Question

If \(\omega\) is a complex cube root of unity, then \[ (1+\omega)(1+\omega^2)(1+\omega^4)(1+\omega^5)(1+\omega^7)(1+\omega^8)\cdots \] (2n factors) is equal to

• A. \(-1\)
• B. \(0\)
• C. \(1\)
• D. \(2\)

Hint:

Whenever powers of cube roots of unity appear, reduce exponents modulo 3. This usually converts long products into simple repeating factors.

Answer 1

The Correct Option is C

Concept: For cube roots of unity, \[ 1+\omega+\omega^2=0 \] and \[ \omega^3=1 \] Hence powers repeat modulo \(3\).

Step 1: Reduce powers modulo \(3\).
\[ \omega^4=\omega,\qquad \omega^5=\omega^2 \] \[ \omega^7=\omega,\qquad \omega^8=\omega^2 \] Thus factors occur in repeating pairs \[ (1+\omega)(1+\omega^2) \]

Step 2: Evaluate one pair.
\[ (1+\omega)(1+\omega^2) \] \[ =1+\omega+\omega^2+\omega^3 \] Using \[ 1+\omega+\omega^2=0 \] and \[ \omega^3=1 \] we get \[ (1+\omega)(1+\omega^2)=1 \]

Step 3: Evaluate the complete product.
The entire expression contains \(n\) such pairs. Hence \[ 1^n=1 \] Therefore \[ \boxed{1} \]