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Math Help - Nth root of a complex number:

  1. #1
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    Nth root of a complex number:

    I need to find the fourth root of -81. Thank you in advance.
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  2. #2
    Moo
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    Hello,
    Quote Originally Posted by Neversh View Post
    I need to find the fourth root of -81. Thank you in advance.
    -81=(-1)*(81)
    The 4th root of 81 is 3 (because 81=9x9=3x3x3x3)

    Now, what is the 4th root of -1 ? Write e^{i\pi+2ik\pi}=-1, where k is an integer
    And then take the fourth root of [tex]e^{i\pi+2ik\pi}... : e^{i\pi/4},e^{i\pi/2},e^{3i\pi/4},e^{i\pi} (k=0,1,2,3) then as k grows up, you'll get back to these solutions...
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  3. #3
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    [QUOTE=Moo;327307]Hello,

    -81=(-1)*(81)
    The 4th root of 81 is 3 (because 81=9x9=3x3x3x3)

    Now, what is the 4th root of -1 ? Write e^{i\pi+2k\pi}=-1, where k is an integer
    And then take the fourth root of e^{i\pi+2k\pi}...
    Typo: This should be either e^{i(\pi+ 2k\pi)}= -1 or e^{i\pi+ 2ki\pi}

    : e^{i\pi/4},e^{i\pi/2},e^{3i\pi/4},e^{i\pi} (k=0,1,2,3) then as k grows up, you'll get back to these solutions...
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  4. #4
    Senior Member pankaj's Avatar
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    Let x be the fourth root of -81.

    x^4=-81=81(\cos \pi +i\sin \pi)=3^4(\cos(2k\pi+\pi)+i\sin(2k\pi+\pi);k=0,1,2,  3

    x=3(\cos(2k\pi+\pi)+i\sin(2k\pi+\pi))^{\frac{1}{4}  }

    By De-Moivre's theorem

     <br />
x=3(\cos(\frac{(2k+1)\pi}{4})+i\sin(\frac{(2k+1)\p  i}{4}))<br />
;k=0,1,2,3

     <br />
k=0;x=3(\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}})<br />

     <br />
k=1;x=3(-\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}})<br />

     <br />
k=2;x=3(-\frac{1}{\sqrt{2}}-\frac{i}{\sqrt{2}})<br />

     <br />
k=3;x=3(\frac{1}{\sqrt{2}}-\frac{i}{\sqrt{2}})<br />
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