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

enter preformatted text here

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...

  • $\begingroup$ 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 $\endgroup$ 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. $\endgroup$ Mar 2 '17 at 1:00

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.

  • $\begingroup$ @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
  • 1
    $\begingroup$ @WillJagy Thanks. Edited to make that point more clear. $\endgroup$
    – dxiv
    Mar 2 '17 at 2:22

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.

  • $\begingroup$ That's not the factored form of my polynomial though... $\endgroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.