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 July 15th, 2009, 06:55 PM #1 Newbie   Joined: Jul 2009 Posts: 6 Thanks: 0 Show that a polynomial has a zero outside unit circle Hi, 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.
 July 16th, 2009, 03:01 AM #2 Newbie   Joined: Jun 2009 Posts: 20 Thanks: 0 Re: Show that a polynomial has a zero outside unit circle Consider $p(z)= (z-b_1)(z-b_2)\cdots (z-b_n).$. So we have $(-1)^nb_1\cdots b_n= a_0$. Notice that multiplying numbers in the unit circle gives another unit. So if $b_1,\cdots,b_n$ were units then $a_0$ would also be a unit. But given is $|a_0| > 1.$
 July 16th, 2009, 07:10 AM #3 Newbie   Joined: Jul 2009 Posts: 6 Thanks: 0 Re: Show that a polynomial has a zero outside unit circle That was quite simple. Thank you very much for you answer.
July 16th, 2009, 12:51 PM   #4
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Re: Show that a polynomial has a zero outside unit circle

No problem

Quote:
 has at least one zero outside the unit circle
These "zero's" are called "roots" by the way

 August 4th, 2009, 10:59 PM #5 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms Re: Show that a polynomial has a zero outside unit circle Equations have roots and expressions (e.g. polynomials) have zeros.
 August 5th, 2009, 03:32 PM #6 Newbie   Joined: Jun 2009 Posts: 20 Thanks: 0 Re: Show that a polynomial has a zero outside unit circle Guess I was wrong then :}
December 3rd, 2009, 06:25 PM   #7
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Re: Show that a polynomial has a zero outside unit circle

Quote:
 Originally Posted by CRGreathouse Equations have roots and expressions (e.g. polynomials) have zeros.
Ummm, I'd have to disagree. Polynomials have roots, that is for sure. Equations have zeros. As in, the eigenvalues are the roots of the characteristic polynomial. Or, I used a variant of Newton's method to find the zeros for this function.

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