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Complex numbers help

  1. Nov 23, 2012 #1
    say we had a complex number [itex] w^4 [/itex] such that [itex] w^4 = -8 +i8\sqrt{3} [/itex] so [itex] w = 2(cos(\frac{\pi}{6} + \frac{k\pi}{2}) + isin(\frac{\pi}{6} + \frac{k\pi}{2})) [/itex] where k is an integer

    in a question I was asked to find the roots of w, as there will be 4 my first assumption is that the roots would be spread throughout the argand diagram, i.e the first root would be [itex] \sqrt{3} + i [/itex] and the second would be [itex] -\sqrt{3} + i [/itex] and so on till I get the 4 roots. However this is not the case, as if I substitute values for k, I get the roots to be [itex] \sqrt{3} + i, -1 + \sqrt{3}i, 1 - \sqrt{3}i, -\sqrt{3} - i [/itex] which is actually correct.

    However in the question below, the roots do seem to be spread evenly:


    Prove that [itex] cos\frac{\pi}{12} = m[/itex] and [itex] sin\frac{\pi}{12} = n, [/itex] where [itex] m = \frac{\sqrt{3} + 1}{2\sqrt{2}} [/itex] and [itex] n = \frac{\sqrt{3} -1}{2\sqrt{2}} [/itex]
    Hence find in terms of m and n, in the form a + ib, where a,b are real, the fourth roots of [itex] 4(cos(\frac{\pi}{3}) + isin(\frac{\pi}{3})) [/itex]

    the first root is [itex] z = \sqrt{2}m + i\sqrt{2}n [/itex] second [itex] -\sqrt{2}m + i\sqrt{2}n [/itex] and so on...

    why is it in one question the 4 roots are not spread evenly and I have to adjust k to find the roots, while in the other they are spread evenly. How do I spot whether or not they will be spread evenly?

    thanks,
     
  2. jcsd
  3. Nov 23, 2012 #2

    tiny-tim

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    hi phospho! :smile:

    the fourth (or nth) roots of any number w4 (or wn)will be w times any fourth (or nth) root of 1 :wink:

    ie w, iw, i2w (= -w), and i3w (= -iw)

    (so this is wrong …)
     
  4. Nov 24, 2012 #3
    I have the answers in front of me, and they have the same roots that I gave above.
     
  5. Nov 24, 2012 #4
    Why don't you simply halve cosine twice? To be more precise, [itex]\cos(\pi/3)=1/2[/itex] and we have the half-angle formula [itex]\cos^2(x)=\frac{\cos(2x)+1}{2}[/itex]. Applying it twice should give you [itex]\cos(\pi/12)[/itex], without the need of complex numbers at all.
     
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