# Real roots of unity

• November 19th 2012, 12:32 PM
algebra123
Real roots of unity
How many real roots does the equation (z-2)^12 = 6 have?

• November 19th 2012, 12:50 PM
Deveno
Re: Real roots of unity
your equation doesn't look quite right.....
• November 19th 2012, 12:53 PM
algebra123
Re: Real roots of unity
Not quite sure what happened, but I've changed it now.
• November 19th 2012, 01:26 PM
Plato
Re: Real roots of unity
Quote:

Originally Posted by algebra123
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.
• November 19th 2012, 02:12 PM
Deveno
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.
• November 19th 2012, 05:56 PM
Prove It
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*}