If z = x - iy and z^{1/3} = p + iq (x, y, p, | vedclass

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(D) Given $z^{1/3} = p + iq$, we have $z = (p + iq)^3$. Expanding this, $z = p^3 + 3p^2(iq) + 3p(iq)^2 + (iq)^3 = p^3 + 3ip^2q - 3pq^2 - iq^3$. Groupi

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$-2$

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(D) Given $z^{1/3} = p + iq$, we have $z = (p + iq)^3$. Expanding this, $z = p^3 + 3p^2(iq) + 3p(iq)^2 + (iq)^3 = p^3 + 3ip^2q - 3pq^2 - iq^3$. Grouping real and imaginary parts, $z = (p^3 - 3pq^2) + i(3p^2q - q^3)$. Since $z = x - iy$, we equate the real and imaginary parts: $x = p^3 - 3pq^2 = p(p^2 - 3q^2)$ $-y = 3p^2q - q^3 \implies y = q^3 - 3p^2q = -q(3p^2 - q^2)$. Now, calculate $\frac{x}{p} + \frac{y}{q}$: $\frac{x}{p} = p^2 - 3q^2$ $\frac{y}{q} = q^2 - 3p^2$ $\frac{x}{p} + \frac{y}{q} = (p^2 - 3q^2) + (q^2 - 3p^2) = -2p^2 - 2q^2 = -2(p^2 + q^2)$. Finally, $\frac{(\frac{x}{p} + \frac{y}{q})}{(p^2 + q^2)} = \frac{-2(p^2 + q^2)}{(p^2 + q^2)} = -2$.

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

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