Question

If $\omega $ is a complex cube root of unity then find ${{\omega }^{17}}$.
A. 0
B. 1
C. $\omega $
D. ${{\omega }^{2}}$

Answer 1

Hint: We first find the speciality about $\omega $, the complex cube root of unity. We use the value of $x=\omega $ as the imaginary cube root of unity. We put the value in the expression of ${{\omega }^{17}}$ and find the value using the identities of \[1+\omega +{{\omega }^{2}}=0;{{\omega }^{3}}=1\].

Complete step by step answer:
We first find the speciality about $\omega $, the complex cube root of unity. As we have $\omega =\dfrac{-1+i\sqrt{3}}{2}$, we can take it as any of the imaginary roots. With the value of $\omega $, we know the identities involved and they are,
\[1+\omega +{{\omega }^{2}}=0;{{\omega }^{3}}=1\]
We now express ${{\omega }^{17}}$ using indices.
${{\omega }^{17}}={{\omega }^{15}}\times {{\omega }^{2}} \\
\Rightarrow {{\omega }^{17}}={{\left( {{\omega }^{3}} \right)}^{5}}\times {{\omega }^{2}}$
We replace the value \[{{\omega }^{3}}=1\].
${{\omega }^{17}}={{\left( {{\omega }^{3}} \right)}^{5}}\times {{\omega }^{2}} \\
\therefore {{\omega }^{17}} ={{\omega }^{2}}$

therefore, the correct option is D.

Note:We can also solve the problem assuming the value of \[{{\omega }^{2}}=\dfrac{-1+i\sqrt{3}}{2}\]. We can take the imaginary value of $x=\dfrac{-1+i\sqrt{3}}{2}$ as any of \[\omega ,{{\omega }^{2}}\] as the square value gives the other imaginary root of $x=\dfrac{-1-i\sqrt{3}}{2}$. Also, we need to remember that as we multiplied the term $x-1$, it gives us the real root.