Question

The product of all the values of \( (\sqrt{3}-i)^{\frac{3}{7}} \) is

• A. 8
• B. -8
• C. 8i
• D. -8i

Hint:

When asked for the product of all values of \( A^{1/n} \), you are essentially finding the product of the roots of \( x^n - A = 0 \). If \( n \) is odd, the product is \( A \). If \( n \) is even, the product is \( -A \).

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:

The expression \( (\sqrt{3}-i)^{\frac{3}{7}} \) represents the 7th roots of the complex number \( (\sqrt{3}-i)^3 \). If we let \( z = \sqrt{3}-i \), we are looking for the product of all roots \( x \) satisfying \( x^7 = z^3 \).
Step 2: Key Formula or Approach:

For a polynomial equation \( x^n - A = 0 \), the product of the roots is given by \( (-1)^n \times (\text{constant term}) = (-1)^n \times (-A) \). Also, De Moivre's Theorem states \( [r(\cos \theta + i \sin \theta)]^n = r^n(\cos n\theta + i \sin n\theta) \).
Step 3: Detailed Explanation:

First, express \( z = \sqrt{3}-i \) in polar form. Modulus \( |z| = \sqrt{(\sqrt{3})^2 + (-1)^2} = \sqrt{3+1} = 2 \). Argument \( \theta \): Since \( \text{Re}(z) \textgreater 0 \) and \( \text{Im}(z) \textless 0 \), \( z \) is in the 4th quadrant. \[ \tan \alpha = \left| \frac{-1}{\sqrt{3}} \right| = \frac{1}{\sqrt{3}} \implies \alpha = \frac{\pi}{6} \] So, \( \theta = -\frac{\pi}{6} \). \[ z = 2 e^{-i\pi/6} \] Now compute \( z^3 \): \[ z^3 = (2 e^{-i\pi/6})^3 = 8 e^{-i\pi/2} = 8(-i) = -8i \] Let the values of the expression be \( x \). Then \( x = (z^3)^{1/7} \), which implies \( x^7 = z^3 \). The equation is \( x^7 - (-8i) = 0 \), or \( x^7 + 8i = 0 \). This is a polynomial of degree 7. Let roots be \( x_1, x_2, \dots, x_7 \). The product of the roots is given by: \[ P = (-1)^n \cdot \frac{\text{Constant term}}{\text{Coefficient of } x^n} \] Here \( n = 7 \) (odd). \[ P = (-1)^7 \cdot \frac{8i}{1} = (-1) \cdot (8i) = -8i \]
Step 4: Final Answer:

The product of all the values is \( -8i \).