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Math Help - A Simple Complex Number Equation

  1. #1
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    Joined
    Jun 2010
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    7

    A Simple Complex Number Equation

    Hello, I'm new here.

    I have a problem with understanding this little equation:
    <br />
z^{4} = i<br />

    According to WolframAlpha, the solutions are:
    <br />
z_{1} = - \sqrt[8]{-1},   <br /> <br />
z_{2} = \sqrt[8]{-1},  <br /> <br />
z_{3} = - (-1)^{\frac{5}{8}},   <br /> <br />
z_{4} =  (-1)^{\frac{5}{8}}<br />

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

    <br />
(z - \sqrt{\sqrt{i}})(z + \sqrt{\sqrt{i}})(z^2 + \sqrt{i}) = 0<br />

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

    Apparently, the solutions are:
    <br />
z_{3} = - \sqrt{- \sqrt{i}}<br />
    and:
    <br />
z_{4} = \sqrt{- \sqrt{i}}<br />

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

    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:

    <br />
- \sqrt{-\sqrt{i}} = - \sqrt{-\sqrt{\sqrt{-1}}} =  =  - \sqrt{-1 * \sqrt{\sqrt{-1}}} = - \sqrt{-1} * \sqrt{\sqrt{i}} =  = - i * i^{\frac{1}{4}} = - i^{\frac{5}{4}} = - (- 1)^{\frac{5}{8}} <br />

    You may delete this topic, sorry for the fuss.
    Thanks,
    Moon
    Last edited by Moon; June 19th 2010 at 10:59 AM.
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  2. #2
    Banned
    Joined
    Oct 2009
    Posts
    4,261
    Thanks
    2
    Quote Originally Posted by Moon View Post
    Hello, I'm new here.

    I have a problem with understanding this little equation:
    <br />
z^{4} = i<br />

    According to WolframAlpha, the solutions are:
    <br />
z_{1} = - \sqrt[8]{-1},   <br /> <br />
z_{2} = \sqrt[8]{-1},  <br /> <br />
z_{3} = - (-1)^{\frac{5}{8}},   <br /> <br />
z_{4} =  (-1)^{\frac{5}{8}}<br />

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

    <br />
(z - \sqrt{\sqrt{i}})(z + \sqrt{\sqrt{i}})(z^2 + \sqrt{i}) = 0<br />

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

    Apparently, the solutions are:
    <br />
z_{3} = - \sqrt{- \sqrt{i}}<br />
    and:
    <br />
z_{4} = \sqrt{- \sqrt{i}}<br />

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

    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:

    <br />
- \sqrt{-\sqrt{i}} = - \sqrt{-\sqrt{\sqrt{-1}}} =  =  - \sqrt{-1 * \sqrt{\sqrt{-1}}} = - \sqrt{-1} * \sqrt{\sqrt{i}} =  = - i * i^{\frac{1}{4}} = - i^{\frac{5}{4}} = - (- 1)^{\frac{5}{8}} <br />

    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:

    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:

    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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