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Let p be a complex valued polynomial of two ral variables: $ \sum {a_{ij} x^i y^j } $ write: $p(z)= \sum {P_j \overline z } ^j $ where each $P_j$ is of the form $ P_j = \sum {b_{ij} z^i } $ Prove that p is an entire function if and only iff $ P_j \equiv 0 $

Clearly I have to consider the "derivate" $ P_{\overline z } = \frac{1} {2}\left( {P_x + iP_y } \right) $ , since the real and imaginary part are $C^1$ functions, being holomorphic it's equivalent of satisfy C.R , or equivalently $ P_{\overline z } = 0 $ in this case $ P_{\overline z } = \sum {jP_j \overline z ^{j - 1} } = 0 $ and now what can I do?

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    which is how you would calculate the regular complex derivative at $z_0$ if you took sequences coming in of the form $z_0 + r_i$, for $r_i$ real. But yeah, I agree that $\bar{z}$ isn't holomorphic for the reason you say.2012-08-24

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Hey Daniel I think your inductive argument makes sense-this is me just writing it out to make sure I understand, more or less :p. Namely, induct on the highest power of $j$ present in $P$. The claim is that $P_{\bar{z}} = 0 \Rightarrow P_j = 0 \in \mathbb{C}[z]$ for $j \geqslant 1$.

For $j = 1$, our equation $P_{\bar{z}}$ reads $P_1 = 0$ as a function, so this is the statement that if $p \in \mathbb{C}[z]$ evaluates to 0 everywhere, then $p$ is the 0 polynomial. Indeed, were it of degree $n$, for $n > 0$, it could have at most $n$ distinct roots, and $\mathbb{C}$ is infinite.

Inductively, to see that $j' < j \Rightarrow j$, suppose we have $\frac{\partial}{\partial \bar{z}} \sum_{i \leqslant j} P_i \bar{z}^i = \sum_{i \leqslant j} i P_i \bar{z}^{i-1} = 0 \quad \in Fun(\mathbb{C}, \mathbb{C})$Pulling the highest power to the other side, we have $-jP_{j} \bar{z}^{j - 1} = \sum_{i < j} i P_i \bar{z}^{i - 1}$Now apply $\frac{\partial^{j - 1}}{\partial \bar{z}^{j - 1}} \; \; \; $ to both sides, we get (up to some factorials $C$) that $C P_j = 0 \Rightarrow P_j = 0$by our inductive hypothesis we deduce that $P_i = 0$ for $i < j$ as well.