# All roots of a polynomial lie on a circle.

I'm stuck in the following problem and I need your help to solve it.

Given a number $\alpha$, $0 < \alpha < 1$. $A_j(x)$ is a sequence of polynomials of $x^{-1}$ such that:

$A_0(x) = 1; \\ A_{j+1}(x) = A_{j}(x) + \alpha^{j+1}x^{-j-1}A_j(x^{-1})\\ (A_1(x) = 1 + \alpha x^{-1}; A_2(x) = 1 + (\alpha + \alpha^3)x^{-1} + \alpha^2x^{-2},...)$

Prove that all roots of the equation $A_j(x) = 0, j \geq 1,$ lie on a circle of radius $\alpha$.

Could anyone give me some hints?

Thank you very much.

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To prove that all the roots lie on a circle of radius $|\alpha|$, find the center $(x_j, y_j)$ as a function of $j$ and $\alpha$. This is an obvious step to me. See what this center is for $j=1$ and $2$, and see if the recurrence can be used to get a recurrence for the center. –  marty cohen Mar 26 '14 at 22:41
It seems to me that all the centers are the origin (0). But what's next? –  Tiep Vu Mar 26 '14 at 22:50
Small hint (for the radius): Look at $p_i(x)=x^iA_i(x)$. –  Steve Kass Mar 26 '14 at 23:31
If the center is at the origin, all the roots are of the form $e^{it}\alpha$ for real $t$. –  marty cohen Mar 27 '14 at 1:44
@SteveKass: I tried your hint but I still can not get the solution. Could you get me some more hints? Thanks. –  Tiep Vu Mar 27 '14 at 14:42