Algebra
Complex Cube Roots
MMTS_Full_Test_06
Grade 12
Question:
If $\omega\ne 1$ is a cube root of unity, then $(1+\omega)(1+\omega^2)(1+\omega^4)(1+\omega^8)\cdots$ to $2n$ factors is
Step-by-Step Solution
Key Concept: $(1+\omega)(1+\omega^2)=1+\omega+\omega^2+\omega^3=1+0+1=$ hmm; pair up: $(1+\omega)(1+\omega^2)=1+\omega+\omega^2+\omega^3=0+1=1$
$(1+\omega)(1+\omega^2)=(1+\omega+\omega^2+\omega^3)=0+1=1$. Each pair of consecutive factors gives 1. Product of $2n$ factors $=1$.
Correct Answer: 1