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Show that all roots of $p(x)$ $=$ $11x^{10} - 10x^9 - 10x + 11$ lie on the unit circle $abs(x)$ = $1$ in the complex plane.

My progress so far enter image description here

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I tried messing around with the polar form complex numbers but my proof turned out to be incorrect. Any help/suggestions would be greatly appreciated. I simply assumed all the roots were complex because the graph of the polynomial shows that there are no real zeroes. The rational root theorem fails and I tried to use the Fundamental Theorem of Algebra. Also, I couldn't LaTeX 11x^10 for some reason...

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  • $\begingroup$ Try 11 x^{10} for $11 x^{10}\,$.Note that $(11x)(11x^9) \ne 11 x^{10}$ if that's what you meant. $\endgroup$ – dxiv Mar 2 '17 at 0:58
  • $\begingroup$ whoops typo thank you @dxiv $\endgroup$ – Sanjoy Kundu Mar 2 '17 at 0:59
  • $\begingroup$ @dxiv but the LaTeX for that still gave me problems. Maybe someone can edit this problem's formatting. $\endgroup$ – Sanjoy Kundu Mar 2 '17 at 1:00
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Hint: $\;p(x)=x^5\left(11(x^5+\frac{1}{x^5}) - 10(x^4+\frac{1}{x^4})\right)$. Let $z=x+\frac{1}{x}\,$, express $x^2+\frac{1}{x^2}=z^2-2$ etc in terms of $z\,$, and derive an equation in $z$.


[ EDIT ]   The resulting equation in $z$ is the quintic $\,11 z^5 - 10 z^4 - 55 z^3 + 40 z^2 + 55 z - 20 = 0\,$ which can be shown to have $5$ real distinct roots in $(-2,2)\,$. For each of those $z$ roots, the corresponding equation $x^2 - z\,x + 1 =0$ will have a pair of complex conjugate $x$ roots because the discriminant $\Delta = z^2-4 \lt 0\,$, and since their product is $1$ by Vieta's relations, they will all have magnitude $1$ i.e. lie on the unit circle.

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  • $\begingroup$ @IntegralBatman The calculations are not the prettiest, but they work out in the end. $\endgroup$ – dxiv Mar 2 '17 at 1:25
  • $\begingroup$ I would emphasize a point I had not known, if $t$ is real and $x + \frac{1}{x}=t,$ we conclude that $x$ is on the unit circle if $-2 \leq t \leq 2,$ otherwise $x$ is real with absolute value $|x| \neq 1.$ $\endgroup$ – Will Jagy Mar 2 '17 at 1:55
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    $\begingroup$ @WillJagy Thanks. Edited to make that point more clear. $\endgroup$ – dxiv Mar 2 '17 at 2:22
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When I look it up on the knowledge engine it says "factorization over infinite fields" is

$$ (x+1)^2(x^4+x^3+x^2+x+1)^2 $$

although neither (x-i) nor (x+i) are regular factors, strangely.

The proof in this one is showing that there are no real factors, I believe.

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  • $\begingroup$ That's not the factored form of my polynomial though... $\endgroup$ – Sanjoy Kundu Mar 3 '17 at 15:22
  • $\begingroup$ If you are allowed to use wolfram, then you can show that all the solutions lie on abs(x)=1. I was just surprised your polynomial didn't have +/- i as a factor. $\endgroup$ – Tim2see Mar 4 '17 at 0:52

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