Question

If \( (r, \theta) \) denotes \( r (\cos \theta + i \sin \theta) \). If \[ x = (1, \alpha), \quad y = (1, \beta), \quad z = (1, \gamma) \] and \( x + y + z = 0 \), then \[ \sum \cos (2\alpha - \beta - \gamma) = \]

• A. \( 3 \)
• B. \( 0 \)
• C. \( 1 \)
• D. \( -1 \)

Hint:

If \(z_1\) and \(z_2\) are complex numbers, then \[\cos (z_1 - z_2) = \cos z_1 \cos z_2 + \sin z_1 \sin z_2.\]

Answer 1

The Correct Option is A

Given that \[ x = e^{i\alpha}, \quad y = e^{i\beta}, \quad z = e^{i\gamma}, \] and \[ x + y + z = 0, \] we want to evaluate \[ \cos(2\alpha - \beta - \gamma) + \cos(2\beta - \alpha - \gamma) + \cos(2\gamma - \alpha - \beta). \]

Step 1. Rewrite each cosine in exponential form. Recall that \[ \cos \theta = \operatorname{Re}\bigl(e^{i\theta}\bigr). \] Hence, \[ \cos(2\alpha - \beta - \gamma) = \operatorname{Re}\bigl(e^{i(2\alpha - \beta - \gamma)}\bigr). \] Similarly, \[ \cos(2\beta - \alpha - \gamma) = \operatorname{Re}\bigl(e^{i(2\beta - \alpha - \gamma)}\bigr), \] and \[ \cos(2\gamma - \alpha - \beta) = \operatorname{Re}\bigl(e^{i(2\gamma - \alpha - \beta)}\bigr). \] Thus the sum we want is \[ S = \operatorname{Re}\Bigl( e^{i(2\alpha - \beta - \gamma)} + e^{i(2\beta - \alpha - \gamma)} + e^{i(2\gamma - \alpha - \beta)} \Bigr). \] In terms of \(x, y, z\), notice that \[ e^{i(2\alpha - \beta - \gamma)} = \frac{x^2}{yz}, \quad e^{i(2\beta - \alpha - \gamma)} = \frac{y^2}{zx}, \quad e^{i(2\gamma - \alpha - \beta)} = \frac{z^2}{xy}. \] Hence \[ S = \operatorname{Re}\Bigl(\frac{x^2}{yz} + \frac{y^2}{zx} + \frac{z^2}{xy}\Bigr) = \operatorname{Re}\Bigl(\frac{x^3 + y^3 + z^3}{xyz}\Bigr). \] 

Step 2. Use the identity when \(x + y + z = 0.\) A classical identity for any complex numbers \(x, y, z\) is \[ x^3 + y^3 + z^3 - 3xyz = (x + y + z)\bigl(x^2 + y^2 + z^2 - xy - yz - zx\bigr). \] Since we are given \(x + y + z = 0\), the right side vanishes, giving \[ x^3 + y^3 + z^3 = 3xyz. \] Thus, \[ \frac{x^3 + y^3 + z^3}{xyz} = \frac{3xyz}{xyz} = 3. \] Taking the real part does not change the value here, so \[ S = \operatorname{Re}\bigl(3\bigr) = 3. \] 

Answer: \[ \cos(2\alpha - \beta - \gamma) + \cos(2\beta - \alpha - \gamma) + \cos(2\gamma - \alpha - \beta) = 3. \]