How do I use DeMoivre's theorem to solve z^3-1=0?

Apr 4, 2015

If ${z}^{3} - 1 = 0$, then we are looking for the cubic roots of unity, i.e. the numbers such that ${z}^{3} = 1$.

If you're using complex numbers, then every polynomial equation of degree $k$ yields exactly $k$ solution. So, we're expecting to find three cubic roots.

De Moivre's theorem uses the fact that we can write any complex number as $\setminus \rho {e}^{i \setminus \theta} = \setminus \rho \left(\setminus \cos \left(\setminus \theta\right) + i \setminus \sin \left(\setminus \theta\right)\right)$, and it states that, if
$z = \setminus \rho \left(\setminus \cos \left(\setminus \theta\right) + i \setminus \sin \left(\setminus \theta\right)\right)$, then
${z}^{n} = \setminus {\rho}^{n} \left(\setminus \cos \left(n \setminus \theta\right) + i \setminus \sin \left(n \setminus \theta\right)\right)$

If you look at $1$ as a complex number, then you have $\setminus \rho = 1$, and $\setminus \theta = 2 \setminus \pi$. We are thus looking for three numbers such that $\setminus {\rho}^{3} = 1$, and $3 \setminus \theta = 2 \setminus \pi$.

Since $\setminus \rho$ is a real number, the only solution to $\setminus {\rho}^{3} = 1$ is $\setminus \rho = 1$. On the other hand, using the periodicity of the angles, we have that the three solutions for $\setminus \theta$ are
$\setminus {\theta}_{1 , 2 , 3} = \setminus \frac{2 k \setminus \pi}{3}$, for $k = 0 , 1 , 2$.

This means that the three solutions are:

1. $\setminus \rho = 1 , \setminus \theta = 0$, which is the real number $1$.
2. $\setminus \rho = 1 , \setminus \theta = \setminus \frac{2 \setminus \pi}{3}$, which is the complex number $- \frac{1}{2} + \setminus \frac{\sqrt{3}}{2} i$
3. $\setminus \rho = 1 , \setminus \theta = \setminus \frac{4 \setminus \pi}{3}$, which is the complex number $- \frac{1}{2} - \setminus \frac{\sqrt{3}}{2} i$
Aug 1, 2017

z=1,ω,ω^2

Explanation:

${z}^{3} - 1 = 0$

${z}^{3} = 1$

We know that any complex number, $a + b i$, can be written in modulus-argument form, $r \left(\cos x + i \sin x\right)$, where $r = \sqrt{{a}^{2} + {b}^{2}}$ and $x$ satisfies $\sin x = \frac{b}{r}$ and $\cos x = \frac{a}{r}$.

$\therefore 1 = 1 \left(\cos 0 + i \sin 0\right)$

So ${z}^{3} = \cos \left(0 + 2 k \pi\right) + i \sin \left(0 + 2 k \pi\right) \rightarrow$ Since the solutions to trig equations aren't unique, we need to consider other possibilities.

$z = {\left[\cos \left(0 + 2 k \pi\right) + i \sin \left(0 + 2 k \pi\right)\right]}^{\frac{1}{3}}$

Use de Moivre's theorem: ${\left(\cos x + i \sin x\right)}^{k} = \cos k x + i \sin k x$

$z = \cos \left(\frac{2}{3} k \pi\right) + i \sin \left(\frac{2}{3} k \pi\right)$

Now we must consider every k such that -pi< 2/3kpi ≤ pi

$k = 0 , z = 1$

$k = 1 , z = \cos \left(\frac{2}{3} \pi\right) + i \sin \left(\frac{2}{3} \pi\right) = - \frac{1}{2} + \frac{\sqrt{3}}{2} i$

$k = - 1 , z = \cos \left(- \frac{2}{3} \pi\right) + i \sin \left(- \frac{2}{3} \pi\right) = - \frac{1}{2} - \frac{\sqrt{3}}{2} i$

These values are called the cubic roots of unity and are usually written as $1 , \omega , {\omega}^{2}$.

The fact that $\omega ' = {\omega}^{2}$ can easily be verified.