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Complex Roots - Not sure I did this right

  1. Mar 27, 2009 #1
    Hello. I'm not sure whether I did this right or messed up somewhere, just need to confirm my results...thanks to anybody who bothers answering.

    1. The problem statement, all variables and given/known data

    Find all the roots of [tex]z^{4}=1-i[/tex]

    2. Relevant equations

    I guess I should state De Moivre's here...

    [tex](r cis(\vartheta))^{n}=r^{n} cis (n\vartheta)[/tex]

    3. The attempt at a solution

    Firstly I re-wrote [tex]z^{4}=1-i[/tex] as

    [tex]z^{4}=\sqrt{2} cis (\frac{-\pi}{4})[/tex]

    Using De Moivre's,

    [tex]z=(2\frac{1}{2})^{\frac{1}{4}} cis (\frac{1}{4}(\frac{-\pi}{4}+k2\pi))[/tex]

    [tex]z=2\frac{1}{8} cis (\frac{1}{4}(\frac{-\pi}{4}+k2\pi))[/tex]

    I found the four roots letting k=0,1,2,3

    [tex]z=2^\frac{1}{8} cis (\frac{-\pi}{16})[/tex]

    [tex]z=2^\frac{1}{8} cis (\frac{1}{4}(\frac{-\pi}{4}+2\pi))=2^\frac{1}{8} cis (\frac{7\pi}{16})[/tex]

    [tex]z=2^\frac{1}{8} cis (\frac{1}{4}(\frac{-\pi}{4}+4\pi))=2^\frac{1}{8} cis (\frac{15\pi}{16})[/tex]

    [tex]z=2^\frac{1}{8} cis (\frac{1}{4}(\frac{-\pi}{4}+6\pi))=2^\frac{1}{8} cis (\frac{23\pi}{16})[/tex]
     
  2. jcsd
  3. Mar 27, 2009 #2
    Hi Atena!

    Your answer is correct. You can check this by taking the fourth powers of the solutions you got (using deMoivre).
     
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