red930 Posted September 7, 2006 Posted September 7, 2006 Alright, I'm supposed to find all the possible solutions for x^5=1. Obviously one answer IS 1, but apparently there are 4 more answers, all of which I assume to include i (or j, to you electrical engineers) as a variable. This has got me stumped and I haven't found anything searching for it yet, so if you've got some pointers I'd really appreciate them. Thanks a bunch guys... Share this post Link to post Share on other sites More sharing options...
UncleDavid218 Posted September 7, 2006 Posted September 7, 2006 But techno... school doesn't start till tomorrow! My brain is still turned off! Naw, I'll look at some books and see if I can help Share this post Link to post Share on other sites More sharing options...
ooztuncer Posted September 7, 2006 Posted September 7, 2006 5(fifth) root 1 Share this post Link to post Share on other sites More sharing options...
ooztuncer Posted September 7, 2006 Posted September 7, 2006 1 / (fifth root 1) Share this post Link to post Share on other sites More sharing options...
ooztuncer Posted September 7, 2006 Posted September 7, 2006 -1 / (fifth root -1) Share this post Link to post Share on other sites More sharing options...
ooztuncer Posted September 7, 2006 Posted September 7, 2006 I believe there are a lot more than 5 solutions (infinite solutions) but lets hear what you come up with... edit: as usual i am wrong probably, hehe edit continues: forget about infinite solutions statement, there are 5 solutions as listed in post #8 Share this post Link to post Share on other sites More sharing options...
red930 Posted September 7, 2006 Posted September 7, 2006 Ah geez..those are so obvious i'm literally beating myself with my qm book right now. Ugh... Yeah I was thinking there was more too, but I have yet to come up with anything better yet... Share this post Link to post Share on other sites More sharing options...
ooztuncer Posted September 7, 2006 Posted September 7, 2006 - fifth root -1: (minus fifth root minus one) - that makes five SUMMARY: 1) x = 1 2) x = fifth root 1 3) x = - fifth root -1 4) x = 1 / (fifth root 1) 5) x = - 1 / (fifth root -1) Share this post Link to post Share on other sites More sharing options...
zkissane Posted September 7, 2006 Posted September 7, 2006 Finding all the solutions to x^5=1 is equivilent to finding the solutions to x^5-1=0. x=1 is obvious. That means you can factor out (x-1), leaving you with a 4th degree polynomial. For education's sake, I won't tell you what it is, but let's call it P(x). P(x) = 0 (which is what you have to solve to find the other solutions) has no real solutions. I can't remember how to find 4th degree solutions, especially when they're complex numbers. EDIT: Also remember that when dealing with complex roots/solutions (roots of the form "a + bi", where a and b are real numbers, and i is the imaginary number), if you find one complex solution, then its complex conjugate is also a solution. So if some complex number (a + bi) is a solution to P(x) = 0, then (a - bi) is also a solution. Share this post Link to post Share on other sites More sharing options...
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