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Exponential complex numbers

  1. Jul 29, 2008 #1
    1. The problem statement, all variables and given/known data
    I've never understood [tex]e^{i\theta}[/tex] very well. I know that [tex]e^{i\theta} = cos \theta + i sin \theta[/tex], but how about [tex]e^{4i}[/tex]? Would this be [tex]cos 1 + 4i sin 1[/tex] or [tex]cos 4 + i sin 4[/tex]? What's the general rule for these kinds of numbers?


    2. Relevant equations
    [tex]e^{i\theta} = cos \theta + i sin \theta[/tex]
    [tex]e^{i\pi} + 1 = 0[/tex]


    3. The attempt at a solution
    None, really. Just wondering how to evaluate numbers like above.
     
  2. jcsd
  3. Jul 29, 2008 #2
    I'm not sure where you got cos(1) + 4i*sin(1) from. For e^(4i), you would simply replace theta with 4, which gives the latter of the two expressions above. I assume we're dealing with radian measure so cos(4) + i*sin(4) can be evaluated using a calculator. On the complex plane, the point would be in the third quadrant since pi < 4 < (3/2)pi.
     
  4. Jul 29, 2008 #3

    HallsofIvy

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    The "rule" is given in the formula you state: [tex]e^{i\theta}= cos(\theta)+ i sin(\theta)[/tex]. In [tex]e^{4i}[/itex], the number multiplying i is "4": [itex]\theta= 4[/itex]. [tex]e^{4i}= cos(4)+ i sin(4)[/tex].


     
  5. Aug 12, 2008 #4
    Hmm... so does this mean that whatever you multiply [tex]i[/tex] by will be the argument of the sine and cosine? But my differential equations book has

    [tex]e^{-ibx} = cos bx - i sin bx[/tex].

    How do I know if this isn't

    [tex]e^{-ibx} = e^{i(-bx)} = cos (-bx) + i sin (-bx)[/tex]?

    How do I tell when it's one and when it's the other?
     
  6. Aug 12, 2008 #5
    OHHHH... never mind. =]

    It just hit me: both are actually the same - just apply the appropriate trig identities.
     
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