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Homework Help: E^2*pi*i, where from?

  1. Sep 19, 2010 #1
    1. The problem statement, all variables and given/known data
    The problem at hand is that I don't understand wherefrom my text book got a certain term(e^(2*pi*i). It doesn't say. At least not as I understand it.
    The book says:

    2. Relevant equations
    e^(z+2*pi*i) = e^z*e^(2*pi*i) = e^z*1 = e^z

    From where does e^(2*pi*i) come? I get the stuff leading to the answer, I just can't seem to understand from where that term comes from.
     
  2. jcsd
  3. Sep 19, 2010 #2

    Borek

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    Staff: Mentor

    Basic property of exponents:

    [tex]e^{a+b} = e^a e^b[/tex]
     
  4. Sep 19, 2010 #3
    I'm not certain as to what exactly your asking, but I hope this helps!

    Recall the identity property of exponents:

    ea+b=eaeb

    In your particular case, let:
    a=z
    b=2*pi*i

    Therefore,
    ez+(2*pi*i) = ez*e2*pi*i
     
  5. Sep 19, 2010 #4
    Oh, no, I am sorry if I was not clear. I simply don't know wherefrom they get the 2*pi*i from in e^(z+2*pi*i).

    The information I get is what I've written. I believe that the 2*pi refers to the period. It just seems kind of abrupt to randomly insert it without any proof or reference to hardly anything..
     
  6. Sep 19, 2010 #5
    From calculus one learns that [tex]e^{i\pi} = -1[/tex]
    So, using a certain property of exponents, [tex]e^{2i\pi} = (e^{i\pi})^2 = (-1)^2 = 1[/tex]
     
  7. Sep 20, 2010 #6

    Borek

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    OK. Could be they just added 2*pi*i at random. Why? Because they can :wink:

    Do you know Euler's formula?

    [tex]e^{ix} = \cos x + i \sin x[/tex]

    If you combine it all you see if you insert 2*pi*i into exponent at random, you will not change the result. Sometimes it can be a useful identity.
     
  8. Sep 20, 2010 #7

    Mentallic

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    Actually they could add any integer multiple of [itex]2\pi i[/itex] and still leave the answer unchanged.

    [tex]e^z=e^{z+2\pi i n}[/tex] where n is any integer.
     
  9. Sep 20, 2010 #8

    Hurkyl

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    We can't directly help you if you don't show us what they were doing up to that point.
     
  10. Sep 20, 2010 #9

    Borek

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    Lol, it must have been a senior moment on my side. I intended to write 2*pi*i*n but looked at 2*pi*i and decided there already is an integer (i) in the formula :grumpy:
     
  11. Sep 20, 2010 #10

    Mentallic

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    People seem to find new uses for i each and every day :wink:
     
  12. Sep 21, 2010 #11
    That's the thing. They aren't doing anything, it has its own little "information box". It says nothing after, nothing before. Just what I've written. All I know is that it has to do with Euler (the chapter is about Euler), if that helps?
     
  13. Sep 21, 2010 #12

    Borek

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    So it must be what I told you earlier - they just show an interesting and important property.

    It is like asking where did the 2*pi came from in sin(x+n*2*pi) = sin(x) :tongue2:
     
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