Q.Find the value of $\omega^{18}$

The cube roots of unity are the three solutions of $x^3=1$, found by factoring $x^3-1=(x-1)(x^2+x+1)=0$: one real root $x=1$, and two complex conjugate roots coming from the quadratic factor, $\omega=-\dfrac12+\dfrac{\sqrt3}{2}i$ and $\omega^2=-\dfrac12-\dfrac{\sqrt3}{2}i$ (so $\omega$ and $\omega^2=\bar\omega$ are complex conjugates of each other). These three roots — $1,\omega,\omega^2$ — satisfy a standing toolkit of identities used constantly in this chapter's exercises: $\omega^3=1$ (so powers of $\omega$ cycle every 3 steps, exactly like powers of $i$ cycle every 4), the sum identity $1+\omega+\omega^2=0$ (equivalently $\omega^2+\omega+1=0$, so $\omega^2=-1-\omega$ and $\omega+1=-\omega^2$), the reciprocal identities $\dfrac1\omega=\omega^2$ and $\dfrac{1}{\omega^2}=\omega$, and the periodicity rules $\omega^{3n}=1$, $\omega^{3n+1}=\omega$, $\omega^{3n+2}=\omega^2$ for any integer $n$. Also, $1=e^{2\pi i}$, $\omega=e^{2\pi i/3}$, and $\omega^2=e^{4\pi i/3}$, linking the cube roots of unity to the exponential form and De Moivre's theorem. These identities let complicated-looking expressions in $\omega$ (like $(1+\omega-\omega^2)^6$ or $(a+b\omega+c\omega^2)$) collapse to small integers or simple multiples of $\omega,\omega^2$, and the same toolkit works for any pair of complex cube roots of unity, often relabelled $\alpha,\beta$ in a problem.

"Cube roots of unity formula and properties of omega" and "cube roots of unity important questions class 11" are heavily searched terms tied to the Complex Numbers chapter of the NCERT/CBSE Class 11 Mathematics curriculum, a near-certain JEE Main topic every year. Fluency with the periodicity rules $\omega^{3n}, \omega^{3n+1}, \omega^{3n+2}$ shown here is what lets students collapse complicated $\omega$-expressions quickly under competitive-exam time pressure.