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 equation.
Given, \[ (x-iy)^{\frac{1}{3}}=2-i\sqrt{3} \] Cubing both sides, \[ x-iy=(2-i\sqrt{3})^3 \]

Step 2: Find \((2-i\sqrt{3})^3\).
First, \[ (2-i\sqrt{3})^2 = 4-4i\sqrt{3}+(-i\sqrt{3})^2 \] Since \[ (-i\sqrt{3})^2=-3, \] we get \[ (2-i\sqrt{3})^2=1-4i\sqrt{3}. \] Now, \[ (2-i\sqrt{3})^3=(1-4i\sqrt{3})(2-i\sqrt{3}). \] \[ =2-i\sqrt{3}-8i\sqrt{3}+4i^2\cdot 3 \] Since \[ i^2=-1, \] we get \[ =2-9i\sqrt{3}-12 \] \[ =-10-9i\sqrt{3}. \] Thus, \[ x-iy=-10-9i\sqrt{3}. \]

Step 3: Compare real and imaginary parts.
Comparing, \[ x=-10 \] and \[ -y=-9\sqrt{3}. \] Therefore, \[ y=9\sqrt{3}. \]

Step 4: Substitute in the line equation.
Given line is \[ \frac{x}{2}+\frac{y}{\sqrt{3}}=k. \] Substitute \[ x=-10,\qquad y=9\sqrt{3}. \] \[ k=\frac{-10}{2}+\frac{9\sqrt{3}}{\sqrt{3}} \] \[ k=-5+9 \] \[ k=4. \]

Step 5: Final conclusion.
Therefore, \[ \boxed{4} \]