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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: $a - b= 2$ $a^6 + b^6= 0$
 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: 7,958 Thanks: 1146 Math Focus: Elementary mathematics and beyond Re: A complex situation $a\,-\,b\,=\,2$ $a^6\,+\,b^6\,=\,0$ $$$a^2\,+\,b^2$$$$b^2\,-\,a^2b^2\,+\,b^6$$\,=\,0$ $a^2\,+\,b^2\,=\,0$ $a^2\,+\,(a\,-\,2)^2\,=\,0$ $2a^2\,-\,4a\,+\,4\,=\,0$ $a^2\,-\,2a\,+\,2\,=\,0\,$ $\Rightarrow\,a_1\,=\,1\,+\,i,\,b_1\,=\,-1\,+\,i,\,a_2\,=\,1\,-\,i,\,b_2\,=\,-1\,-\,i$ $a^4\,-\,a^2b^2\,+\,b^4\,=\,0$ $x\,=\,a^2,\,y\,=\,b^2$ $x^2\,-\,xy\,+\,y^2\,=\,0\,\Rightarrow\,x\,=\,\frac{y\,\p m\,\sqrt{3}iy}{2}$ $a^2\,=\,b^2$$\frac{1\,\pm\,\sqrt{3}i}{2}$$$ $2b^2\,+\,8b\,+\,8\,=\,$$1\,\pm\,\sqrt{3}i$$b^2$ $$$1\,\pm\,\sqrt{3}i$$b^2\,+\,8b\,+\,8\,=\,0$ $b_3\,=\,\frac{-8\,+\,\sqrt{64\,-\,32(1\,+\,\sqrt{3}i)}}{2(1\,+\,\sqrt{3}i)},\,a_3\ ,=\,b_3\,+\,2$ $b_4\,=\,\frac{-8\,+\,\sqrt{64\,-\,32(1\,-\,\sqrt{3}i)}}{2(1\,-\,\sqrt{3}i)},\,a_4\,=\,b_4\,+\,2$ $b_5\,=\,\frac{-8\,-\,\sqrt{64\,-\,32(1\,+\,\sqrt{3}i)}}{2(1\,+\,\sqrt{3}i)},\,a_5\ ,=\,b_5\,+\,2$ $b_6\,=\,\frac{-8\,-\,\sqrt{64\,-\,32(1\,-\,\sqrt{3}i)}}{2(1\,-\,\sqrt{3}i)},\,a_6\,=\,b_6\,+\,2$
 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: 7,958 Thanks: 1146 Math Focus: Elementary mathematics and beyond Re: A complex situation $a_3\,=\,1\,-\,(2\,-\,\sqrt{3})i,\,b_3\,=\,-1\,-\,(2\,-\,\sqrt{3})i$ $a_4\,=\,1\,+\,(2\,-\,\sqrt{3})i,\,b_4\,=\,-1\,+\,(2\,-\,\sqrt{3})i$ $a_5\,=\,1\,+\,(2\,+\,\sqrt{3})i,\,b_5\,=\,-1\,+\,(2\,+\,\sqrt{3})i$ $a_6\,=\,1\,-\,(2\,+\,\sqrt{3})i,\,b_4\,=\,-1\,-\,(2\,+\,\sqrt{3})i$

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