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I am working on a project regarding the number of zeros of a harmonic polynomial and am stuck with the proof of this: The zero set of the harmonic polynomial h(z)= $z^n$ - $\bar{z}^n$ consists of n equally spaced lines through the origin. Any suggestion is appreciated. Thanks.

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You might want to write $z = re^{i \theta}$ with $r, \theta \in \mathbb{R}$; this should help. –  Akhil Mathew Mar 15 '11 at 17:22
    
thanks Akhil! :-) –  user8301 Mar 15 '11 at 17:40

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if $z=re^{i\theta}$ then this reduces to $e^{2ni\theta}=1$. the solutions to this are $\theta=\pi k/n$. so the lines $\{re^{k\pi i/n} : r\in(-\infty,\infty)\}, k=0,1,...n-1$ are the zero set.

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Hi yoyo, thanks for breaking it down. For some reason, the modulus and phase conversion didn't strike me. Thanks again. –  user8301 Mar 15 '11 at 17:41

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