Show that a polynomial has a zero outside unit circle

[Jone]

+polynomial +unit circle

I'm stuck in proving the following:

Show that if p(z) = z^n + a_n-1*z^n-1 + ... + a_0 is a polynomial of degree n ≥ 1 and |a_0| > 1, then p(z) has at least one zero outside the unit circle. Notice that the leading coefficient a_n = 1. There is also a hint given: Consider the factored form of p(z).

This problem is from the book Fundamentals of Complex Analysis by Saff and Snider.

 

[doraemon]

zeros of polynomial on a unit circle

Hello!

Your polynom degree is n. This means that there is set
of (pn, qn) ...... (p1, q1) pairs that can factorize your polynom.

Note: in all the following, pn shall mean p at the index n,
pn-1 shall mean p at the index n-1, and similarily for q.

Therefore p(z) can be written:

p(z) = (pn z + qn) * (pn-1 z + qn-1) ...... * (p2 z + q2) * (p1 z + q1)

Your an is 1, and it is also the product of all pns:
an = pn * pn-1 * pn-2 ... *p2 * p1 = 1

Similarily, a0 is greater than 1, but it is also the product of all qns

a0 = qn * qn-1 * qn-2 ... * q2 * q1 > 1

NB: I know that the purists would start yelling when there are = and < signs
on the same line...

Now if we want to demonstrate that there is at least one zero out
of the unit circle, this is equivalent to demonstrate that there is at
least one rank k where |z| > 1, therefore |qk| / |pk| > 1, therefore there
is at least one rank k where |qk| > |pk|

Let's suppose that it's not the case. Let's suppose that in all the ranks
we have qk <= pk.

In this case, the product of all the qks will be less than the product of all
the pks. This would therefore indicate that a0 cannot be greater than an.

But at the beginning we said that a0 > 1 and an = 1, therefore we can
conclude that there is at least one rank k where |qk| > |pk|, therefore
there is one rank k where |qk| / |pk| > 1.
Therefore there is at least one root outside of the unit circle.

That's it!

Dora.

I'm stuck in proving the following:

Show that if p(z) = z^n + a_n-1*z^n-1 + ... + a_0 is a polynomial of degree n ≥ 1 and |a_0| > 1, then p(z) has at least one zero outside the unit circle. Notice that the leading coefficient a_n = 1. There is also a hint given: Consider the factored form of p(z).

This problem is from the book Fundamentals of Complex Analysis by Saff and Snider.

 

[Jone]

polynomial zeros on unit circle

I found a somewhat simpler way:

p(z) = z^n + a_n-1*z^n-1 + ... + a0 = (z-z1)*...*(z-zn) = z^n + ... + (-1)^n*z1*...*zn
So a0 = (-1)^n*z1*...*zn and we have that |a0| = |z1*...*zn| > 1. If the product of zeros is larger in magnitude than 1, then all zeros can't be smaller than 1, or, at least one of them need to be larger than 1.

 

[horzcat]

polynomial unit circle

Do any stability test. Take reference from any control system book.