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From: John Silver on 15 Jul 2010 12:00
Let f be a polynomial in the complex plane with no roots on S^1. Show
that the number of roots z, with |z| < 1, is equal to the degree of f/|
f| as a function on S^1.

How to proceed?
From: Robert Israel on 15 Jul 2010 19:20
John Silver <johnsilveremail(a)gmail.com> writes:

> Let f be a polynomial in the complex plane with no roots on S^1. Show
> that the number of roots z, with |z| < 1, is equal to the degree of f/|
> f| as a function on S^1.
>
> How to proceed?

Hint: translate the Argument Principle into the language of degrees.
--
Robert Israel israel(a)math.MyUniversitysInitials.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
From: John Silver on 27 Jul 2010 08:35
On Thu, 15 Jul 2010 18:20:39 -0500, Robert Israel wrote:

> John Silver <johnsilveremail(a)gmail.com> writes:
>
>> Let f be a polynomial in the complex plane with no roots on S^1. Show
>> that the number of roots z, with |z| < 1, is equal to the degree of f/|
>> f| as a function on S^1.
>>
>> How to proceed?
>
> Hint: translate the Argument Principle into the language of degrees.

Dear professor Israel

My complex analysis is a bit rusty so I am not able to do that
translation. The proof of the argument principle also uses derivatives
and I think the pure algebrotopological proof is not suppose to do that.
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