1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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?

  2. jcsd
  3. Nov 23, 2012 #2


    User Avatar
    Science Advisor
    Homework Helper

    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook