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Why does (e^ix)(e^-ix) = 1?

  1. Oct 3, 2012 #1
    I was doing some math, typed in (eix)(e-ix) and it came out as 1.

    I was expecting it to come out as simply ex. Explain pleaseeee thanks =).
     
  2. jcsd
  3. Oct 3, 2012 #2

    dextercioby

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    Going to complex numbers doesn't change the rules for the exponential function you (should have) learnt way back in high-school.
     
  4. Oct 3, 2012 #3
    oHHHHHH LOOOOOL

    yeah I'm dumb. Thanks lol. THat's what I get for a four month summer
     
  5. Oct 3, 2012 #4

    HallsofIvy

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    For those who are a bit concerned about applying exponential formulas to complex numbers, we can use [itex]e^{ix}= cos(x)+ i sin(x)[/itex] so that [itex]e^{-ix}= cos(-x)+ i sin(-x)= cos(x)- i sin(x)[/itex] because sine is an odd function and cosine is an even function.

    Then [itex]e^{ix}e^{-ix}= (cos(x)+ i sin(x))(cos(x)- i sin(x))= cos^2(x)- isin(x)cos(x)+ icos(x)sin(x)- i^2sin^2(x)= cos^2(x)+ sin^2(x)= 1[/itex]
     
  6. Oct 4, 2012 #5
    And for those that didn't catch the simpler nature of the problem, as dextercioby said, you can use basic laws of exponents:
    eixe-ix=eix-ix=e0=1.

    Alternatively:
    eixe-ix=eix/eix=1.

    :)
     
  7. Oct 4, 2012 #6

    micromass

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    Yes, but there is no reason that basic laws of exponents should apply to complex numbers. That requires a proof and Halls provided it.
     
  8. Oct 4, 2012 #7
    Well, micromass, actually there is a BIG reason why the law of exponents should apply to complex numbers. It's because the exp(z) is an analytic continuation of the natural exponent ex that preserves the functional identity:
    [tex]
    f(z_1 + z_2) = f(z_1) f(z_2)
    [/tex]
    This is the rule of exponents. Of course, being analytic, the derivative of this function always exists. Differentiating the above identity w.r.t. z2 and setting z2 = 0 after that, we get the ODE:
    [tex]
    f'(z) = f'(0) f(z)
    [/tex]
    with the initial condition [itex]f(0) = 1[/itex] which follows from the above functional identity by taking z2 = 0 in the beginning.

    The solution of the above initial value problem is [itex]f(z) = \exp(k \, z)[/itex], where [itex]k \equiv f'(0)[/itex], and the exponential function is given by its Taylor series expansion (Frobenius method):
    [tex]
    \exp(z) = 1 + \sum_{n = 1}^{\infty}{\frac{z^n}{n!}}
    [/tex]

    From this expansion, and the Taylor series expansions of the sine and the cosine, which are to be understood as definitions of the trig functions, we can derive Euler's identity for purely imaginary arguments.
     
  9. Oct 4, 2012 #8

    micromass

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    Well, without proof, it isn't clear why such an analytic continuation should exist. You still need to actually define

    [tex]e^z=\sum_{n=0}^{+\infty} \frac{z^n}{n!}[/tex]

    and prove that [itex]e^{z+z^\prime}=e^ze^{z^\prime}[/itex]. After all, a lot of real-analytic functions do not extend on the complex plane such that the normal rules remain true.

    This FAQ seems nice for interested readers: https://www.physicsforums.com/showthread.php?p=4080308#post4080308 [Broken]
     
    Last edited by a moderator: May 6, 2017
  10. Oct 4, 2012 #9
    I did not go through the FAQ you had linked, but here's a proof of the sufficient condition.

    Suppose a function satisfies the initial value problem (IVP):
    [tex]
    f'(z) = k \, f(z), \ f(0) = 1
    [/tex]
    We already showed what the solution to this problem is:
    [tex]
    f(z) = \exp(k \, z)
    [/tex]
    where [itex]\exp(z)[/itex] stands for the Taylor's series.

    By induction you may prove that:
    [tex]
    f^{(n)}(z) = k^n \, f(z), \ n \ge 0
    [/tex]

    The Taylor's series converges on the whole complex plane. According to Taylor's Theorem, we would have:
    [tex]
    f(z_1 + z_2) = f(z_1) + \sum_{n = 1}^{\infty}{\frac{f^{(n)}(z_1)}{n!} \, z^n_2}
    [/tex]
    [tex]
    f(z_1 + z_2) = f(z_1) \, \left[ 1 + \sum_{n = 1}^{\infty}{\frac{(k \, z_2)^n}{n!}} \right]
    [/tex]
    [tex]
    f(z_1 + z_2) = f(z_1) \, \exp(k \, z)
    [/tex]
    [tex]
    f(z_1 + z_2) = f(z_1) \, f(z_2)
    [/tex]
    Therefore, we proved that this form of a function is also a sufficient, apart from being a necessary condition for the validity of the Laws of exponentials.
     
  11. Oct 4, 2012 #10

    micromass

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    Yep. So you can't say that the basic laws of exponents hold without going through such an argument first. That was what I said.
     
  12. Oct 4, 2012 #11
    Sure, but show me a proof of Euler's identity that was used in post #4!
     
  13. Oct 4, 2012 #12

    micromass

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    Why do you need a proof? It can also be accepted as definition of the complex exponential. It is easy to show that

    [tex]e^{x+iy}=e^x (cos(y)+i sin(y))[/tex]

    also is an analytic continuation of the real exponential. So in this case, Euler's identity becomes a definition. There is nothing mathematically wrong with this approach.
     
  14. Oct 4, 2012 #13
    So, then, poster #2 was not wrong.
     
  15. Oct 4, 2012 #14

    micromass

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    No, he wasn't.
     
  16. Oct 5, 2012 #15
    Glad you asked :P

    Also, I wish I had replied yesterday. Micromass is completely correct. We needed to define how exponentiation works when it is extended to complex powers. Until that point, we cannot simply assume how it works. In my post, all I had to say was that complex numbers are closed under addition and multiplication. Then, you can derive my solution.
     
  17. Oct 5, 2012 #16
    Why did you define the exponential through the Maclaurin series in step 1?
     
  18. Oct 5, 2012 #17
    Since that is a valid representation of eu, you can see how it leads to representing an imaginary power of e directly to a function of cosine and sine. I think this is how Euler showed the identity, too.
     
  19. Oct 5, 2012 #18
    But, that is called an ANALYTIC CONTINUATION of the exponential. In this case, it is done through the Taylor's (Maclaurin) series expansion, which has the same form for real and complex arguments, but the radius of convergence is infinite, and so the continuation is valid on the whole complex plane.

    I showed in an earlier post that such a function satisfies the exponential rule directly, without invoking any cosines or sines.
     
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