Question

If $w$ is the complex cube root of unity, then show that: ${(1 - w + {w^2})^5} + {(1 + w - {w^2})^5} = 32$.

Answer 1

This question is based on complex numbers. A complex number is the sum of a real number and pure imaginary number. The standard form of a complex number is $a + bi$, where $a$ and $b$ are real numbers, $a$ is the real part and $b$ is the imaginary part. If a number is a complex number and it gives $1$ if raised to the power of a positive integer is called cube root of unity. The three cube roots of unity are: $1,\dfrac{{ - 1 + i\sqrt 3 }}{2}{\text{ and }}\dfrac{{ - 1 - i\sqrt 3 }}{2}$ one cube root being real and the other two complex. In this question we are going to prove that the cube root of unity in L.H.S. is equal to the real number in the R.H.S., by using the values: If $\omega$ is the cube root of unity then, ${\omega ^3} = 1$, $1 + \omega + {\omega ^2} = 0$.

Complete answer:
In this problem,
We are given that,
${(1 - w + {w^2})^5} + {(1 + w - {w^2})^5} = 32$
L.H.S, $= {(1 - w + {w^2})^5} + {(1 + w - {w^2})^5}$
We can simply rearrange the terms in ${(1 - w + {w^2})^5}$ as ${(1 + {w^2} - w)^5}$ for the sake of simplification.
$= {(1 + {w^2} - w)^5} + {(1 + w - {w^2})^5}$
We know that $1 + \omega + {\omega ^2} = 0$, thus $1 + {\omega ^2} = - \omega$ and $1 + \omega = - {\omega ^2}$
$= {((1 + {w^2}) - w)^5} + {((1 + w) - {w^2})^5}$
Here the brackets inside the brackets is in the form, $1 + {\omega ^2} = - \omega$ and $1 + \omega = - {\omega ^2}$, applying this we will get,
$= {( - w - w)^5} + {( - {w^2} - {w^2})^5}$
$= {( - 2w)^5} + {( - 2{w^2})^5}$
Multiplying the power inside the bracket,
$= ( - {2^5}{w^5}) + ( - {2^5}{w^{2(5)}})$
${w^{2(5)}}$ is in the form ${a^{mn}}$ , we know that ${a^{mn}} = {a^{m + n}}$
$= ( - {2^5}{w^5}) + ( - {2^5}{w^{10}})$
Here ${2^5}$ is common, so let take it outside,
$= {2^5}( - {w^5} - {w^{10}})$
We know that ${\omega ^3} = 1$, in order to get this form let write ${w^5}$ and ${w^{10}}$ in terms of ${w^3}$,
$= {2^5}( - ({w^3}.{w^2}) - ({w^3}.{w^3}.{w^3}.w))$
We know that ${\omega ^3} = 1$, substituting this value,
$= {2^5}( - 1.{w^2} - (1.1.1.w))$
Multiplying the $1$’s inside the bracket we will get $1$ and anything multiplied by $1$ is the thing itself,
$= {2^5}( - {w^2} - w)$
We know that $1 + \omega + {\omega ^2} = 0$, by altering this equation let try to find the value of $( - {w^2} - w)$.
$1 = - {\omega ^2} - \omega$.
We get the value $- {\omega ^2} - \omega = 1$, substituting this,
$= {2^5}(1)$
$= {2^5}$
Expanding the power,
$= 2 \times 2 \times 2 \times 2 \times 2$
$= 32$
$=$ R.H.S.
Hence proved.

Note:
Each complex cube root of unity is the square of the other, i.e., ${\left( {\dfrac{{ - 1 + i\sqrt 3 }}{2}{\text{ }}} \right)^2} = \dfrac{{ - 1 - i\sqrt 3 }}{2}{\text{ }}$ and ${\left( {\dfrac{{ - 1 - i\sqrt 3 }}{2}{\text{ }}} \right)^2} = \dfrac{{ - 1 + i\sqrt 3 }}{2}{\text{ }}$.
If we denote one of the complex cube roots of unity by $\omega$(omega), then the other one is ${\omega ^2}$.
The sum of the three cube roots of unity is zero, i.e., .$1 + \omega + {\omega ^2} = 0$.
The product of the three cube roots of unity is one, i.e., $1.\omega .{\omega ^2} = 1.{\omega ^3} = {\omega ^3} = 1$.
In mathematics multiplication is also represented by dot ($.$) .
Remember that: ${\omega ^3} = 1$
$1 + \omega + {\omega ^2} = 0$
$1 + \omega = - {\omega ^2}$
$1 + {\omega ^2} = - \omega$
$\omega + {\omega ^2} = - 1$