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Complex, exponential problem

  1. Nov 12, 2011 #1
    Prove that cos z = 0 if and only if z = π/2 + kπ, where k is an integer.

    The solution is :
    cos z = (eiz+e-iz)/2 = 0
    <=>2iz=πi +2kπi
    <=>z=π/2 + kπ

    But i dont understand how does these 2 steps work
    <=>2iz=πi +2kπi

    how can a exp function suddenly become πi +2kπi

  2. jcsd
  3. Nov 12, 2011 #2
    i know there is an equation :
    ez1=ez2 holds if and only if z1=z2 + 2kπi

    But how can i know in exam -1 = eπi
    is it by try an error
  4. Nov 12, 2011 #3
    It's hard to explain why those things hold because I don't know what I can use and what I cannot use.

    How did you define the exponential?? What properties have you already seen about the exponential??
    Is there a book you're reading where this appears in?? Lecture notes??
  5. Nov 12, 2011 #4
    Yes, this is what you use.

    Knowing this depends on what you know about the exponential. I can define for real x:


    if you put in [itex]x=\pi[/itex] in the above, then it follows.
  6. Nov 12, 2011 #5


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    Multiply [itex]e^{iz}+e^{-iz}[/itex] by [itex]e^{iz}\,.[/itex]

    [itex]\displaystyle e^{iz}\left(e^{iz}+e^{-iz}\right)=e^{2iz}+1[/itex]

    For the second question: [itex]\displaystyle -1=e^{(\pi i)}e^{2\pi k i}\,.[/itex]
  7. Nov 12, 2011 #6
    thanks, i know eπi=-1

    but the problem is , when i was in exam i was looking at the -1 , how can i know e?, what about the question is -2 , do i need to try it one by one like π/2,π/3....
  8. Nov 12, 2011 #7
    So you want to know what x solves [itex]e^{z}=-2[/itex]? Without going into complex logarithms, this can be solved as


    So [itex]\log(2)+i\pi[/itex] will be a solution (but it will not be the only one).

    This can of course also be found by complex logarithms, but you shouldn't use those until you've seen it.
  9. Nov 12, 2011 #8


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    Polar representation is the key. So lets say in exam, you're given some number like [itex]a+ib[/itex] then this is equal to [itex]r(cos(x)+i \ sin(x))[/itex] which is also equal to [itex]r \ e^{ix}[/itex] So this is how you convert from one form to the other.

    So your example of -2 clearly has the same x as for -1, but the only difference is that r is twice as great.
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