Question

If $\omega \neq 1$ is a cube root of unity, then one root among the $7^{th}$ roots of $(1+\omega)$ is

• A. $1+\omega$
• B. $1-\omega$
• C. $\omega-\omega^2$
• D. $\omega - \omega^2$

Hint:

Use the identities $1+\omega+\omega^2=0$ and $\omega^3=1$ to simplify expressions involving cube roots of unity.

Answer 1

The Correct Option is A

Step 1: Fundamental properties of cube roots of unity.
For $\omega \neq 1$, a cube root of unity: \[ 1 + \omega + \omega^2 = 0, \quad \omega^3 = 1. \] Step 2: Express $1+\omega$ in simpler form.
From $1 + \omega + \omega^2 = 0$, we have \[ 1+\omega = -\omega^2. \] Step 3: Verify the 7th root.
We want $z$ such that \[ z^7 = 1+\omega = -\omega^2. \] Testing $z = 1+\omega$: \[ (1+\omega)^7 = (-\omega^2)^7 = (-1)^7 (\omega^2)^7 = -\omega^{14}. \] Step 4: Simplify using $\omega^3 = 1$.
\[ \omega^{14} = \omega^{12} \cdot \omega^2 = (\omega^3)^4 \cdot \omega^2 = 1 \cdot \omega^2 = \omega^2. \] Thus, \[ (1+\omega)^7 = -\omega^2, \] confirming $z = 1+\omega$ is indeed a 7th root.