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Finding solution of equation in complex domain?

  1. Sep 22, 2015 #1
    (1+a)n = (1-a)n

    I tried following:

    (1+a)n = (1-a)n

    [(1+a)/(1-a)]n=1 but what can i do next?
     
  2. jcsd
  3. Sep 22, 2015 #2

    Mark44

    Staff: Mentor

    The first thing I did was to look at this equation using a couple of small values for n: n = 3 and n = 4, to get some idea of solutions.

    Next, I would expand each side using the Binomial Theorem.
     
  4. Sep 22, 2015 #3

    mathman

    User Avatar
    Science Advisor
    Gold Member

    [itex](\frac{1+a}{1-a})^n=1[/itex] Has n solutions the n nth roots of unity [itex]e^{\frac {2\pi ki}{n}}=c_k[/itex] for k=0,1,...,n-1.
    This leads to [itex]a_k=\frac{c_k-1}{c_k+1}[/itex].
     
  5. Sep 23, 2015 #4
    But how can i do that if i don't know the value of a, i mean if i have n=3 then i should first represent both sides in parentheses as cosx + isinx and then find (a+1) and (a-1) to the third degree, but i can't find that trigonometric expression because i don't know the value of a.
     
  6. Sep 23, 2015 #5

    mathman

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    Gold Member

    When n = 3, expanding by the binomial leads to [itex]3a^2+1=0[/itex].
     
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