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Equation with complex number

  1. Dec 7, 2015 #1
    1. The problem statement, all variables and given/known data
    Solve the following equation: ## (1+a)^n=(1-a)^n## where a is complex number and n is natural number

    2. Relevant equations
    Euler's formula

    3. The attempt at a solution
    I've tried something like this

    ##
    (1+a)^n=(1-a)^n \\

    (\frac{1+a}{1-a})^n=1 ##

    But i really have no idea what to do next, since that n-th degree complicates things, i mean i could write a as a= x+iy where x and y are real numbers but it won't help much.
     
  2. jcsd
  3. Dec 7, 2015 #2

    mfb

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    What do you know about numbers z that satisfy zn = 1? In particular, how many are there?
     
  4. Dec 7, 2015 #3
    Well, there are n solutions to such equation, ##z=cos\frac{2\pi +2k\pi}{n} + isin\frac{2\pi +2k\pi}{n}##
     
  5. Dec 7, 2015 #4

    mfb

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    Right. Using the ##r e^{i \phi}## notation is probably easier in the next steps.
     
  6. Dec 8, 2015 #5
    Ok, so, i have ##\frac{1+a}{1-a}=e^{i\frac{2k\pi}{n}}## , but i should find a, what should i do next? Since a is complex number a=x +iy (x, y are real numbers) and then if i multiply and divide LHS by by 1-x +iy it doesn't seem like it gets me any closer to the solution.
     
  7. Dec 8, 2015 #6

    HallsofIvy

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    Why not? [itex]e^{i\frac{2k\pi}{n}}[/itex] is a constant. If it is bothering you, call it "k". Then [itex]\frac{1+ a}{1- a}= k[/itex] which we can write 1+ a= k(1- a)= k- ka. Can you solve that equation for a? Then, if you need to, write a as x+ iy to solve for x and y.
     
    Last edited by a moderator: Dec 8, 2015
  8. Dec 8, 2015 #7
    Well, i think it could go this way:

    ## 1+a=k-ka \\ a+ka=k-1 \\ a=\frac{k-1}{k+1} \\ a=\frac{e^{i\frac{2k\pi}{n}}-1}{e^{i\frac{2k\pi}{n}}+1} ##

    Is this correct?
     
  9. Dec 8, 2015 #8

    Samy_A

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    Looks on track, but ...
    What happens when ##\frac{k}{n}=\frac{1}{2}##?
     
  10. Dec 8, 2015 #9
    I think i know what are you trying to say, denominator can't be zero, so this works for every k and n except when n=2k.
     
  11. Dec 8, 2015 #10

    PeroK

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    Yes, but can you simply that expression you have?
     
  12. Dec 8, 2015 #11

    Samy_A

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    Indeed, because in your derivation you at a point divided by ##e^{i\frac{2k\pi}{n}}+1##, and that doesn't work if that value is 0.
    As @PeroK suggested, you can simplify your result.
     
  13. Dec 8, 2015 #12
    Let's see, i guess i could multiply numerator and denominator by ## cos\frac{2k\pi}{n} + 1 - sin\frac{2k\pi}{n} ## is this the thing i should do?
     
  14. Dec 8, 2015 #13

    mfb

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    There is no need to start with cosine and sine. How can you expand every fraction to make its denominator real?
     
  15. Dec 8, 2015 #14
    Well, if it's complex number, i can do that by multiplying with it's conjugate, that's what i did, i can't remember something else.
     
  16. Dec 8, 2015 #15

    mfb

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    Right, multiply both sides with its complex conjugate (but don't use sine and cosine, they just make it messy).
     
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