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Just a thought that came up-- consider a polynomial $P(y)=a_ni^n(x+iy)^n+a_{n-1}i^{n-1}(x+iy)^{n-1}+...+a_0.$

Is there a way of finding some value of $y$ for which the polynomial is nonzero for any given $x$? I think a closed form expression might be possible, but after some bashing with the binomial theorem I have $P(y)=\sum_{m=0}^n\left(a_mi^m\left(\sum_{k=0}^m \frac{m!}{(m-k)!k!}x^k(iy)^{m-k}\right)\right),$ and solving for $y$ here doesn't seem too fun. Perhaps someone would know a better way to go about this? Thanks in advance.

Edit: I'm thinking this is identical to solving a polynomial of order $n$, in which case explicit solutions for the roots are impossible if $n>4$.

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    Sorry, should have been more explicit: $x,y\in \mathbb{R}$.2012-03-22

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Assuming the polynomial is not identically $0$, just take any $n+1$ different $y$'s. Since such a polynomial can't have more than $n$ roots, at least one of them is guaranteed to give you a nonzero value.

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    Thanks, I get that now. I think your method works and is completely right.2012-03-22