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

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.