Question

If $z = x - iy$ and $z^{1/3} = p + iq$ ($x, y, p, q \in \mathbb{R}$), then $\frac{\left( \frac{x}{p} + \frac{y}{q} \right)}{p^2 + q^2}$ is equal to

• A. 2
• B. -1
• C. 1
• D. -2

Answer 1

The Correct Option is D

We are given that \( z = x - iy \) and \( z^{1/3} = p + iq \). Let's first write down the equation in terms of complex numbers: \[ z = (p + iq)^3 \] Expanding \( (p + iq)^3 \) using the binomial theorem: \[ (p + iq)^3 = p^3 + 3p^2(iq) + 3p(iq)^2 + (iq)^3 \] Since \( i^2 = -1 \) and \( i^3 = -i \), we get: \[ (p + iq)^3 = p^3 + 3p^2(iq) + 3p(-q^2) - iq^3 \] This simplifies to: \[ (p + iq)^3 = p^3 - 3pq^2 + i(3p^2q - q^3) \] So, comparing the real and imaginary parts with \( z = x - iy \), we have the equations: \[ x = p^3 - 3pq^2 \quad \text{(real part)} \] \[ y = -(3p^2q - q^3) \quad \text{(imaginary part)} \] Step 1: Express \( \frac{x}{p} + \frac{y}{q} \) Now, substitute the expressions for \( x \) and \( y \) into \( \frac{x}{p} + \frac{y}{q} \): \[ \frac{x}{p} + \frac{y}{q} = \frac{p^3 - 3pq^2}{p} + \frac{-(3p^2q - q^3)}{q} \] Simplifying each term: \[ \frac{x}{p} + \frac{y}{q} = p^2 - 3q^2 - (3p^2 - q^2) \] \[ = p^2 - 3q^2 - 3p^2 + q^2 \] \[ = -2p^2 - 2q^2 \] \[ = -2(p^2 + q^2) \] Step 2: Simplifying the expression Now, we need to evaluate the full expression: \[ \frac{\left( \frac{x}{p} + \frac{y}{q} \right)}{p^2 + q^2} = \frac{-2(p^2 + q^2)}{p^2 + q^2} \] This simplifies to: \[ -2 \]

Answer:

\[ \boxed{-2} \]