Q32P - Mathematical Methods in Physical Sciences

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.

Step-by-Step Solution

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 1: Given Information

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

Step 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 the coefficient of the nth term  c and  is a constant.

Step 3: 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:

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

Step 4: 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.