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Given:

$p(x)=z^{2012}-z^{1010}+2z^{1006}+20243z^8-2z^4+1$

I need to prove the polynomial has a root $|z_0|<1$.

What I have so far: Plugging $p(0)=1$ we get (from the fundamental theorem of algebra) that $|z_0|...|z_n|=1$, were $z_0,...,z_n$ are the roots of the polynomial. So this means the polynomial's roots are either all on the unit circle, or at least one of them is inside of it. (Maybe an 'assume for the sake of contradiction' is good here?)

Also, because $p(-1), p(1)$ aren't roots we know all the roots are complex.

I was thinking of using the intermediate value theorem somehow here, but it doesn't seem applicable (for one, I'd need to prove $p(A)$ is open for some choice of $A$, but we haven't learned any theorem that would make this very obvious).

Also, this polynomial's exponent coefficients are all even, which kind of feels like it should mean something, but I don't know.

This being homework, I'd appreciate hints moreso than answers, but I'll accept either. Thanks!

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@BrettFrankel: A quick search here brought up a number of questions that use this theorem, but we've just begun studying this stuff so I don't think it's applicable... (e.g. I don't know what a holomorphic/analytic function is yet!) –  ro44 Nov 4 '12 at 1:03
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1 Answer

Hint: Show that if a polynomial with real coefficients has all of its roots on the unit circle and no real roots, then it must be palindromic, i.e. if $p(z) = a_0 + a_1z + \cdots + a_nz^n$ then $a_i = a_{n-i}$ for all $i = 0,1,\ldots,n$.

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Consider the polynomial $p(z)=z^n-1$ for a counterexample to your claim. –  Cameron Buie Nov 4 '12 at 1:11
    
Sorry, forgot another condition: after factoring out all real roots (i.e. $\pm 1$) –  Alan Guo Nov 4 '12 at 1:12
    
Ah! That's better –  Cameron Buie Nov 4 '12 at 1:17
    
I'll give this some thought now, thank you! –  ro44 Nov 4 '12 at 1:21
    
@AlanGuo: I'm having a hard time with this. Could you give me a hint for the proof? Thanks! –  ro44 Nov 4 '12 at 3:47
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