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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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Are you assuming the $x,y$ to be real or complex? If it is the latter, there is no real (no pun intended) point in writing $x+iy$: for any fixed $x$, the value of $x+iy$ could be any complex number, and one can easily convert between $y$ and $x+iy$ and back. –  Marc van Leeuwen Mar 22 '12 at 8:02
    
Sorry, should have been more explicit: $x,y\in \mathbb{R}$. –  Peter C. Mar 22 '12 at 8:48

1 Answer 1

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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Yes, that's certainly true, but I'm thinking if it would be possible to come up with a closed form expression for $y$... –  Peter C. Mar 22 '12 at 7:39
    
Well, $$|P(y)| \ge |a_n| |x+iy|^n - \sum_{j=1}^{n-1} |a_j| |x+iy|^j \ge |a_n| |x + i y|^n - \left(\sum_{j=0}^{n-1} |a_j|\right) |x + iy|^{n-1}$$ if $|x+iy|>1$, so take any $y > \min\left(1, \sum_{j=0}^{n-1} |a_j/a_n|\right)$ –  Robert Israel Mar 22 '12 at 15:21
    
Shouldn't the value of $y$ be somehow dependent on $x$ as well? $x$ can be arbitrary in the polynomial, so conceivably we can choose an $x$ that just exactly cancels out everything in the equation... –  Peter C. Mar 22 '12 at 15:51
    
$|x+iy|\ge |y|$ –  Robert Israel Mar 22 '12 at 15:55
    
Thanks, I get that now. I think your method works and is completely right. –  Peter C. Mar 22 '12 at 16:03

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