Question

Find the value of
${\omega ^{18}} + {\omega ^{21}} + {\omega ^{ - 30}} + {\omega ^{ - 105}}$ where $\omega $ is a complex cube root of unity.

Answer 1

Hint: We have to proceed by using the real cube root of unity and not the imaginary values . We will further try to use the property ${\omega ^3} = 1$ with ${\omega ^{18}} + {\omega ^{21}} + {\omega ^{ - 30}} + {\omega ^{ - 105}}$ to get to the desired answer.

Complete Step-by-Step solution:
We know that
 $\begin{gathered}
  {\omega ^3} = 1 \\
   \Rightarrow {\omega ^{3n}} = {1^n}{\text{ }}\left( {{\text{raising to the power n on both sides}}} \right) \\
   \Rightarrow {\omega ^{3n}} = 1 \\
\end{gathered} $
Hence,
${\omega ^{18}} + {\omega ^{21}} + {\omega ^{ - 30}} + {\omega ^{ - 105}}$
$ = {\omega ^{18}} + {\omega ^{21}} + \dfrac{1}{{{\omega ^{30}}}} + \dfrac{1}{{{\omega ^{105}}}}{\text{ }}\left( {{\text{since }}{{\text{a}}^{ - n}} = \dfrac{1}{{{{\text{a}}^n}}}} \right)$
$\begin{gathered}
   = {\omega ^{3\left( 6 \right)}} + {\omega ^{3\left( 7 \right)}} + \dfrac{1}{{{\omega ^{3\left( {10} \right)}}}} + \dfrac{1}{{{\omega ^{3\left( {35} \right)}}}} \\
   = 1 + 1 + \dfrac{1}{1} + \dfrac{1}{1} = 4 \\
\end{gathered} $
Thus the value of ${\omega ^{18}} + {\omega ^{21}} + {\omega ^{ - 30}} + {\omega ^{ - 105}}$ is 4.

Note: It is important to keep in mind that the cube roots of unity are $1, - \dfrac{1}{2} + i\dfrac{{\sqrt 3 }}{2}{\text{ and - }}\dfrac{1}{2} - i\dfrac{{\sqrt 3 }}{2}$ but we use only the real value 1 to solve this particular type of question . It becomes important to recall the properties of the cube root of unity and use them to find the desired result.