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Complex Numbers(2)

  1. Apr 13, 2007 #1
    Hey again;
    It doesn't seem to be my day for spotting the obvious;

    I've come accross this problem now:

    [tex]2^{8n}\exp^{2ni\Pi} = 2^{8n}[/tex]

    This imples that [tex]exp^{2ni\Pi} = 1[/tex], i know that [tex]exp^{i\Pi} =-1[/tex], but is that related?

    hmm, Latex issues...i'll try and fix them :) - sorted
    Craig
     
    Last edited: Apr 13, 2007
  2. jcsd
  3. Apr 13, 2007 #2

    cristo

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    Well, using the exponential form: [tex]e^{2\pi i}=\cos(2\pi)+i\sin(2\pi)=1[/tex]. So, [tex]e^{2\pi ni}=\cos(2n\pi)+i\sin(2n\pi)[/tex] and since cos(2n pi)=cos(2 pi)=1, and sin(2n pi)=sin(2 pi)=0, the result exp(2ni pi)=1 is obtained
     
  4. Apr 13, 2007 #3
    The relation to that famous identity is
    [tex]e^{2i\pi}=(e^{i\pi})^2=(-1)^2=1.[/tex]
     
  5. Apr 13, 2007 #4
    Hi craig100,

    you can visualize the complex number [tex]e^{i \theta}[/tex] as a vector of length 1 that is rotated by an angle [tex]\theta[/tex] with respect to the positive x-axis,
    see [URL [Broken].[/URL]

    Also, see the section "Polar Form" here

    The form [tex]z=r \cdot e^{i \theta}[/tex] is also called polar form of the complex number z. The vector has the length r and an angle [tex]\theta[/tex] with respect to the positive x axis.

    If you visualize the complex number in that way, you can immediately see what value the complex number z will have for certain values such as
    [tex]\theta = \pi/4, \pi/2, \pi, 2 \pi[/tex].

    This example shows [tex]z=5 \cdot e^{i \pi /3}[/tex]
     
    Last edited by a moderator: May 2, 2017
  6. Apr 15, 2007 #5

    Gib Z

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    Best way to show how its related is Eighty's post.
     
  7. Apr 15, 2007 #6
    Thank guys, quite simple afterall :smile:
     
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