# Methods for determining which roots of a polynomial are inside of the unit circle?

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.

• Are $b$ and $c$ real or complex? – rogerl Oct 19 '14 at 22:53
• 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 $< 4$. – snar Oct 19 '14 at 22:55
• 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?" – Semiclassical Oct 19 '14 at 22:57
• @rogerl In my specific case they are real, but I am interested in both cases. – user157227 Oct 19 '14 at 23:04
• 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. – J. M. ain't a mathematician Aug 4 '16 at 5:36

$p\left(\frac{1}{x}\right) = \frac{1}{x^4}p(x)$. 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.
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.