Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

E to an imaginary power, equivalent expressions, inequal outcomes?

  1. Feb 25, 2012 #1
    I'm completely stumped. So is my high-school calculus teacher, but he hasn't done imaginary powers for forty-five years. Hopefully somebody can explain this...

    To clarify, I understand the reasoning between the following equation:

    [itex]e^{i x}=cos(x)+i sin(x)[/itex]

    Now, I need to put some things on the table

    First, do you agree that the following is true:

    [itex]a^{b c d}=(a^{b c})^d=(a^{b d})^c=(a^{c d})^b=(a^b)^{c d}=(a^c)^{b d}=(a^d)^{b c}[/itex]

    for all values of a, b, c, and d where the original expression is defined?

    If not, tell me why...

    If yes, let's continue.

    I will now define a, b, c, and d.

    [itex]a=e,b=\pi,c=i,d=\frac{1}{3}[/itex]

    Now, we'll go through some of the above equal expressions

    [itex]a^{b c d}=e^{\frac{\pi}{3} i}=cos(\frac{\pi}{3})+i sin(\frac{\pi}{3})=\frac{1}{2}+i\frac{\sqrt{3}}{2}=\frac{1+i\sqrt{3}}{2}[/itex]

    [itex](a^{b c})^d=(e^{\pi i})^{\frac{1}{3}}=(cos(\pi)+i sin(\pi))^{\frac{1}{3}}=((-1)+i*(0))^{\frac{1}{3}}=(-1)^{\frac{1}{3}}=-1[/itex]

    [itex](a^{b d})^c=(e^{\frac{\pi}{3}})^i=(2.849653908\ldots)^i=\frac{1+i\sqrt{3}}{2}[/itex]

    [itex](a^{c d})^b=(e^{\frac{i}{3}})^\pi=(cos(\frac{1}{3})+i sin(\frac{1}{3}))^\pi=(0.944956946\ldots+0.327194697\ldots i)^\pi=\frac{1+i\sqrt{3}}{2}[/itex]

    I won't do any of the form (a^x)^(y z), because my problem is already present...

    If the first expressions I mentioned are indeed equivalent, then why is the second one that I evaluated negative one?

    It is completely confusing.

    Oh, and hello to the forum for the first time :P
     
  2. jcsd
  3. Feb 25, 2012 #2

    pwsnafu

    User Avatar
    Science Advisor

    No it's not. It is true for positive reals, but just because it is true for that doesn't mean it is true for all complex numbers.
    It's easy to show that it fails because we have a counterexample: for an integer n
    [itex]e^{1+2ni\pi} = e[/itex]
    [itex](e^{1+2ni\pi})^{1+2ni\pi} = e[/itex]
    Multiply the exponents out and divide by e
    [itex]e^{-4\pi^2 n^2} = 1[/itex]
    which is nonsense. I've left out some steps so its a good exercise to put them back in.
     
  4. Feb 25, 2012 #3
    Okay, that's what I was looking for. The last time I did anything with proofs was two years ago in 10th grade geometry, and it mostly fill-in-the-blank. Thanks though!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: E to an imaginary power, equivalent expressions, inequal outcomes?
Loading...