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Thread: A Simple Complex Number Equation

  1. #1
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    Jun 2010
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    A Simple Complex Number Equation

    Hello, I'm new here.

    I have a problem with understanding this little equation:
    $\displaystyle
    z^{4} = i
    $

    According to WolframAlpha, the solutions are:
    $\displaystyle
    z_{1} = - \sqrt[8]{-1},

    z_{2} = \sqrt[8]{-1},

    z_{3} = - (-1)^{\frac{5}{8}},

    z_{4} = (-1)^{\frac{5}{8}}
    $

    I can understand the first two solutions, since this equation can be written like this:

    $\displaystyle
    (z - \sqrt{\sqrt{i}})(z + \sqrt{\sqrt{i}})(z^2 + \sqrt{i}) = 0
    $

    And I just don't know how to deal with the following equation:
    $\displaystyle
    (z^2 + \sqrt{i}) = 0
    $

    Apparently, the solutions are:
    $\displaystyle
    z_{3} = - \sqrt{- \sqrt{i}}
    $
    and:
    $\displaystyle
    z_{4} = \sqrt{- \sqrt{i}}
    $

    But how do I go from that to this?:
    $\displaystyle
    z_{3} = - (-1)^{\frac{5}{8}} $
    $\displaystyle
    z_{4} = (-1)^{\frac{5}{8}}
    $

    Please give me a step by step solution, if you can.

    Uhmm... I just got it, somehow seeing it on screen made it simpler, I'm sorry to bother you.
    It is of course:

    $\displaystyle
    - \sqrt{-\sqrt{i}} = - \sqrt{-\sqrt{\sqrt{-1}}} = $ $\displaystyle = - \sqrt{-1 * \sqrt{\sqrt{-1}}} = - \sqrt{-1} * \sqrt{\sqrt{i}} =$ $\displaystyle = - i * i^{\frac{1}{4}} = - i^{\frac{5}{4}} = - (- 1)^{\frac{5}{8}}
    $

    You may delete this topic, sorry for the fuss.
    Thanks,
    Moon
    Last edited by Moon; Jun 19th 2010 at 09:59 AM.
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  2. #2
    Banned
    Joined
    Oct 2009
    Posts
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    3
    Quote Originally Posted by Moon View Post
    Hello, I'm new here.

    I have a problem with understanding this little equation:
    $\displaystyle
    z^{4} = i
    $

    According to WolframAlpha, the solutions are:
    $\displaystyle
    z_{1} = - \sqrt[8]{-1},

    z_{2} = \sqrt[8]{-1},

    z_{3} = - (-1)^{\frac{5}{8}},

    z_{4} = (-1)^{\frac{5}{8}}
    $

    I can understand the first two solutions, since this equation can be written like this:

    $\displaystyle
    (z - \sqrt{\sqrt{i}})(z + \sqrt{\sqrt{i}})(z^2 + \sqrt{i}) = 0
    $

    And I just don't know how to deal with the following equation:
    $\displaystyle
    (z^2 + \sqrt{i}) = 0
    $

    Apparently, the solutions are:
    $\displaystyle
    z_{3} = - \sqrt{- \sqrt{i}}
    $
    and:
    $\displaystyle
    z_{4} = \sqrt{- \sqrt{i}}
    $

    But how do I go from that to this?:
    $\displaystyle
    z_{3} = - (-1)^{\frac{5}{8}} $
    $\displaystyle
    z_{4} = (-1)^{\frac{5}{8}}
    $

    Please give me a step by step solution, if you can.

    Uhmm... I just got it, somehow seeing it on screen made it simpler, I'm sorry to bother you.
    It is of course:

    $\displaystyle
    - \sqrt{-\sqrt{i}} = - \sqrt{-\sqrt{\sqrt{-1}}} = $ $\displaystyle = - \sqrt{-1 * \sqrt{\sqrt{-1}}} = - \sqrt{-1} * \sqrt{\sqrt{i}} =$ $\displaystyle = - i * i^{\frac{1}{4}} = - i^{\frac{5}{4}} = - (- 1)^{\frac{5}{8}}
    $

    You may delete this topic, sorry for the fuss.
    Thanks,
    Moon
    That's a weird way to write down complex numbers...I'd do the following:

    $\displaystyle z^4=i=e^{\pi i/2+2k\pi i}=e^{\frac{\pi i}{2}(1+4k)}\,,\,\,k=0,1,2,3$ , so now taking the fourth rooth on both sides and using De Moivre's theorem we get:

    $\displaystyle z_k=e^{\frac{\pi i}{8}(1+4k)}$ , and the four different roots are obtained by taking the four different values of k.

    Tonio
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