Complex Numbers
Properties of Cube Roots of Unity
Grade 11
Question:
<p><strong>Paragraph for Question nos. 674 to 675</strong></p><p>Let \(\omega\) be the complex number representing the point \(M\left(\dfrac{-1}{2}, \dfrac{\sqrt{3}}{2}\right)\) and \(a, b, c, \alpha, \beta, \gamma\) be non-zero complex numbers such that</p><p>\(a + b + c = \alpha\)</p><p>\(a + b\omega + c\omega^2 = \beta\)</p><p>\(a + b\omega^2 + c\omega = \gamma\).</p><p><strong>Q674.</strong> If \(|\alpha|^2 + |\beta|^2 + |\gamma|^2 = \lambda(|a|^2 + |b|^2 + |c|^2)\) then \(\lambda\) is equal to:</p>
<p>1</p>
<p>2</p>
<p>3</p>
<p>4</p>
Step-by-Step Solution
Key Concept: ω is a primitive cube root of unity (ω³=1, 1+ω+ω²=0), so the three equations form an orthogonal system. Use the property that the sum of squared moduli of linear combinations with cube roots of unity preserves a constant multiple of the sum of squared moduli of the original variables.
<p><strong>Step 1:</strong> Recognize that $\omega = -\frac{1}{2} + i\frac{\sqrt{3}}{2}$ is a primitive cube root of unity, so $\omega^3 = 1$ and $1 + \omega + \omega^2 = 0$.</p><p><strong>Step 2:</strong> Write the system in matrix form:
$$\begin{bmatrix}\alpha\\\beta\\\gamma\end{bmatrix} = \begin{bmatrix}1 & 1 & 1\\1 & \omega & \omega^2\\1 & \omega^2 & \omega\end{bmatrix}\begin{bmatrix}a\\b\\c\end{bmatrix}$$</p><p><strong>Step 3:</strong> The matrix is a DFT-type matrix. Calculate $|\alpha|^2 + |\beta|^2 + |\gamma|^2$ using the orthogonality relations:
$$|\alpha|^2 = |a+b+c|^2 = |a|^2 + |b|^2 + |c|^2 + 2\text{Re}(a\overline{b} + b\overline{c} + c\overline{a})$$
$$|\beta|^2 = |a+b\omega+c\omega^2|^2$$
$$|\gamma|^2 = |a+b\omega^2+c\omega|^2$$</p><p><strong>Step 4:</strong> Add all three using $\omega^3=1$ and orthogonality: when we sum $|\alpha|^2+|\beta|^2+|\gamma|^2$, the cross terms involving different pairs cancel due to $1+\omega+\omega^2=0$, leaving:
$$|\alpha|^2 + |\beta|^2 + |\gamma|^2 = 3(|a|^2 + |b|^2 + |c|^2)$$
$$\therefore \lambda = 3$$</p>
Correct Answer: C