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Solving exp(z) = 1

  1. Mar 12, 2014 #1
    1. The problem statement, all variables and given/known data

    solving the equation ez = 1 , where z is complex



    3. The attempt at a solution


    This is my attempt at the solution

    ez = ea+ib

    and then i can write it as ea(cos(b)+isin(b))= 1
    ea has to be larger than 0 and therefor i*sin(b) = 0 → b can either be n2[itex]\pi[/itex] or [itex]\pi[/itex] + n2[itex]\pi[/itex]
    and cos(b) has to be positive because ea is positive, and the only posible b is n2[itex]\pi[/itex]. And because a is the real part a has to be equal to 0. I really can't see where I'm missing something because in the solution z = in2[itex]\pi[/itex]
     
  2. jcsd
  3. Mar 12, 2014 #2

    Dick

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    All fine. So b has to be 2*pi*n and a has to be 0. So z=a+i*b=i*2*pi*n. What's the problem?
     
  4. Mar 12, 2014 #3
    Ok i got it now thx. I forgot to factor in the i with b. Wow really simple misstake
     
  5. Mar 13, 2014 #4

    SteamKing

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    misstake is a mistake
     
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