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    Solve the equation z^(3/4) = root6 + (root2)i, giving your answers in the form re^itheta, where r>0 and theta is more than -pi and less than or equal to pi.
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    Can you guys reply with a fully worked solution?
    I dont understand when to add the 2kpi...
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    (Original post by Pokepal03)
    Solve the equation z^(3/4) = root6 + (root2)i, giving your answers in the form re^itheta, where r>0 and theta is more than -pi and less than or equal to pi.
    Demoirves theorum

    https://youtu.be/N0Y8ia57C24
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    (Original post by Kvothe the arcane)
    Demoirves theorum

    https://youtu.be/N0Y8ia57C24
    Thank you! I understand everything, but when there isnt an integer above z, i dont get how to solve it... can you please elaborate?
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    (Original post by Pokepal03)
    Thank you! I understand everything, but when there isnt an integer above z, i dont get how to solve it... can you please elaborate?
    Do it as normal and you'll have z^{1/4}=this for k=1, 2 etc...
    Then you can say z=this for k=1, 2 etc.. via DeMoirves
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    (Original post by Pokepal03)
    Can you guys reply with a fully worked solution?
    I dont understand when to add the 2kpi...
    Full solutions aren't encouraged on this forum - it doesn't help learning.

    As for the 2k\pi part, you know that cosine and sine are periodic functions, hence we can say that \displaystyle \cos\theta \equiv \cos(2k\pi + \theta)
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    Write the complex number in exponential form (or polar form if you'd prefer - although arguably exponential form makes the algebra more intuitive -  \sqrt{6}+\sqrt{2}i \equiv \sqrt{8}e^{i\arctan{1/ \sqrt{3}}} \equiv \sqrt{8}e^{\pi i/6}
    Does this perhaps make it easier to solve?

    So  z^{3/4}=\sqrt{8}e^{\pi i/6} .
 
 
 
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