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Could really use a hand with a complex exponential and integral problem

  1. Oct 22, 2006 #1
    I have two homework problems that have been driving me nuts:

    1.) evaluate the indefinite integral:

    integral(dx(e^ax)cos^2(2bx))

    where a and b are real positive constants. I just don't know where to start on it.

    2.) Find all values of i^(2/3)

    So far I have:

    i^(2/3)
    = e^(2/3*ln(i))
    = e^(2/3*i*(Pi/2 + 2*n*Pi))
    = e^(i*Pi/3)*e^(i*n*4Pi/3)

    I know my three solutions should end up being (1+i*sqrt(3))/2, (1-i*sqrt(3))/2, -1. But I can't seem to get there. I'd really appreciate any help. Sorry if my shorthand is confusing.
     
  2. jcsd
  3. Oct 22, 2006 #2

    Hurkyl

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    Can't you write it as nothing but complex exponentials?

    Have you tried plugging in values for n?
     
    Last edited: Oct 22, 2006
  4. Oct 22, 2006 #3
    When I plugged in values for n, I drew out my

    e^(i*Pi/3)* +/- (1/2 + i*sqrt(3)/2)

    I don't see how I get any of my three answers out of that.
     
  5. Oct 22, 2006 #4

    Hurkyl

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    (1) You know the value of e^(i*Pi/3).
    (2) n ranges over all integers... not just 0 and 1.
     
  6. Oct 22, 2006 #5
    I am assuming that it is just -1 as that is the only way it will end up as one of my answers, when n=0. I don't understand why though.

    -Got that, I was having a brain fart and thinking that for n=2, theta was 10Pi/4
     
  7. Oct 22, 2006 #6

    Hurkyl

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    But it's not -1... why do you think the n=0 solution has to be the one that's -1?
     
  8. Oct 22, 2006 #7
    it should be 1/2 + i*sqrt(3)/2, but then how do I ever get -1 for a value?
     
  9. Oct 22, 2006 #8
    because i'm an idiot... got it, thanks a bunch. Also, with the integral problem (I haven't done calc in a long long time), once I substitute

    cosx = (e^ix - e^-ix)/2

    how do I solve?
     
  10. Oct 22, 2006 #9

    Hurkyl

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    Well, what did you get when you simplified after substituting?
     
  11. Oct 22, 2006 #10
    something really ugly that doesn't seem to cancel out:

    integral(dx(e^ax)cos^2(2bx))

    =(e^ax)*(2e^(-4*b^2*x^2)-2e^(4*b^2*x^2))
     
  12. Oct 22, 2006 #11

    Hurkyl

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    Aha! There shouldn't be any x²'s in there. Remember your exponent laws: what is [itex](a^b)^c[/itex]?
     
  13. Oct 22, 2006 #12
    but I have:

    (e^ax)cos^2(2bx) = (e^ax)* [((e^(i*2*b*x)-e^(-i*2*b*x)/2)^2]

    when I square [""] how do I not end up with x^2?
     
  14. Oct 22, 2006 #13

    Hurkyl

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    Remember your basic arithmetic.

    First off, remember how to compute (a + b)².
    Secondly, as I said in my previous post, remember your exponent laws: what is
    [tex]\left( a^b \right)^c?[/tex]
     
  15. Oct 22, 2006 #14
    I know that (a + b)² = a² + 2ab + b² and (a^b)^c = a^b*c

    so if:

    (e^ax)cos^2(2bx) = (e^ax) [(e^(i*2*b*x) - e^(-i*2*b*x))/2]

    because: cosx = (e^ix - e^-ix)/2

    why when I square inside the brackets don't I get

    (e^ax)* [((e^(i*2*b*x)-e^(-i*2*b*x)/2)^2] ?

    I'm just really not getting something here
     
  16. Oct 23, 2006 #15
    [tex] \cos 2bx = \frac{e^{2ibx}+e^{-2ibx}}{2} [/tex]
    [tex] \cos 2bx = \frac{1}{2} \left[ (e^{2ibx}+e^{-2ibx})(e^{2ibx}+e^{-2ibx}) \right] [/tex]
    [tex] \cos^2 2bx = \frac{e^{4bix}}{4} + \frac{e^{-4bix}}{4} + \frac{1}{2} [/tex]

    right?
     
  17. Oct 23, 2006 #16

    quasar987

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    Provided your second line reads instead,

    [tex] \cos^2 2bx = \frac{1}{4} \left[ (e^{2ibx}+e^{-2ibx})(e^{2ibx}+e^{-2ibx}) \right] [/tex]
     
  18. Oct 23, 2006 #17
    Woops. That line was all kinds of messed up.
     
  19. Oct 23, 2006 #18
    yes, that makes sense, but how is that any easier to integrate when multiplied by e^ax?
     
  20. Oct 23, 2006 #19
    What's [tex] (e^{ax})(e^{bix}) [/tex] ?

    Remember how nice exponentials play with multiplication:

    [tex] (e^{ax})(e^{bix}) = e^{ax + bix} = e^{(a+bi)x} [/tex]


    So,
    [tex] \int e^{(a+bi)x} \, dx = \frac{1}{a+bi} e^{(a+bi)x} [/tex]
     
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