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May 17th, 2013, 12:25 PM  #1 
Senior Member Joined: Feb 2012 Posts: 628 Thanks: 1  A complex situation
Find all solutions to the system of equations: 
May 17th, 2013, 01:17 PM  #2 
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: A complex situation
You're asking for the zeros of (dividing out the factor of 2) a^6  6a^5 + 30a^4  80a^3 + 120a^2  96a + 32 which factors as (a^2  2a + 2)(a^4  4a^3 + 20a^2  32a + 16) Finding these zeros is routine since their degrees are less than 5. 
May 17th, 2013, 02:53 PM  #3 
Senior Member Joined: Feb 2012 Posts: 628 Thanks: 1  Re: A complex situation
How did you factor the sixth degree polynomial and how are you finding the zeros of that quartic polynomial?

May 17th, 2013, 06:04 PM  #4 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,930 Thanks: 1124 Math Focus: Elementary mathematics and beyond  Re: A complex situation 
May 17th, 2013, 06:38 PM  #5 
Senior Member Joined: Feb 2012 Posts: 628 Thanks: 1  Re: A complex situation
Interesting solution, greg1313. I think I see now where CRG's factorization comes from. The last four pairs look a little different in your solution than they do in mine, but I'm sure they are equivalent. I wrote all of the numbers in the typical form x + yi, where x and y are real numbers.

May 17th, 2013, 07:00 PM  #6 
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: A complex situation
The four solutions coming from the quartic have a very nice form, something like a + i sqrt(b + sqrt(c)) if I recall correctly.

May 18th, 2013, 05:42 AM  #7 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,930 Thanks: 1124 Math Focus: Elementary mathematics and beyond  Re: A complex situation 

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