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Eulers Formula

  1. May 13, 2009 #1
    dc4d000103c1293c5cde00fac04850d6.png

    Is this true:

    [tex]e^{i(a+bx)}=cos(a+bx)+i sin(a+bx)[/tex] ?
     
    Last edited by a moderator: Apr 24, 2017
  2. jcsd
  3. May 13, 2009 #2

    chroot

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    Yes -- that's the beauty of function notation. You can replace 'x' with any expression you like.

    - Warren
     
  4. May 13, 2009 #3
    Thanks
     
  5. May 15, 2009 #4
    I know Daves question is answered, but instead of making a new thread I just ask another one here as it is somewhat related. Unfortunately I don't have a mathtype program so you'll have to imagine.
    So if you have the e^xi expression and set x = pi, this will equal -1. Adding 2Pi equals 1. So far so good? But here comes what I don't understand. 1 can also be written e^0, so using logarithms we say: e^i2Pi = e^0 -> i2Pi = 0, but this can't be true as i2Pi = 6,28i. I got one possible solution from my math teacher but it did not quite tell me how it could equal 0. Anyone care to give an explenation a shot?
     
  6. May 15, 2009 #5
    My best guess would be that the exponential function is not a bijection over the complex numbers, while it is so over the real numbers. That is, for real numbers,

    e^x = e^y <=> x = y is true.

    For complex numbers, it must be the case that

    e^x = e^y <=> x = y is false.

    This shouldn't be so surprising. For the real numbers, you can't say things like...

    x^2 = y^2 <=> x = y
    sin(x) = sin(y) <=> x = y
    ...

    I suppose it's just an accident that you can't do that for the exponential function over complex numbers.
     
  7. May 15, 2009 #6

    dx

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    You mean injection.
     
  8. May 15, 2009 #7
    Well, injection may be true too, but it's definitely not a bijection either.

    I tend to talk in terms of bijection or not bijection. I usually don't delve into things like injectivity and surjectivity. I think that the exponential is neither bijective nor injective over the complex numbers... then again, what do I know?
     
  9. May 15, 2009 #8

    dx

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    You said exp was a bijection over the reals which it isn't. It's an injection, but not a bijection.
     
  10. May 15, 2009 #9
    The problem is not with the exponential function, but with the logarithmic function. You need to define a branch cut to define the logarithmic function. This then means that the imaginary part of the Log of a complex number (the so called argument) is a unique number in some interval of length 2 pi.
     
  11. May 15, 2009 #10
    Yeah, what Count Ibis said:
    For the less mathematically versed, this is a little like taking the square root of a number and having to remember that there are actually two roots (+/-). When you take the log of a complex number, there are (potentially) multiple correct results, so you have to specify which one you're choosing.

    (At least, I think that's a correct statement; someone please correct me if it's not.)
     
  12. May 15, 2009 #11
    "You said exp was a bijection over the reals which it isn't. It's an injection, but not a bijection."

    Oops. You're right. I guess that's the price you pay when you get too used to using the general terminology too often.
     
  13. May 15, 2009 #12
    The trouble is that the complex exponential function is periodic, so saying that e^x = e^y iff x=y for complex x and y is akin to saying that cos x = cos y iff x=y.
     
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