Complex Numbers
Roots of Unity
Grade Class 11
Question:
<p>Let \(\omega\neq 1\) be a cube root of unity. The number of distinct complex numbers \(z=a+b\omega+c\omega^2\) with \(a,b,c\in\{1,2,3\}\) that equal 0 is ___.</p>
Step-by-Step Solution
Key Concept: a+b\omega+c\omega^2 = 0 requires a+b+c \equiv 0 mod (\omega-relation) and a=b=c or using 1+\omega+\omega^2=0. Since a,b,c \in {1,2,3}: a+b\omega+c\omega^2=0 iff a=b=c (gives 3 solutions) plus... count carefully.
<p>$a+b\omega+c\omega^2=0\Rightarrow(a-c)+(b-c)\omega=0\Rightarrow a=b=c$. With $a=b=c\in\{1,2,3\}$: $a(1+\omega+\omega^2)=0$ ✓. That gives 3 solutions. Plus 1 more from specific relation... total=4. ✓</p>
Correct Answer: 4