Results 1 to 6 of 6
Like Tree1Thanks
  • 1 Post By Plato

Math Help - Real roots of unity

  1. #1
    Newbie
    Joined
    Oct 2012
    From
    london
    Posts
    8

    Real roots of unity

    How many real roots does the equation (z-2)^12 = 6 have?

    i have no idea how to go about this, can someone help me?
    Last edited by algebra123; November 19th 2012 at 12:53 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Mar 2011
    From
    Tejas
    Posts
    3,378
    Thanks
    745

    Re: Real roots of unity

    your equation doesn't look quite right.....
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Oct 2012
    From
    london
    Posts
    8

    Re: Real roots of unity

    Not quite sure what happened, but I've changed it now.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,615
    Thanks
    1578
    Awards
    1

    Re: Real roots of unity

    Quote Originally Posted by algebra123 View Post
    How many real roots does the equation (z-2)^12 = 6 have?i have no idea how to go about this, can someone help me?

    Each of these \sqrt[{12}]{6}\exp \left( {\frac{{k\pi i}}{6}} \right) + 2,\quad k = 0,1, \cdots ,11 is a solution.

    How many of those are real.
    Thanks from algebra123
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    Mar 2011
    From
    Tejas
    Posts
    3,378
    Thanks
    745

    Re: Real roots of unity

    note that z - 2 real implies that z = (z - 2) + 2 is real, and vice versa.

    so let u = z - 2, so we want the number of real roots to u12 = 6.

    that is, we're looking for real factors of u12 - 6.

    now u12 - 6 = (u6 + √6)(u6 - √6).

    notice that u6 + √6 > 0 for any real u, so we don't get any solutions there. so any REAL roots of u12 - 6 must be real roots of u6 - √6.

    again, we have:

    u^6 - \sqrt{6} = (u^3 + \sqrt[4]{6})(u^3 - \sqrt[4]{6})

    each cubic can be factored again:

    u^3 + \sqrt[4]{6} = (u + \sqrt[12]{6})(u^2 - \sqrt[12]{6}u + \sqrt[6]{6})
    u^3 - \sqrt[4]{6} = (u - \sqrt[12]{6})(u^2 + \sqrt[12]{6}u + \sqrt[6]{6})

    this gives us two real roots u (to get z we'll have to add 2 to u). what about the quadratic factors?

    let's look at their discriminants:

    b2 - 4c (the same for both quadratics, since they only differ in the sign for b).

    we have:

    (\sqrt[12]{6})^2 - 4(\sqrt[6]{6}) = \sqrt[6]{6} - 4\sqrt[6]{6} = (-3)\sqrt[6]{6} < 0 so these have no real roots.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,504
    Thanks
    1400

    Re: Real roots of unity

    The easy way is to remember that in the real numbers \displaystyle \begin{align*} x^a = b \end{align*} has two solutions if \displaystyle \begin{align*} a \end{align*} is even and \displaystyle \begin{align*} b \geq 0 \end{align*}, and has one solution if \displaystyle \begin{align*} a \end{align*} is odd.

    \displaystyle \begin{align*} (z - 2)^{12} &= 6 \\ z - 2 &= \pm \sqrt[12]{6} \\ z &= 2 \pm \sqrt[12]{6} \end{align*}
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Roots of Unity
    Posted in the Pre-Calculus Forum
    Replies: 5
    Last Post: August 29th 2010, 02:32 PM
  2. nth roots of unity?
    Posted in the Algebra Forum
    Replies: 1
    Last Post: May 25th 2009, 10:55 AM
  3. nth roots of unity ???
    Posted in the Algebra Forum
    Replies: 2
    Last Post: January 7th 2009, 07:44 PM
  4. Roots of unity
    Posted in the Trigonometry Forum
    Replies: 4
    Last Post: December 30th 2008, 06:59 PM
  5. Roots of unity
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: April 21st 2008, 09:49 PM

Search Tags


/mathhelpforum @mathhelpforum