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Math Help - Solutions to this equation - complex numbers

  1. #1
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    Solutions to this equation - complex numbers

    1. x^3-y^3=27
    The question says, what is the number of solutions to this equation in the field of complex numbers?

    The choices are

    a) 1
    b) 2
    c) 3
    d) infinitely many

    I have no clue to this one.

    2. And here's another one:

    For any real number x, \sqrt{\sqrt{-x}} is equal to

    a) +x
    b) -x
    c) complex
    d) pure imaginary

    Writing it in the form ({-x}^{1/2})^{1/2}, we can simplify it to \sqrt[4]{-x}

    Which shows that the number is not real (since no number to the power four would give -x). So +x and -x are eliminated. But what about the last two choices? What is the difference between complex and pure imaginary?
    Last edited by cosmonavt; January 7th 2012 at 02:45 AM.
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  2. #2
    MHF Contributor Siron's Avatar
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    Re: Solutions to this equation - complex numbers

    2) You can write: (-x)^{\frac{1}{4}}=(-1)^{\frac{1}{4}}\cdot x^{4}
    What about (-1)^{4}?

    You know every complex number has a real part and an imaginary part, if the real part is equal to zero (so if there's no real part, only an imaginary part) then we call it pure imaginary.
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  3. #3
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    Re: Solutions to this equation - complex numbers

    Quote Originally Posted by Siron View Post
    2) You can write: (-x)^{\frac{1}{4}}=(-1)^{\frac{1}{4}}\cdot x^{4}
    What about (-1)^{4}?

    You know every complex number has a real part and an imaginary part, if the real part is equal to zero (so if there's no real part, only an imaginary part) then we call it pure imaginary.
    Thanks for the reply!!

    Complex numbers are of the form a + b i (plus sign in between the complex and imaginary part). The number (-x)^{\frac{1}{4}}=(-1)^{\frac{1}{4}}\cdot x^{4} has a multiply sign in between so even if x^{4} is real, this does not qualify as a complex. Does this mean that this would be pure imaginary?

    Also, any clue about the first question?
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  4. #4
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    Re: Solutions to this equation - complex numbers

    Every imaginary number is a complex number: 0+ bi.
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  5. #5
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    Re: Solutions to this equation - complex numbers

    2. As HallsofIvy has already pointed out, the imaginary numbers are in fact a subset of the complex numbers.

    We can easily rewrite this expression as the following:
    \sqrt{\sqrt{-x}} = \sqrt{i\sqrt{x}} = \sqrt[4]{x}\sqrt{i}

    Sadly I was unable to come up with a solution of my own after two attempts, however a nice solution is presented here.

    Your answer is C.
    Last edited by Charney; January 13th 2012 at 12:53 PM.
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