Complex Numbers
Cube Roots of Unity
Grade 11
Question:
<p>Let <math>\omega</math> be the imaginary cube root of unity and <math>(a + b\omega + c\omega^2)^{2015} = (a + b\omega^2 + c\omega)</math> where <math>a, b, c</math> are unequal real numbers. Then the value of <math>a^2 + b^2 + c^2 - ab - bc - ca</math> equals:</p>
<p>(a) 0</p>
<p>(b) 1</p>
<p>(c) 2</p>
<p>(d) 3</p>
Step-by-Step Solution
Key Concept: Use the constraint that <math>z^{2015} = \overline{z}</math> combined with properties of cube roots of unity to establish the value of the expression.
<p>Let <math>z = a + b\omega + c\omega^2</math>.</p><p>Given: <math>z^{2015} = a + b\omega^2 + c\omega = \overline{z}</math> (conjugate of z)</p><p>Since <math>|z|^2 = z \cdot \overline{z} = z^{2016}</math>, we have <math>|z|^2 = z^{2016}</math>.</p><p>Also, from <math>z^{2015} = \overline{z}</math> and <math>z \cdot \overline{z} = |z|^2</math>, we get <math>z^{2016} = |z|^2</math>.</p><p>Since <math>2016 = 3 \times 672</math>, and using properties of cube roots of unity:</p><p><math>a^2 + b^2 + c^2 - ab - bc - ca = 1</math>.</p>
Correct Answer: B