Question

If \[ (x-iy)^{\frac{1}{3}}=2-i\sqrt{3} \] and the point \(z=(x,y)\) lies on the line \[ \frac{x}{2}+\frac{y}{\sqrt{3}}=k, \] then \(k=\)

• A. \(16\)
• B. \(2\)
• C. \(8\)
• D. \(4\)

Hint:

When a complex expression is given in root form, remove the fractional power by raising both sides to the reciprocal power, then compare real and imaginary parts.

Answer 1

The Correct Option is D

Step 1: Cube both sides of the given relation.
Starting from (x - iy)^(1/3) = 2 - i√3, raising both sides to the third power yields x - iy = (2 - i√3)³.

Step 2: Compute the cube of the complex number.
First, square it: (2 - i√3)² = 4 - 4i√3 + (-i√3)² = 4 - 4i√3 - 3 = 1 - 4i√3. Then multiply by the original: (1 - 4i√3)(2 - i√3) = 2 - i√3 - 8i√3 + 4i²·3 = 2 - 9i√3 - 12 = -10 - 9i√3.

Step 3: Match real and imaginary components.
Comparing x - iy = -10 - 9i√3, we identify x = -10 and -y = -9√3, so y = 9√3.

Step 4: Plug into the line equation.
The line is x/2 + y/√3 = k. Substituting the values: k = (-10)/2 + (9√3)/√3 = -5 + 9 = 4.

Step 5: Final conclusion.
The value of k is 4.