Results 1 to 7 of 7

Thread: Finding the rootsof complex number

  1. #1
    Super Member
    Joined
    Dec 2008
    Posts
    509

    Finding the rootsof complex number

    Hi

    How would i do the following question:

    Find the roots of the equation: $\displaystyle z^4-1$

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

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


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

    P.S
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Unknown008's Avatar
    Joined
    May 2010
    From
    Mauritius
    Posts
    1,260
    Thanks
    1
    I would have used the difference of two squares twice:

    $\displaystyle z^4 - 1 = (z^2+1)(z^2 -1) = (z^2+1)(z+1)(z-1)$

    Then, z = 1 and -1.

    For the other two complex roots, we get:

    $\displaystyle z^2 = -1$

    Which you should be able to solve.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Banned
    Joined
    Oct 2009
    Posts
    4,261
    Thanks
    3
    Quote Originally Posted by Unknown008 View Post
    I would have used the difference of two squares twice:

    $\displaystyle z^4 - 1 = (z^2+1)(z^2 -1) = (z^2+1)(z+1)(z-1)$

    Then, z = 1 and -1.

    For the other two complex roots, we get:

    $\displaystyle z^2 = -1$

    Which you should be able to solve.

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

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

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

    Tonio
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Apr 2005
    Posts
    19,773
    Thanks
    3028
    Quote Originally Posted by Paymemoney View Post
    Hi

    How would i do the following question:

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

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

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



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

    P.S
    Last edited by HallsofIvy; Nov 15th 2010 at 03:11 AM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    Apr 2005
    Posts
    19,773
    Thanks
    3028
    Quote Originally Posted by Unknown008 View Post
    I would have used the difference of two squares twice:

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

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

    Then, z = 1 and -1.

    For the other two complex roots, we get:

    $\displaystyle z^2 = -1$

    Which you should be able to solve.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor Unknown008's Avatar
    Joined
    May 2010
    From
    Mauritius
    Posts
    1,260
    Thanks
    1
    Quote Originally Posted by HallsofIvy View Post
    almost
    $\displaystyle z^4+ 1= (z^2+ i)(z^2- i)= (z-(\sqrt{2}{2}+ i\sqrt{2}{2})(z+(\sqrt{2}{2}+ i\sqrt{2}{2})(z-(\sqrt{2}{2}- i\sqrt{2}{2})(z+(\sqrt{2}{2}- i\sqrt{2}{2})$
    Um... I don't understand why you look for $\displaystyle z^4+1$ while it's $\displaystyle z^4-1$...

    EDIT: Ah, okay. I understand now.
    Last edited by Unknown008; Nov 14th 2010 at 07:54 PM.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Super Member
    Joined
    Dec 2008
    Posts
    509
    yes my bad, its meant to be $\displaystyle z^4+1=0$ srry for the confusion
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Complex Number - Finding the Modulus
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: Nov 8th 2010, 06:51 PM
  2. Finding the roots of a complex number
    Posted in the Pre-Calculus Forum
    Replies: 5
    Last Post: Aug 14th 2010, 01:49 PM
  3. Finding roots of complex number
    Posted in the Pre-Calculus Forum
    Replies: 4
    Last Post: May 19th 2009, 01:55 PM
  4. finding powers of a complex number
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: Dec 10th 2008, 07:35 PM
  5. Replies: 3
    Last Post: Aug 21st 2006, 10:57 AM

Search Tags


/mathhelpforum @mathhelpforum