Question

Among the solutions of the equation \( x^6 = 64 \), the sum of all those solutions whose real part is negative is:

• A. \( 0 \)
• B. \( -2 \)
• C. \( -4 \)
• D. \( -3 \)

Hint:

For \( x^n = a \), the roots are symmetric about the origin. The sum of all roots is 0. Subtract the roots with positive real parts from the total sum (which is 0) to find the sum of negative real parts.

Answer 1

The Correct Option is C

Concept: Roots of \( x^6 = 2^6 \) are \( x = 2(\cos \frac{2k\pi}{6} + i\sin \frac{2k\pi}{6}) \) for \( k=0,1,2,3,4,5 \).

Step 1: List the roots.
\( x_k = 2(\cos(k\pi/3) + i\sin(k\pi/3)) \). \( k=0: 2(1) = 2 \) (Real part 2) \( k=1: 2(1/2 + i\sqrt{3}/2) = 1 + i\sqrt{3} \) (Real part 1) \( k=2: 2(-1/2 + i\sqrt{3}/2) = -1 + i\sqrt{3} \) (Real part -1) \( k=3: 2(-1) = -2 \) (Real part -2) \( k=4: 2(-1/2 - i\sqrt{3}/2) = -1 - i\sqrt{3} \) (Real part -1) \( k=5: 2(1/2 - i\sqrt{3}/2) = 1 - i\sqrt{3} \) (Real part 1)

Step 2: Sum the roots with real part < 0.
Roots are \( -1 + i\sqrt{3} \), \( -2 \), and \( -1 - i\sqrt{3} \). Sum \( = (-1-2-1) + i(\sqrt{3} + 0 - \sqrt{3}) = -4 \). -4