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Math Help - please help me

  1. #1
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    please help me

    x^2+2i = 0

    Can anyone show me how to get the answer?

    answer: -1+i , 1-i
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  2. #2
    Member Glaysher's Avatar
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    Quote Originally Posted by stupid_kid View Post
    x^2+2i = 0

    Can anyone show me how to get the answer?

    answer: -1+i , 1-i
    Use quadratic formula

    or

    x^2 = -2i

    Let x = a + bi (a,b real numbers)

    (a + bi)^2 = -2i

    a^2 - b^2 + 2abi = -2i

    2ab = -2 gives ab = -1 and a^2 - b^2 = 0

    a = -1/b so (-1/b)^2 - b^2 = 0

    1/b^2 - b^2 = 0

    Multiply by b^2

    1 - b^4 = 0

    b^4 = 1 gives b = 1 or b = -1

    b = 1 gives a = -1/1 = -1 giving solution x = -1 + i

    b = -1 gives a = -1/-1 = 1 giving solution x = 1 - i

    This methos should be used if you haven't done De Moivre's theorem
    Last edited by Glaysher; April 5th 2007 at 10:48 PM.
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by stupid_kid View Post
    x^2+2i = 0

    Can anyone show me how to get the answer?

    answer: -1+i , 1-i
    I don't have time to do the thing properly, I'll just have to hope you know that:
    e^{I*x} = cos(x) + I*sin(x) where I^2 = -1.

    We know that
    -I = e^{-I*(pi)/2}, so:

    x^2 + 2I = 0

    x^2 = -2I

    x = sqrt{-2I} = sqrt{2}*sqrt[e^{-I*(pi)/2}]

    x = sqrt{2}*[e^{-I*(pi)/2}]^{1/2} <-- Since sqrt(a) = a^{1/2}

    x = sqrt{2}*e^{-I*(pi)/4}

    I'll have to let you take the rest from here.

    -Dan
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  4. #4
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    Quote Originally Posted by stupid_kid View Post
    x^2+2i = 0

    Can anyone show me how to get the answer?

    answer: -1+i , 1-i
    The solutions to the equation,

    x^2 = -i

    Are,

    x= sqrt(2)/2 +/- sqrt(2)/2 i

    (You can confirm this with de Moiver's theorem. But I just happen to know this fact.

    Now,

    x^2 = -2i

    When you take the square root (or attempt to) you get the square root of -i, which is shown above. Times the sqrt(2).

    Thus,

    sqrt(2) * (sqrt(2)/2 +/- sqrt(2)/2) i) = 1+/- i
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