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Homework Help: Integration, complex roots and radians

  1. Mar 30, 2006 #1
    The first one, integration, I just want to check my answer.

    [tex]\int \frac{1}{64} (\cos6\theta + 6\cos4\theta + 15\cos2\theta + 20) = \frac{1}{64} (\frac{\sin6\theta}{6} + \frac{6\sin4\theta}{4} + \frac{15\sin2\theta}{2} + 20\theta + c[/tex]

    I just wasn't sure if the integral of a constant wrt theta was constant*theta.

    The complex roots question:
    [tex]z^5 - i = (z-w1)(z-w2)(z-w3)(z-w4)(z-w5) = 0[/tex]
    Show that w1 + w2 + w3 + w4 + w5 = 0

    From what our lecturer told us, [tex]-\prod_{1}[/tex] = 0 so the sum of the roots = 0. I don't really understand this though :confused:

    The radians problem... is basically because I don't know how to express radian in decimals into pi radians. [As part of an answer for coshz = 2i] I want to express the argument of [tex]2 \pm \sqrt{5}[/tex] which would be [tex]\tan^{-1} \sqrt{5}/2[/tex].
    In case you want to know, I found [tex]z = log(2 \pm i\sqrt{5}) = ln|2 \pm i\sqrt{5}| + i arg(2 \pm i\sqrt{5})[/tex]
    I just need help finishing up the question :)
    Last edited: Mar 30, 2006
  2. jcsd
  3. Mar 30, 2006 #2


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    Do you mean...?

    [tex] \int 0 \ d\theta = \mbox{const}_{1} \cdot \theta + \mbox{const}_{2}[/tex]

    If so, then it's simply wrong, unless [itex] \mbox{const}_{1}=0 [/itex] which doesn't make any sense to write in the first place...

    For the second problem, couldn't u use a formula due to Viète...?

  4. Mar 30, 2006 #3
    For the roots one, try writing i as [tex]i = e^{\frac{i\pi}{2}+2\pi ni}[/tex] and taking the fifth root of that. Then you can sum a geometric series to give you the answer.
  5. Mar 30, 2006 #4
    oops sorry, I missed out a 20 in the integral so I'm asking if

    [tex]\int 20 d\theta = 20\theta[/tex]

    I have never heard of this forumla?

    And also, AlphaNumeric, could you explain why you suggested [tex]i = e^{\frac{i\pi}{2}+2\pi ni}[/tex]??
  6. Mar 30, 2006 #5
    [tex]z^{5} = i[/tex]

    You're looking for the 5 solutions of this equation, you're [tex]\omega_{1}[/tex] to [tex]\omega_{5}[/tex]

    They are clearly of modulus 1, and complex, so you'll have [tex]\omega_{n} = e^{it}[/tex] for some t you need to find. They are easy to find if you write i in for form I suggest.

    [tex]z^{5} = e^{\frac{i\pi}{2}+2\pi n i}[/tex]
    Take 5th roots to get
    [tex]z = e^{\frac{i\pi}{10}+\frac{2\pi n i}{5}[/tex]

    Usually express arguments in the range 0 to 2pi, so find the values of n which give you an argument in that range. n=0,1,2,3,4 will give you that, giving you roots

    [tex]w_{1} = e^{\frac{i\pi}{10}}[/tex]
    [tex]w_{2} = e^{\frac{i\pi}{10}+\frac{2\pi i}{5}} = e^{\frac{5i\pi}{10}} = e^{\frac{i\pi}{2}} = i[/tex]

    There's your 5 roots. Now you want to show they sum to zero. Well there's 5 of them and they are in geometric series, and you know how to sum those. The general formula is

    [tex]S_{n} = a\frac{r^{n}-1}{r-1} [/tex]

    In this case [tex]r = e^{\frac{2\pi i}{5}}[/tex], so [tex]r^{5} = 1[/tex], and so [tex]S_{n} = 0[/tex].
    Last edited: Mar 30, 2006
  7. Mar 31, 2006 #6
    ok, thanks for your help alphanumeric, it's a little easier to see why it is true that way.
  8. Mar 31, 2006 #7
    For a more graphical representation for why it's true, think about an Argand diagram. If you've the equation [tex]z^{n} = k[/tex] for any complex k, then the roots will be centred about the origin in a circle spread at an angle [tex]\frac{2\pi}{n}[/tex] radians apart from one another. You can see this from writing [tex]k = re^{i\theta}[/tex] (as I did for i) and repeating that method.

    The average value of the n roots will be their sum divided by n. Geometrically, this is the centre of the circle, the origin. Therefore they sum to zero.

    As an example, consider [tex]z^{n} = 16[/tex]. Solutions are 2, -2, 2i, -2i. Plot them in the complex plane and you'll see they lie on a circle radius 2, with the circle centred on the origin.

    I hope that helps you're understanding a bit :smile:
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