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Math Help - quick complex number question

  1. #1
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    quick complex number question

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    Last edited by yellow4321; March 6th 2008 at 05:13 AM.
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    There are three possible values of z.
    z = 2e^{\frac{{\pi i}}{6}} \,,\,2e^{\frac{{ - \pi i}}{2}} \,\mbox{or} \,2e^{\frac{{5\pi i}}{6}}
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    Quote Originally Posted by yellow4321 View Post
    if z^3=8i
    i know very little about complex numbers.
    i used euler to get rt(3) +i
    but im having doubts would the answer be +/- 2i ? i thought that too easy.
    Let \zeta = e^{2\pi i/3}.
    And \sqrt[3]{8i} = |8i|^{1/3} e^{i\arg(8i)/3} = 2 e^{i\pi/6} = 2 \cos \frac{\pi}{6} + 2 i\sin \frac{\pi}{6} = \sqrt{3} + i.

    Thus, the solution to z^3 = 8i are: (\sqrt{3}+i), (\sqrt{3}+i)\zeta, (\sqrt{3}+i)\zeta^2.
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  4. #4
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    ok thanks i understand how to get 2e^i*(pi/6) but the over two solutions i do not understand how to achieve?
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    Quote Originally Posted by yellow4321 View Post
    ok thanks i understand how to get 2e^i*(pi/6) but the over two solutions i do not understand how to achieve?
    Theorem: The polynomial x^n - 1 has n roots over \mathbb{C}. Furthermore, if \zeta = e^{2\pi i/n} then \zeta, \zeta^2, ... ,\zeta^n are these solutions.

    Proof: We use a fact that the polynomial f(x) = x^n - 1 can have at most n zeros. If we let \zeta = e^{2\pi i/n} then it is evident that \zeta^n =  e^{2\pi i} = 1 and so f(\zeta) = 0. In fact if 1\leq k \leq n then (\zeta^k)^n = (\zeta^n)^k = 1^k = 1. Thus, f(\zeta^k) = 0. Now we show that all \zeta,...,\zeta^n are distinct. Suppose that \zeta^a = \zeta^b then \zeta^{(a-b)} = 1 and so \cos \frac{2\pi (a-b)}{n} + i\sin \frac{2\pi (a-b)}{n} = 1 + 0i that means \frac{2\pi (a-b)}{n} = 0 because 1\leq a,b\leq n so a=b. Thus, the set \{ \zeta , ... ,\zeta^n =1\} is a complete set of zeros for f(x).

    Theorem: If a\not = 0 is a complex number then f(x) = x^n - a has exactly n zeros and those are a^{1/n}\zeta, a^{1/n}\zeta^2, ... ,a^{1/n}\zeta^n.

    Proof: First the polynomial f(x) can have at most n zeros so if we can show that these are all zeros (and that they are distinct) it will complete the proof. First a^{1/n} is the principle n-root defined at |a|^{1/n} e^{i\arg(a)/n}, and so \left( a^{1/n} \right)^n = 1. Now if 1\leq k \leq n it means \left( a^{1/n} \zeta^k \right)^n = \left( a^{1/n}\right)^n \left( \zeta^k \right)^n = 1. Which means a^{1/n} \zeta^k are all zeros. Finally to show there are all distinct it is easy. Suppose a^{1/n} \zeta^r = a^{1/n} \zeta^s where 1\leq r,s \leq n then since a\not = 0 it means \zeta^r = \zeta^s \implies r=s, as in the previous proof. And so this means that all these zeros are distinct.
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  6. #6
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    Quote Originally Posted by yellow4321 View Post
    ok thanks i understand how to get 2e^i*(pi/6) but the over two solutions i do not understand how to achieve?
    The three cube roots are located on a circle with center at the origin and radius 2.
    On that circle the roots are equally ‘spread’ out.
    Because you know one is at {\frac{\pi}{6}}, see how the other two are spread out.
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  7. #7
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    ok i sort of understand, and now know the fact that you have n-1 solutions if n is the power so solve for k=0,1,2..n-1 but could you show me how to get to the 5pi/6 and -pi/2 why arent i just added pi to pi/6 to get 7pi/6 or something.


    arrrr i see so if i worked in degrees i could just do 360/3 which is 120 then add and minus this 30, then il get the radian form.
    you guys are my heros
    Last edited by yellow4321; January 15th 2008 at 02:31 PM. Reason: brainwave!
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