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Thread: 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 $\displaystyle e^{i\pi+2ik\pi}=-1$, where k is an integer
    And then take the fourth root of [tex]e^{i\pi+2ik\pi}... : $\displaystyle e^{i\pi/4},e^{i\pi/2},e^{3i\pi/4},e^{i\pi}$ (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 $\displaystyle e^{i\pi+2k\pi}=-1$, where k is an integer
    And then take the fourth root of $\displaystyle e^{i\pi+2k\pi}...$
    Typo: This should be either $\displaystyle e^{i(\pi+ 2k\pi)}= -1$ or $\displaystyle e^{i\pi+ 2ki\pi}$

    : $\displaystyle e^{i\pi/4},e^{i\pi/2},e^{3i\pi/4},e^{i\pi}$ (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.

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

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

    By De-Moivre's theorem

    $\displaystyle
    x=3(\cos(\frac{(2k+1)\pi}{4})+i\sin(\frac{(2k+1)\p i}{4}))
    $;k=0,1,2,3

    $\displaystyle
    k=0;x=3(\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}})
    $

    $\displaystyle
    k=1;x=3(-\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}})
    $

    $\displaystyle
    k=2;x=3(-\frac{1}{\sqrt{2}}-\frac{i}{\sqrt{2}})
    $

    $\displaystyle
    k=3;x=3(\frac{1}{\sqrt{2}}-\frac{i}{\sqrt{2}})
    $
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