Question

If $\omega$ is complex cube root of unity and $(1+\omega)^7=A+B\omega$, then values of $A$ and $B$ are, respectively

• A. 0, 1
• B. 1, 1
• C. 1, 0
• D. -1, 1

Hint:

Whenever you see $(1+\omega)$ or $(1+\omega^2)$, immediately replace them with $-\omega^2$ and $-\omega$, respectively. This single substitution rapidly collapses high-power polynomials into easily manageable terms.

Answer 1

The Correct Option is B

Step 1: Recall the cube root facts.
For the complex cube root of unity $\omega$: $1 + \omega + \omega^2 = 0$ and $\omega^3 = 1$.

Step 2: Replace $1 + \omega$.
From the first fact, $1 + \omega = -\omega^2$.

Step 3: Raise to the 7th power.
\[ (1+\omega)^7 = (-\omega^2)^7 = (-1)^7 \omega^{14} = -\omega^{14} \]
Step 4: Reduce the power of $\omega$.
Since $\omega^3 = 1$, divide 14 by 3 to get remainder 2.
\[ \omega^{14} = (\omega^3)^4 \cdot \omega^2 = \omega^2 \] So the value is $-\omega^2$.

Step 5: Write in the form $A + B\omega$.
Again use $-\omega^2 = 1 + \omega$.
\[ (1+\omega)^7 = 1 + \omega \]
Step 6: Read off $A$ and $B$.
Comparing with $A + B\omega$ gives $A = 1$ and $B = 1$. \[ \boxed{A = 1,\ B = 1 \text{ (Option 2)}} \]