. If $F\in \mathbb C[x,y]$ the condition is that $F\in \mathbb C[z]$, remembering that $ \mathbb C[z] \subsetneq \mathbb C[x,y]$. For example $x^2-iy^2\in \mathbb C[x,y] \setminus \mathbb C[z]$, while $x^2-y^2+2ixy=z^2\in C[z]$ – Georges Elencwajg Sep 12 '16 at 14:07
  • @Georges Elencwajg : I guess it depends on how you interpret polynomial. From the context I had inferred it meant being in $\mathbb{C} [z]$ (which is why I said they are all holomorphic). – Joel Cohen Sep 12 '16 at 14:46
  • @Joel: once and for all whatever interpretation you choose (for example that $F(z)\in \mathbb C[z]$) you will not be able to conclude that $F$ is constant.Is it so difficult to realize that $\frac{\partial}{\partial \bar{z}} z^2=0$, but that $z$ is not constant? – Georges Elencwajg Sep 12 '16 at 19:28
  • @Georges Elencwajg : my comment was about a part of the question that has since been edited. The question used to say that F was anti-holomorphic, to which I replied that this would imply that $F$ is constant. In any case, I don't see any reason to be aggressive here. – Joel Cohen Sep 12 '16 at 23:20

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If $g:\ {\mathbb C}\to{\mathbb C}$ is a smooth function then ${\partial g\over\partial\bar z}\equiv0$ is tantamount to the fact that the real and imaginary parts of $g$ satisfy the CR equations, hence $g$ is an analytic function of $z$.

In the example at hand we now know that the polynomial $F^2$ is an entire function. The following question:

If $f^2$ is an analytic function then so is $f$

on MSE deals with the problem we are facing now. Check Georges Elencwajg's answer there.

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Christian Blatter
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