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May 17th, 2013, 12:25 PM   #1
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A complex situation

Find all solutions to the system of equations:



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May 17th, 2013, 01:17 PM   #2
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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.
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May 17th, 2013, 02:53 PM   #3
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Re: A complex situation

How did you factor the sixth degree polynomial and how are you finding the zeros of that quartic polynomial?
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May 17th, 2013, 06:04 PM   #4
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May 17th, 2013, 06:38 PM   #5
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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.
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May 17th, 2013, 07:00 PM   #6
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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.
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May 18th, 2013, 05:42 AM   #7
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