Showing a polynomial has at least one zero outside the unit circle.

[jdinatale]

The first thing that we should notice is that the leading coefficient $a_n = 1$. I was thinking about considering the factored form of p.

I googled, and there is an algorithm called the "Schur-Cohn Algorithm" that is suppose to answer exactly this, but I can't find any information on it or how to use it. Besides, this question is asked in an introduction to complex variables class, so we shouldn't have to use that algorithm.

Also, I found this questioned answered elsewhere, but I can't understand their solutions

http://www.edaboard.com/thread154228.html

 

[SammyS]

The first thing that we should notice is that the leading coefficient $a_n = 1$. I was thinking about considering the factored form of p.

I googled, and there is an algorithm called the "Schur-Cohn Algorithm" that is suppose to answer exactly this, but I can't find any information on it or how to use it. Besides, this question is asked in an introduction to complex variables class, so we shouldn't have to use that algorithm.

Also, I found this questioned answered elsewhere, but I can't understand their solutions

http://www.edaboard.com/thread154228.html

The second solution in that link looks reasonable.

What is p(0) ?

Suppose we write the zeros of p(z) as z₁, z₂, z₃, ..., z_n.

Write p(z) in factored form, and from that, set z=0 to find a different expression for p(0).

 

[jdinatale]

The second solution in that link looks reasonable.

What is p(0) ?

Suppose we write the zeros of p(z) as z₁, z₂, z₃, ..., z_n.

Write p(z) in factored form, and from that, set z=0 to find a different expression for p(0).

That makes total sense now! Thanks, I have the problem solved now!