Question

The three cube roots of +1 are often called 1,$\mathbf{\omega }$, and ${\mathbf{\omega }}^{\mathbf{2}}$. Show that this is reasonable, that is, show that the cube roots of +1 are +1 and two other numbers, each of which is the square of the other.

Short answer

The three cube roots of +1 are:$1=e^{\left(2\pi i\right)},\omega =e^{\left(2\pi i/3\right)},{\omega }^2={\left(e^{\left(2\pi i/3\right)}\right)}^2$,and cube roots of +1 are +1 and two other numbers, each of which is the square of the other.

Step-by-step solution

Given Information

It has been given that three cube roots are 1,$\omega$and${\omega }^2$ .

Definition of Power series

A power series is an infinite series that looks like :

$\mathbf{\sum }_{\mathbf{n}\mathbf{=}\mathbf{0}}^{\mathbf{\infty }}{\mathit{a}}_{\mathbf{n}}{\left(x-c\right)}^{\mathbf{n}}\mathbf{=}{\mathit{a}}_{\mathbf{0}}\mathbf{+}{\mathit{a}}_{\mathbf{1}}\left(x-c\right)\mathbf{+}{\mathit{a}}_{\mathbf{2}}{\left(x-c\right)}^{\mathbf{2}}\mathbf{+}\mathbf{.}\mathbf{.}\mathbf{.}$
Where${\mathit{a}}_{\mathbf{n}}$ represents thecoefficient of the nthterm c andis a constant.

Find the exponential form of z =1

Find the exponential form of z = 1 as:

$$\begin{gathered} z=e^{\left(\theta i\right)} \\ z=e^{\left(2\pi i\right)} \end{gathered}$$

It has three roots.

Write the general term of the of the roots as:

$z_k=r^{1/n}{e}^{\left({\theta }_{k}i\right)}$

Write the angle in a general term as:

${\theta }_k=\frac{2\pi +2\pi k}{n}$

Substitute the value

Put k= 0,1,2, and n= 3.

$$\begin{gathered} {\theta }_o=\frac{2\pi }{3} \\ {z}_o=e^{\left(2\pi i/3\right)} \\ {\theta }_1=\frac{4\pi }{3} \\ {z}_1=e^{\left(4\pi i/3\right)} \\ {\theta }_2=2\pi \\ {z}_2=e^{\left(2\pi i\right)} \end{gathered}$$

Take$z_o=\omega$.

Therefore, $\omega =e^{\left(2\pi /3\right)}$.

$$\begin{gathered} z_1=e^{\left(4\pi /3\right)} \\ {z}_1=e^{{\left(2\pi /3\right)}^2} \\ {z}_1={\left(\omega \right)}^2 \end{gathered}$$

The angle of $z_2=1$because the angle is $2\pi$.

Therefore, the three-cube roots of 1 are:$1=e^{\left(2\pi i\right)},\omega =e^{\left(2\pi i/3\right)},{\omega }^2={\left(e^{\left(2\pi i/3\right)}\right)}^2$ and cube roots of +1 are +1 and two other numbers, each of which is the square of the other.