Question

If $w$ is the complex cube root of unity, find the value of
i. $w + \dfrac{1}{w}$
ii. ${w^2} + {w^3} + {w^4}$
iii. ${\left( {1 + {w^2}} \right)^3}$
iv. ${\left( {1 - w - {w^2}} \right)^3} + {\left( {1 - w + {w^2}} \right)^3}$

Answer 1

Hint:Before solving the complex cube root questions, we need to know the basic properties of the complex cube root of the unity. Which tells that ${w^3} = 1$ which can also be written in the form of $1 + w + {w^2} = 0$ using these two equations, we can find the value of all the above options.

Complete step by step solution:
In the above question, we are given that $w$ is the complex cube root of unity which mean ${w^3} = 1$
or can be written as $1 + w + {w^2} = 0$
Or we can use it as $1 + w = - {w^2}$ or $1 + {w^2} = - w$
Now, finding the
i. $w + \dfrac{1}{w}$ $ = \dfrac{{{w^2} + 1}}{w}$ (cross-multiplying)
$ = \dfrac{{ - w}}{w} = - 1$ (using the above written properties $1 + {w^2} = -
w$)
ii. ${w^2} + {w^3} + {w^4}$
taking common and then using the above property $1 + w + {w^2} = 0$
$ = {w^2}\left( {1 + w + {w^2}} \right)$
$ = 0$
iii. ${\left( {1 + {w^2}} \right)^3}$
Using the property $1 + {w^2} = - w$
$
= {\left( { - w} \right)^3} = - {w^3} \\
= - 1 \\
$ (using ${w^3} = 1$ )
iv. ${\left( {1 - w - {w^2}} \right)^3} + {\left( {1 - w + {w^2}} \right)^3}$
Using $w + {w^2} = - 1$ in first term and $1 + {w^2} = - w$ in the second term
$
= {\left( {1 + 1} \right)^3} + {\left( { - w - w} \right)^3} \\
= {\left( 2 \right)^3}\left[ {1 - {w^3}} \right] \\
= {\left( 2 \right)^3}\left[ {1 - 1} \right] \\
= 0 \\
$
Note: Cube root of units means cube root of one. The roots are $w,{w^2},{w^3}( = 1)$ and the value of w is $\dfrac{{ - 1 + \sqrt 3 }}{2}$. We can also learn the properties of ${n^{th}}$ root of unity, which gives us a general formula for any root of unity. This basically means that finding the root of ${1^{\dfrac{1}{n}}}$. Most of the results are analogous to the cube root of unity and these results can be useful to solve complex problems.