Question

If \( (x - iy)^{1/3} = a - ib \), then the value of \( \frac{x}{2a} + \frac{y}{2b} \) is:

• A. \( 2(a^2 - b^2) \)
• B. \( 4(a^2 - b^2) \)
• C. \( a^2 - b^2 \)
• D. \( \frac{1}{2}(a^2 - b^2) \)

Hint:

When given a root of a complex number, cube or square the expression as needed, then equate real and imaginary parts to extract variables.

Answer 1

The Correct Option is A

We are given: \[ (x - iy)^{1/3} = a - ib \Rightarrow x - iy = (a - ib)^3 \] Compute \( (a - ib)^3 \): \[ = a^3 - 3a^2 ib + 3a(ib)^2 - (ib)^3 = a^3 - 3a^2 ib - 3ab^2 i^2 + b^3 i^3 \] \[ = a^3 - 3a^2 ib - 3ab^2(-1) + b^3(-i) = a^3 + 3ab^2 - i(3a^2b + b^3) \] Thus, comparing with \( x - iy \), we get: \[ x = a^3 + 3ab^2, \quad y = 3a^2b + b^3 \] Now compute: \[ \begin{align} \frac{x}{2a} + \frac{y}{2b}
= \frac{1}{2a}(a^3 + 3ab^2) + \frac{1}{2b}(3a^2b + b^3)
= \frac{a^2 + 3b^2}{2} + \frac{3a^2 + b^2}{2}
= \frac{(a^2 + 3b^2 + 3a^2 + b^2)}{2}
= \frac{4a^2 + 4b^2}{2} = 2(a^2 - b^2)
\]