Can you find its roots?

Algebra Level 2

How many roots does the equation $x^{ 8 }- 1=0$ have?

Bonus : find the exact values of all these roots.

Image Credit: Wikimedia László Németh

The answer is 8.

1 solution

Rmflute Shrivastav
Apr 4, 2015

For anything to the power x, there will be x roots. That is the "rule". This is only true however, if you are looking for all roots, including imaginary.

Lets find the 8 roots.

Basically we have: $x^8 - 1 = 0 \equiv x^8 = 1 \equiv x^8 = 1 + 0i$

Therefore based on De Moivre's formula we can find the roots using:

$r^\frac{1}{n}\cdot (cos (\frac {\theta + 2k\pi}{n}) + i\cdot sin (\frac {\theta + 2k\pi}{n})); k = 0, 1, 2, \ldots , n-1.$

Where $r$ is the modulus of the complex number and $\theta$ is its argument.

For our problem we have:

$r = \sqrt {1^2 + 0^2} = 1$

$tan (\theta) = \frac {0}{1} = 0 \equiv \theta = 0$ , since $cos (\theta) > 0$ .

And then:

$1^\frac{1}{8}\cdot(cos (\frac {2k\pi}{8}) + i\cdot sin (\frac {2k\pi}{8})); k = 0, 1, 2, \ldots, 7.$

And our roots are:

$k = 0 \rightarrow cos (0) + i\cdot sin (0) = 1$

$k = 1 \rightarrow cos (\frac {\pi}{4}) + i\cdot sin (\frac {\pi}{4}) = \frac {\sqrt {2}}{2} + \frac {\sqrt {2}}{2}i$

$k = 2 \rightarrow cos (\frac {\pi}{2}) + i\cdot sin (\frac {\pi}{2}) = i$

$k = 3 \rightarrow cos (\frac {3\pi}{4}) + i\cdot sin (\frac {3\pi}{4}) = -\frac {\sqrt {2}}{2} + \frac {\sqrt {2}}{2}i$

$k = 4 \rightarrow cos (\pi) + i\cdot sin (\pi) = -1$

$k = 5 \rightarrow cos (\frac {5\pi}{4}) + i\cdot sin (\frac {5\pi}{4}) = -\frac {\sqrt {2}}{2} - \frac {\sqrt {2}}{2} i$

$k = 6 \rightarrow cos (\frac {3\pi}{2}) + i\cdot sin (\frac {3\pi}{2}) = -i$

$k = 7 \rightarrow cos (\frac {7\pi}{4}) + i\cdot sin (\frac {7\pi}{4}) = \frac {\sqrt {2}}{2} - \frac {\sqrt {2}}{2}i$

Done!

Carlos Revés - 6 years, 2 months ago

There’s another way to find the roots by decomposing the polynomial $x^8 - 1 = 0$ .

$x^8 - 1 = 0 \equiv (x^4 + 1)(x^4 - 1) = 0 \equiv (x^4 + 1)(x^2 + 1)(x^2 - 1) = 0 \equiv$

$\equiv (x^4 + 1)(x^2 + 1)(x + 1)(x - 1) = 0$

Now lets find our roots (from right to left):

$x - 1 = 0 \equiv x = 1$ (#1)

$x + 1 = 0 \equiv x = -1$ (#2)

$x^2 + 1 = 0 \equiv x^2 = -1 \equiv x = \pm\sqrt {-1} \equiv x = \pm i$ (#3 and #4)

And for the last one we use the method of my previous post:

$x^4 + 1 = 0 \equiv x^4 = -1 \equiv x^4 = -1 + 0i$

$r = \sqrt {(-1)^2 + 0^2} = 1$

$tan (\theta) = \frac {0}{-1} = 0 \equiv \theta = \pi$ , since $cos (\theta) < 0$ .

$1^\frac{1}{4}\cdot(cos (\frac {\pi + 2k\pi}{4}) + i\cdot sin (\frac {\pi + 2k\pi}{4})); k = 0, 1, 2, 3.$

$k = 0 \rightarrow cos (\frac {\pi}{4}) + i\cdot sin (\frac {\pi}{4}) = \frac {\sqrt {2}}{2} + \frac {\sqrt {2}}{2}i$ (#5)

$k = 1 \rightarrow cos (\frac {3\pi}{4}) + i\cdot sin (\frac {3\pi}{4}) = -\frac {\sqrt {2}}{2} + \frac {\sqrt {2}}{2}i$ (#6)

$k = 2 \rightarrow cos (\frac {5\pi}{4}) + i\cdot sin (\frac {5\pi}{4}) = -\frac {\sqrt {2}}{2} - \frac {\sqrt {2}}{2}i$ (#7)

$k = 3 \rightarrow cos (\frac {7\pi}{4}) + i\cdot sin (\frac {7\pi}{4}) = \frac {\sqrt {2}}{2} - \frac {\sqrt {2}}{2}i$ (#8)

And done!

Carlos Revés - 6 years, 2 months ago

This is correct. I would like to add that sometimes, roots are not distinct, but are counted separately for the purpose of this theorem (for example, $x^2 - 2x + 1$ has two roots: $1$ and $1.$ ) So if they are the same, they are counted multiple times. In fact, the roots are all distinct in this case anyway.

Caleb Townsend - 6 years, 2 months ago

So is there a way to know if the roots are same?

Rmflute Shrivastav - 6 years, 2 months ago

Yes, I think the best way in general (and I think by definition) is to factor the polynomial, for example $x^2 - 2x + 1 = (x-1)^2.$ The exponent of each root is the algebraic multiplicity of that root, i.e. the number of times that root occurs.

Alternatively, if you have a graph of the polynomial showing all roots, you can use the "bounce, wiggle, cross" theorem. If a root has multiplicity 1, the graph crosses the x axis smoothly at that root. If a root has multiplicity 2, 4, 6, 8, ... Then the graph bounces off the x axis at that root. And finally if a root has multiplicity 3, 5, 7, 9, ... Then the graph crosses the x axis at that root, but it flattens out first, i.e. it wiggles across. So if it bounces or wiggles, it's a repeated root.