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Math Help - Equations on the set of complex numbers

  1. #1
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    Equations on the set of complex numbers

    Hello,
    Can someone help, how to solve
    x ^ 3 = 1
    and/or
    x ^ 4 = -4
    equations on the set of complex numbers?

    Thanks in advance!
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    I'll do the first, I leave figuring out the second to you
    Quote Originally Posted by Fujiwara View Post
    Hello,
    Can someone help, how to solve
    x ^ 3 = 1
    We will solve equation in polar form, and then we can convert to the more familiar \displaystyle a + ib form using Euler's equation, \displaystyle e^{i \theta} = \cos \theta + i \sin \theta

    Let \displaystyle x = Re^{i \theta}, then we have

    \displaystyle R^3 e^{i 3 \theta} = 1

    But by Euler's equation, \displaystyle 1 = e^{i 2k \pi}, hence

    \displaystyle R^3 e^{i3 \theta} = e^{i 2k \pi}

    \displaystyle \Rightarrow R^3 = 1 \implies R = 1, and

    \displaystyle e^{i 3 \theta} = e^{i 2k \pi}

    \displaystyle \Rightarrow x = e^{i \theta} = e^{i \frac {2 k \pi}3} for \displaystyle k = 0,~1,~2 (after that, the solutions repeat)

    Hence, \displaystyle x = e^{0},~e^{i \frac {2 \pi}3},~e^{i \frac {4 \pi}3}

    Which, by Euler's equation, gives:

    \displaystyle x = 1, ~-\frac 12 + i \frac {\sqrt 3}2,~-\frac 12 - i \frac {\sqrt 3}2

    those are the three roots in the complex numbers.
    x ^ 4 = -4
    equations on the set of complex numbers?

    Thanks in advance!
    Now, off you go. Try what you've learned.
    Last edited by Jhevon; October 31st 2010 at 11:28 AM.
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  3. #3
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    See also this discussion at Dr. Math Forum.
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  4. #4
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    Thanks for your help, now I understand and I can solve the problem.
    Last edited by mr fantastic; November 1st 2010 at 01:53 AM. Reason: Moved a question to a new thread.
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  5. #5
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    Another way to solve x^3= 1, which is equivalent to x^3- 1= 0 is to factor it as (x- 1)(x^2 +x+ 1)= 0. You can then solve x^2+ x+ 1= 0 with the quadratic formula.

    x^4= -4 is the same as x^4- (-4)= 0 and you can think of that as the difference of two squares: a^2- b^2= (a- b)(a+ b) with a= x^2 and b= 2i. x^4-(-4)= (x^2- 2i)(x^2+ 2i)= 0. But you will now need to change to polar form to find \sqrt{2i} and \sqrt{-2i}.
    Last edited by HallsofIvy; November 1st 2010 at 08:39 AM.
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