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Exponential law and complex numbers

  1. Sep 25, 2007 #1
    I was playing around with complex exponentials and came to this result:

    [tex]$\begin{eqnarray*}
    e^{\frac{2\pi i}{5}}&=&e^\left(\frac{2}{5}\right)\left(\pi i\right)\\
    &=&\left(e^{\pi i}\right)^{\frac{2}{5}}\\
    &=&\left(-1\right)^{\frac{2}{5}}\\
    &=&1\end{eqnarray*}$[/tex]

    But obviously [tex]e^{\frac{2\pi i}{5}}=\mathrm{cos}\frac{2\pi}{5}+i \mathrm{sin}\frac{2\pi}{5}\approx 0.309+0.951i\neq 1[/tex]

    So after some research I found that the exponential law [tex]a^{mn}=\left(a^{m}\right)^{n}[/tex] is only true when [tex]a,m,n\in\mathbb{R}[/tex] and not otherwise.

    My question now is WHY does the index law fail for imaginary base/exponents?

    Thanks!

    *PS how can I get rid of that [tex](0)[/tex] appearing after the eqnarray in my [tex]$\LaTeX$[/tex] code above?? :P
     
    Last edited: Sep 25, 2007
  2. jcsd
  3. Sep 25, 2007 #2

    Gib Z

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    Homework Helper

    Because dealing only in real numbers, questions like "the fifth root of 1" is easy, its just 1. However with complex numbers we know that there are 5 solutions, and 1 is the only real one, the rest are imaginary. I think you will find that exp( 2*pi*i /5) is one of the roots =]
     
  4. Sep 25, 2007 #3
    In fact, the rule does work for all [tex]a,m,n\in\mathbb{C}[/tex] but only by extending the complex plane using an infinite number of branch cuts and planes. The logarithm rule works in this way as well. The rule will fail if you are only using one "copy" of the complex plane.

    The rule does work in your example as well. [tex](-1)^2 = 1[/tex], and one of the fifth roots of 1 is indeed [tex]e^{\frac{2\pi i}{5}}[/tex]. In fact, in the branced complex plane, this is the only fifth root of the 1 in question, as in the branched plane, roots are no longer multivalued.

    It's all very confusing at first, but you'll get used to it.
     
  5. May 13, 2008 #4
    Positronized, the awnser to your question is simple. The first expression you wrote, is equivalent to the second one. The only error you did was not to take the "magnitude" of your complexe number, this is the awnser you are looking for. All you do is:

    Magnitude = ((Real)^2 + (Img)^2)^0.5

    So, in your case, you find:

    Magnitude = (Cos(2*pi/5)^2 + Sin(2*pi/5)^2)^0.5 = 1^0.5 = 1 hence the two methods you used are equivilent.

    I suggest you go read wiki for any basic questions on complexe numbers.
     
    Last edited: May 13, 2008
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