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Math Help - Finding the rootsof complex number

  1. #1
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    Finding the rootsof complex number

    Hi

    How would i do the following question:

    Find the roots of the equation: z^4-1

    This is what i have done:
    z^4 = -1

    z^4 = cis(\frac{\pi}{4}) , cis(\frac{5\pi}{4}), cis(\frac{9\pi}{4}), cis(\frac{13\pi}{4}


    z^4 = cis(\frac{\pi}{16}) , cis(\frac{5\pi}{16}), cis(\frac{9\pi}{16}), cis(\frac{13\pi}{16}

    P.S
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  2. #2
    MHF Contributor Unknown008's Avatar
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    I would have used the difference of two squares twice:

    z^4 - 1 = (z^2+1)(z^2 -1) = (z^2+1)(z+1)(z-1)

    Then, z = 1 and -1.

    For the other two complex roots, we get:

    z^2 = -1

    Which you should be able to solve.
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  3. #3
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    Quote Originally Posted by Unknown008 View Post
    I would have used the difference of two squares twice:

    z^4 - 1 = (z^2+1)(z^2 -1) = (z^2+1)(z+1)(z-1)

    Then, z = 1 and -1.

    For the other two complex roots, we get:

    z^2 = -1

    Which you should be able to solve.

    I think from what she/he wrote, though, that the equation she's trying to solve is

    z^4=-1=cis(\pi i+2k\pi i)=cis(\pi i(2k+1))\,,\,\,k=0,1,2,3 ,

    with solutions \displaystyle{z_k=cis\left(\frac{\pi}{4}i(2k+1)\ri  ght)}\,,\,\,k=01,2,3

    Tonio
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  4. #4
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    Quote Originally Posted by Paymemoney View Post
    Hi

    How would i do the following question:

    Find the roots of the equation: z^4-1
    This isn't an equation. I suppose you mean z^4- 1= 0

    This is what i have done:
    z^4 = -1

    z^4 = cis(\frac{\pi}{4}) , cis(\frac{5\pi}{4}), cis(\frac{9\pi}{4}), cis(\frac{13\pi}{4}
    I wrote what I did about what you had not being an equation because you seem to have a problem with writing exactly what you mean- and it hurt you here. Those four numbers are the fourth roots- you have already solved the problem at this point:
    z^4= cis(\pi) and
    z = cis(\frac{\pi}{4}) , cis(\frac{5\pi}{4}), cis(\frac{9\pi}{4}), cis(\frac{13\pi}{4})



    z^4 = cis(\frac{\pi}{16}) , cis(\frac{5\pi}{16}), cis(\frac{9\pi}{16}), cis(\frac{13\pi}{16}

    P.S
    Last edited by HallsofIvy; November 15th 2010 at 04:11 AM.
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  5. #5
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    Quote Originally Posted by Unknown008 View Post
    I would have used the difference of two squares twice:

    z^4 - 1 = (z^2+1)(z^2 -1) = (z^2+1)(z+1)(z-1)
    almost:

    z^4+ 1= (z^2+ i)(z^2- i)= (z-(\frac{\sqrt{2}}{2}+ i\frac{\sqrt{2}}{2})(z+(\frac{\sqrt{2}}{2}+ i\frac{\sqrt{2}}{2}))(z-(\frac{\sqrt{2}}{2}- i\frac{\sqrt{2}}{2})(z+(\frac{\sqrt{2}}{2}- i\frac{\sqrt{2}}{2})

    Then, z = 1 and -1.

    For the other two complex roots, we get:

    z^2 = -1

    Which you should be able to solve.
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  6. #6
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    Quote Originally Posted by HallsofIvy View Post
    almost
    z^4+ 1= (z^2+ i)(z^2- i)= (z-(\sqrt{2}{2}+ i\sqrt{2}{2})(z+(\sqrt{2}{2}+ i\sqrt{2}{2})(z-(\sqrt{2}{2}- i\sqrt{2}{2})(z+(\sqrt{2}{2}- i\sqrt{2}{2})
    Um... I don't understand why you look for z^4+1 while it's z^4-1...

    EDIT: Ah, okay. I understand now.
    Last edited by Unknown008; November 14th 2010 at 08:54 PM.
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  7. #7
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    yes my bad, its meant to be z^4+1=0 srry for the confusion
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