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Let's say I have a polynomial such as $$p(x) = x^4 + bx^3 + cx^2 + bx + 1.$$

I strongly suspect that, for any parameters, there are always two roots inside the unit circle and two roots outside of the unit circle.

What tools can I use to determine whether or not this is correct?

I am not necessarily looking for a solution to this problem but any answer that solves this problem would naturally contain tools that can be used in the general case.

Thank you.

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  • $\begingroup$ Are $b$ and $c$ real or complex? $\endgroup$ – rogerl Oct 19 '14 at 22:53
  • $\begingroup$ Do you mean $(-1,1)$ or $\{|z| < 1\}$? You could try Rouche's theorem, en.wikipedia.org/wiki/Rouch%C3%A9's_theorem. You could also just solve for the roots explicitly since the degree of the polynomial is $< 4$. $\endgroup$ – snar Oct 19 '14 at 22:55
  • $\begingroup$ A Mobius transformation that takes the unit circle to the real line would be one approach, since then the scenario becomes "How many roots are in the upper/lower half plane?" $\endgroup$ – Semiclassical Oct 19 '14 at 22:57
  • $\begingroup$ @rogerl In my specific case they are real, but I am interested in both cases. $\endgroup$ – user157227 Oct 19 '14 at 23:04
  • $\begingroup$ Look up the Schur-Cohn or Jury-Marden criteria. If you want to take @Semiclassical's approach, then look up the Routh-Hurwitz criterion. See this as well. $\endgroup$ – J. M. is a poor mathematician Aug 4 '16 at 5:36
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$p\left(\frac{1}{x}\right) = \frac{1}{x^4}p(x)$. So unless there are roots on the unit circle (which is not ruled out in the problem as stated), there are two inside and two outside the unit circle.

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As rogerl said, by symmetry the number of roots inside the unit circle is equal to the number outside. Now, how many are on the unit circle? Let's suppose $1$ is not a root, i.e. $2 + 2 b + c \ne 0$. The Möbius transformation $ w = i(1+z)/(1-z)$ ($z = (w-i)/(w+i)$) takes the unit circle (except for the point $1$) to the real line, and $p(z) = 0$ becomes $g(w) = (2b+c+2) w^4+(2c-12)w^2-2b+c+2 = 0$. The real solutions of $g(w) = 0$ correspond (in pairs, when nonzero) to nonnegative solutions of $(2b+c+2) t^2 + (2c-12) t - 2b + c + 2 = 0$. The roots of that quadratic are ${\dfrac {-c+6\pm 2\,\sqrt {{b}^{2}-4\,c+8}}{2\,b+c+2}}$. So for example when $b = 1$ and $c = 2$, the quadratic has two positive roots ($1$ and $1/3$), and all four of the roots of your quartic are on the unit circle.

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