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Conference rusure::math

Title:Mathematics at DEC
Moderator:RUSURE::EDP
Created:Mon Feb 03 1986
Last Modified:Fri Jun 06 1997
Last Successful Update:Fri Jun 06 1997
Number of topics:2083
Total number of notes:14613

795.0. "USENET: Polynomials with complex coefficients" by ZFC::DERAMO (Daniel V. D'Eramo) Wed Dec 02 1987 21:19

Newsgroups: sci.math
Path: decwrl!ucbvax!cartan!chex!mccarthy
Subject: Polynomials
Posted: 30 Nov 87 21:04:08 GMT
Organization: UC Berkeley Math Department
 
 
   I am interested in the following problem:
    When can a polynomial (over the complex numbers) be written
in the form  [p(z)]^2 + z*[q(z)]^2 ,
where p and q are also polynomials?
Always?
  If you don't know the answer, any suggestions for lines of attack
would be welcome.
        Thanks,
           John McCarthy
T.RTitleUserPersonal
Name
DateLines
795.1A hint from the USENET solutionZFC::DERAMODaniel V. D'EramoMon Dec 07 1987 13:248
     A hint from the USENET solution:


     What are the closure properties of the class of polynomials,
     with complex coefficients, that can be wriiten in the form
     (p(z))^2 + z(q(z))^2?

     Dan
795.2all of themZFC::DERAMOCan I take your personal name?Mon Jan 18 1988 22:3847
    Let P1 and P2 be polynomials of the requested form:
    
         P1(z) = [p1(z)]^2 + z*[q1(z)]^2
         P2(z) = [p2(z)]^2 + z*[q2(z)]^2
    
    or, more briefly,
    
         P1 = p1^2 + z q1^2
         P2 = p2^2 + z q2^2
    
    Then
    
         P1 P2 = (p1^2 + z q1^2)(p2^2 + z q2^2)
               = p1^2 p2^2 + z (q1^2 p2^2 + p1^2 q2^2) + z^2 q1^2 q2^2
               = (p1 p2 + z q1 q2)^2 - 2 z p1 p2 q1 q2
                 +  z (q1^2 p2^2 + p1^2 q2^2)
               = (p1 p2 + z q1 q2)^2
                 + z (q1^2 p2^2 - 2 p1 p2 q1 q2 + p1^2 q2^2)
               = (p1 p2 + z q1 q2)^2 + z (q1 p2 - p1 q2)^2
    
    Therefor, if P1 and P2 are have the desired form, so does the
    product P1 P2.
    
    Now, any zero-degree polynomial P(z) = a can be written as
    
         P(z) = a = (sqrt(a))^2 + z * 0^2
    
    because every complex value a has a square root.  Likewise, every
    first degree polynomial P(z) = a + bz can be written as
    
         P(z) = a + bz = (sqrt(a))^2 + z * (sqrt(b))^2
    
    So every zeroth or first degree polynomial has the desired form.
    Now an arbitrary nth degree polynomial with leading coefficient
    a and roots z1, ..., zn can be written as
    
         P(z) = a (z - z1) ... (z - zn)
    
    Since this is a product of polynomials of the desired form, it
    follows that P(z) is of the desired form.
    
    So every polynomial in z with complex coefficients can be written
    in the form (p(z))^2 + z * (q(z))^2.
    
    Dan
    
    P.S.  This is not my solution, it was in a USENET article.