Key Concepts and Formulas

• Complex number representation: $z = x + iy$, where $x$ is the real part and $y$ is the imaginary part.
• Equality of complex numbers: If $a + ib = c + id$, then $a = c$ and $b = d$.
• Binomial theorem: $(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$
• Powers of $i$: $i^2 = -1$, $i^3 = -i$, $i^4 = 1$

Step-by-Step Solution

Step 1: Expressing $z$ in terms of $p$ and $q$

We are given that $z^{1/3} = p + iq$. To eliminate the fractional exponent and obtain an expression for $z$, we cube both sides of the equation. This allows us to work with a polynomial expansion. $(z^{\frac{1}{3}})^3 = (p + iq)^3$ $z = (p + iq)^3$

Step 2: Expanding $(p + iq)^3$ using the Binomial Theorem

We expand the right-hand side using the binomial theorem. This will allow us to separate the real and imaginary parts of $z$. $z = p^3 + 3p^2(iq) + 3p(iq)^2 + (iq)^3$ Now, we simplify using the properties of $i$: $z = p^3 + 3ip^2q + 3p(i^2q^2) + i^3q^3$ Recall $i^2 = -1$ and $i^3 = -i$, so we have: $z = p^3 + 3ip^2q - 3pq^2 - iq^3$

Step 3: Grouping Real and Imaginary Parts of $z$

To compare with $z = x - iy$, we group the real and imaginary terms: $z = (p^3 - 3pq^2) + i(3p^2q - q^3)$

Step 4: Equating Real and Imaginary Parts

We are given $z = x - iy$. Comparing this with the expression we derived in Step 3, we can equate the real and imaginary parts. This allows us to establish relationships between $x, y, p,$ and $q$. $x - iy = (p^3 - 3pq^2) + i(3p^2q - q^3)$ Equating the real parts, we get: $x = p^3 - 3pq^2$ Equating the imaginary parts, we get: $-y = 3p^2q - q^3$ Which simplifies to: $y = q^3 - 3p^2q$

Step 5: Deriving Expressions for $\frac{x}{p}$ and $\frac{y}{q}$

To evaluate the target expression, we need to find $\frac{x}{p}$ and $\frac{y}{q}$. This simplifies the subsequent substitutions. From the equation $x = p^3 - 3pq^2$, we divide both sides by $p$: $\frac{x}{p} = p^2 - 3q^2$ From the equation $y = q^3 - 3p^2q$, we divide both sides by $q$: $\frac{y}{q} = q^2 - 3p^2$

Step 6: Substituting into the Target Expression and Simplifying

We now substitute the expressions we found for $\frac{x}{p}$ and $\frac{y}{q}$ into the target expression: $\frac{\frac{x}{p} + \frac{y}{q}}{p^2 + q^2} = \frac{(p^2 - 3q^2) + (q^2 - 3p^2)}{p^2 + q^2}$ Simplifying the numerator: $\frac{p^2 - 3q^2 + q^2 - 3p^2}{p^2 + q^2} = \frac{-2p^2 - 2q^2}{p^2 + q^2}$ Factoring out $-2$ from the numerator: $\frac{-2(p^2 + q^2)}{p^2 + q^2}$ Assuming $p^2 + q^2 \neq 0$ (otherwise $z=0$), we can cancel out the term $(p^2 + q^2)$: $= -2$

Common Mistakes & Tips

• Be extremely careful with the signs when equating imaginary parts. Remember that if $x - iy = a + ib$, then $-y = b$.
• Double-check your algebraic manipulations, especially when dealing with multiple terms and negative signs.
• Keep the target expression in mind throughout the solution. This helps in deciding which expressions to isolate.

Summary

We started with the given relationship $z^{1/3} = p + iq$ and $z = x - iy$. By cubing the first equation, expanding using the binomial theorem, and equating the real and imaginary parts, we found expressions for $x$ and $y$ in terms of $p$ and $q$. We then calculated $\frac{x}{p}$ and $\frac{y}{q}$, substituted these into the target expression, and simplified to obtain the final answer.

Final Answer

The final answer is \boxed{-2}, which corresponds to option (A).