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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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    Suppose some $P_j$ weren't 0 everywhere; how could you still get your last line?2012-08-07
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    intuitively I think that it's imposible to have that equality for all the "z" except in the case that I want to prove , but How can I prove it?2012-08-07
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    OK, as a start, suppose WLOG $P_1\neq 0$. Then $P_1\equiv -\sum_{j=2}^n jP_j \bar{z}^{j-1}$. So we've got a polynomial in $z$ equal to a polynomial in $z,\bar{z}$. Any thoughts on how we can rule this out?2012-08-07
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    @Kevin Carlson Maybe I'm bad , but thinking again, I can derivate a lot of times with respect to $ {\overline z } $ in the equality with zero, to deduce what I want to prove for example if $$ \eqalign{ & P\left( z \right) = P_0 \left( z \right) + P_1 \left( z \right)\overline z + P_2 \left( z \right)\overline z ^2 \cr & P_{\overline z } = P_1 \left( z \right) + 2P_2 \left( z \right)\overline z = 0 \cr & \left( {P_{\overline z } } \right)_{\overline z } = 0 = 2P_2 \left( z \right) \cr} $$ and using an inductive argument I have what I want2012-08-07
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    Hmm...the issue there is that a polynomial in $z$ doesn't have derivative 0 relative to $\bar{z}$: rather, that derivative doesn't exist. This is what I was going for before-that the left-hand side I'd gotten to was complex differentiable, while the right-hand side wasn't, unless both were constant.2012-08-07
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    @KevinCarlson Wait $$\frac{\partial}{\partial \bar{z}} := \frac{1}{2}( \; \frac{\partial}{\partial x} \; + \; i \frac{\partial}{\partial y}\; )$$so isn't $\frac{\partial}{\partial \bar{z}} (z) = \frac{\partial}{\partial \bar{z}} (x + iy) = \frac{1}{2}(1 - 1) = 0$? For example, that's how (I think) you get $P_{\bar{z}} = \sum_j jP_j \bar{z}^{j-1}$?2012-08-07
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    Don't be differentiating real functions with respect to a complex variable, now. If you write down the definition of the derivative of $\bar{z}$ with respect to $z$, which is more natural, you'll see it doesn't exist, simply because I can approach 0 along the real and the imaginary axes and get different results.2012-08-07
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    @KevinCarlson Thanks for responding-I think there's a slight ambiguity in notation. I'm interpreting stuff as smooth functions $\mathbb{R}^2 \rightarrow \mathbb{C}$, as in: http://en.wikipedia.org/wiki/Wirtinger_derivative Note that for an actually holomorphic function $f = u(x,y) + i v(x,y)$, it recovers the usual notion of differentiation with respect to a complex variable, namely: $$\frac{\partial}{\partial z} := \frac{1}{2}(\frac{\partial}{\partial x} \; - \; i \frac{\partial}{\partial y}) =^{C-R} \frac{\partial u}{\partial x} + i \frac{\partial v}{\partial x}$$2012-08-24
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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.