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Note this is a problem 1.31 for Robert E Green, Steven G.Krantz, Function theory of one complex varialbe, 3rd edition : I guess their definition for polynomial is not restricted on holomorphic function.

Let $F: \mathbb{C} \rightarrow \mathbb{C}$ is a polynomial then, what is $ \frac{\partial^2}{\partial z^2} F=0$ means?

One can i simply guess from calculus is that $F$ is a linear function of $z$ and any arbitrary function for $\bar{z}$, but i couldn't guess the proof for complex variable theory.

My guess \begin{align} \frac{\partial^2}{\partial x^2} f=0, \quad \Rightarrow \quad f= ax+b \end{align} and since $\bar{z}$ parts doesn't contribute i can just multiply overall factor, $g(\bar{z})$ and guess the solution as \begin{align} f = (az+b) g(\bar{z}) = z g(\bar{z}) + h(\bar{z}) \end{align}

But i am not sure about its mathematical rigours

phy_math
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    For me, a polynomial function on $\mathbb{C}$ does not contain terms of the form $\bar{z}$, but only $z$. – b00n heT Sep 12 '16 at 14:10
  • True, check Cauchy-Riemann for holomorphic functions. By definition $\delta_z=\frac 1 2 (\delta_x-i\delta_y)$, so $\delta²_z=\Delta$, and your function is harmonic, it can't use $\bar z$. – marmouset Sep 12 '16 at 14:13

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