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Homework Help: Complex numbers

  1. Mar 13, 2010 #1

    Mentallic

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    1. The problem statement, all variables and given/known data
    Show that if [tex]\theta[/tex] is not a multiple of [tex]2\pi[/tex] then

    [tex]Im\left(\frac{1-e^{i(n+1)\theta}}{1-e^{i\theta}}\right)=\frac{sin\left(\frac{1}{2}(n+1)\theta\right)sin(\frac{1}{2}n\theta)}{sin\frac{1}{2}\theta}[/tex]


    2. Relevant equations
    [tex]e^{i\theta}=cos\theta+isin\theta[/tex]


    3. The attempt at a solution

    I noticed that [tex]\frac{1-e^{i(n+1)\theta}}{1-e^{i\theta}}\right)[/tex] is a geometric summation with [tex]e^{i\theta}=r[/tex] then we have:

    [tex]1+e^{i\theta}+e^{i2\theta}+...+e^{in\theta}[/tex]

    So,

    [tex]Im\left(1+e^{i\theta}+e^{i2\theta}+...+e^{in\theta}\right)=sin\theta+sin2\theta+...+sin(n\theta)[/tex]

    I have no idea how to show this summation is equal to what I have to show. Most likely I'm not even headed in the right direction.
     
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  3. Mar 13, 2010 #2

    Dick

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    If you can express your quantity in the form a+ib, then it's easy to find Im(a+ib)=b. First multiply numerator and denominator by the complex conjugate of the denominator and then split stuff into real and imaginary parts.
     
  4. Mar 13, 2010 #3

    Mentallic

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    Ok then I have [tex]
    \frac{1-e^{i(n+1)\theta}}{1-e^{i\theta}}\right)=\frac{1-cos(n+1)\theta-isin(n+1)\theta}{1-cos\theta-isin\theta}
    [/tex]

    Multiplying through by the conjugate and expanding:


    [tex]\frac{1-cos\theta+isin\theta-cos(n+1)\theta+isin(n+1)\theta+cos\theta cos(n+1)\theta +icos\theta sin(n+1)\theta +isin\theta cos(n+1)\theta -sin\theta sin(n+1)\theta}{2-2cos\theta}[/tex]


    Just the imaginary part now, and simplifying since [tex]sin\theta cos(n+1)\theta +cos\theta sin(n+1)\theta=sin(n+2)\theta[/tex]


    [tex]\frac{sin\theta+sin(n+1)\theta+sin(n+2)\theta}{2(1-cos\theta)}[/tex]


    Seems interesting, because of the pattern. But I'm unsure what I can do from here...
     
  5. Mar 13, 2010 #4

    Mentallic

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    If the expression I showed is supposed to be equal to what I'm trying to show, then I must've failed to expand correctly.
    Using a simple case of n=1, [tex]\frac{sin\left(\frac{1}{2}(n+1 )\theta\right)sin(\frac{1}{2}n\theta)}{sin\frac{1} {2}\theta}[/tex] is supposed to be [tex]sin\theta[/tex]. I graphed my expression and it is definitely equal to [tex]sin\theta[/tex]...

    I'll check over my expanding again.
     
  6. Mar 13, 2010 #5

    Dick

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    Use some trig identities. I can see right away that your denominator is sin(theta/2)^2.
     
  7. Mar 13, 2010 #6

    Mentallic

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    Ok I did the algebra slowly and got the correct result now.

    It's

    [tex]\frac{sin\theta+sin n\theta-sin(n+1)\theta}{2-2cos\theta}[/tex]

    I have to somehow show this is equal to

    [tex]
    \frac{sin\left(\frac{1}{2}(n+1 )\theta\right)sin(\frac{1}{2}n\theta)}{sin\frac{1} {2}\theta}
    [/tex]
     
  8. Mar 13, 2010 #7

    Mentallic

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    It's been a while since I used half-angle trig identities so I don't remember them off-hand. I'll see if I can derive them in that case...

    But really, this topic is on complex numbers. I don't see why it would go into such great depths to test my algebra and trigonometry skills.
     
  9. Mar 13, 2010 #8

    Dick

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    Yes, I suppose. Might have been better to have you prove the result in the form you got first and skip the tricky trig.
     
  10. Mar 13, 2010 #9

    Mentallic

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    Well, we've already started on this method, let's not lose the momentum thus far :smile:

    So [tex]2-2cos\theta=4sin^2(\theta/2)[/tex]

    and to get anywhere with the numerator, I had to cheat a little bit. Using the answer that I need to find, by the double-angle formula:

    [tex]sin\left(\frac{1}{2}(n+1)\theta\right)sin\left(\frac{n\theta}{2}\right)=sin\left(\frac{\theta}{2}\right)cos\left(\frac{n\theta}{2}\right)sin\left(\frac{n\theta}{2}\right)+cos\left(\frac{\theta}{2}\right)sin^2\left(\frac{n\theta}{2}\right)[/tex]

    and

    [tex]2sinxcosx=sin(2x)[/tex]

    so it can be simplified to

    [tex]\frac{1}{2}sin\left(\frac{\theta}{2}\right)sin\left(n\theta\right)+cos\left(\frac{\theta}{2}\right)sin^2\left(\frac{n\theta}{2}\right)[/tex]


    Any other identities I can exploit?
     
  11. Mar 13, 2010 #10

    Dick

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    Arrrgh. Finally got it. I was going in circles for while. I'll write t for theta.
    i) Expand sin(nt+t) in the numerator using the sin addition formula.
    ii) Combine with the other two terms looking to get (1-cos(something*t)) factors.
    iii) Change the (1-cos(something*t)) stuff into sin(something*t/2)^2.
    iv) Change the remaining t stuff into t/2 stuff using sin(x)=2*sin(x/2)*cos(x/2).
    v) Pull out the common factors and recognize what's left as another sin addition formula.

    Let's not do that again soon. And I totally agree, not much to do with complex numbers here.
     
  12. Mar 13, 2010 #11

    vela

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    One trick you can use is to note that

    [tex]1-e^{ix} = e^{ix/2}(e^{-ix/2}-e^{ix/2})=-2ie^{ix/2}\sin(x/2)[/tex]
     
  13. Mar 13, 2010 #12

    Dick

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    Ah! That helps. I was looking for an easier way to do it once I started messing with the trig. But I didn't look back far enough. Thanks!
     
  14. Mar 14, 2010 #13

    Mentallic

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    Well this question is unfair for those few people that don't know that equality off by heart :biggrin:
     
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