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 (√3 - i)3/7 represents the 7th roots of the complex number (√3 - i)3. If we let z = √3 - i, we are looking for the product of all roots x satisfying x7 = z3.

Step 2: Key Formula or Approach:

For a polynomial equation xn - A = 0, the product of the roots is given by the formula based on coefficients.
Also, De Moivre's Theorem states:

[r(cos θ + i sin θ)]n = rn(cos nθ + i sin nθ)

Step 3: Detailed Explanation:

First, express z = √3 - i in polar form.

Modulus of z:
|z| = √[(√3)² + (-1)²] = √(3 + 1) = 2

Argument of z: Since Re(z) > 0 and Im(z) < 0, z lies in the fourth quadrant.

tan α = |-1 / √3| = 1 / √3
α = π/6

So,
θ = -π/6

Hence,
z = 2e-iπ/6

Now compute z3:

z3 = (2e-iπ/6)3 = 8e-iπ/2 = 8(-i) = -8i

Let the values of the expression be x. Then:
x = (z3)1/7
which implies
x7 = z3

So the equation becomes:
x7 - (-8i) = 0
or
x7 + 8i = 0

This is a polynomial of degree 7. Let the roots be x1, x2, ..., x7.

The product of the roots is given by:

P = (-1)n × (constant term / coefficient of xn)

Here n = 7, so:

P = (-1)7 × (8i / 1) = -8i

Step 4: Final Answer:

The product of all the values is -8i.