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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!

ro44
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  • http://en.wikipedia.org/wiki/Rouch%C3%A9's_theorem – Brett Frankel Nov 04 '12 at 01:01
  • @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 04 '12 at 01:03

1 Answers1

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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 $i = 0,1,\ldots,n$.

Alan Guo
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  • Consider the polynomial $p(z)=z^n-1$ for a counterexample to your claim. – Cameron Buie Nov 04 '12 at 01:11
  • Sorry, forgot another condition: after factoring out all real roots (i.e. $\pm 1$) – Alan Guo Nov 04 '12 at 01:12
  • Ah! That's better – Cameron Buie Nov 04 '12 at 01:17
  • I'll give this some thought now, thank you! – ro44 Nov 04 '12 at 01:21
  • @AlanGuo: I'm having a hard time with this. Could you give me a hint for the proof? Thanks! – ro44 Nov 04 '12 at 03:47
  • Here's how to show that $a_1 = a_{n-1}$. First of all, $n$ has to be even because the polynomial has real coefficients and therefore its roots must come in complex conjugate pairs. Let $r_1,\ldots,r_n$ be the roots of $p$ (and notice that $a_0 = \prod_i r_i = 1$. Then $-a_{n-1} = r_1 + \cdots + r_n$ and $-a_1 = 1/r_1 + \cdots + 1/r_n$. I claim these two sums are the same. Pair up the roots by conjugate pairs, so $r_1,r_2$ are complex conjugates, and so on. Then $1 = r_1r_2 = r_3r_4 = \cdots$ and so $1/r_1 + 1/r_2 = r_2 + r_1$, so $a_{n-1} = a_1$. Use a similar argument for the general case. – Alan Guo Nov 04 '12 at 03:57
  • Whew, finally have it. Thank you very much. I'll leave this question open for a bit since I'm curious if there are 'easier' answers. – ro44 Nov 04 '12 at 05:11