Question

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

• A. $1$
• B. $-1$
• C. $2$
• D. $-2$

Answer 1

The correct option is D

$-2$

The explanation for the correct option:

The given equation, $z=x-iy$.

It is also given that, $z^{\frac{1}{3}}=p+iq$.

$$\begin{gathered} \Rightarrow {\left(z^{\frac{1}{3}}\right)}^3={\left(p+iq\right)}^3 \\ \Rightarrow z={\left(p+iq\right)}^3 \\ \Rightarrow x-iy=p^3+{\left(iq\right)}^3+3p^2\left(iq\right)+3p{\left(iq\right)}^2 \\ \Rightarrow x-iy=p^3+i^3{q}^3+3i{p}^{2}q+3p{i}^2{q}^2 \\ \Rightarrow x-iy=p^3-i{q}^3+3i{p}^{2}q-3p{q}^2 \\ \Rightarrow x-iy=p^3-3p{q}^2-i\left(q^3-3p^{2}q\right) \end{gathered}$$

Compare both sides of the equation, $x=p^3-3p{q}^2$ and $y=\left(q^3-3p^{2}q\right)$.

Thus, $\frac{{\displaystyle \frac{x}{p}}+{\displaystyle \frac{y}{q}}}{p^2+q^2}=\frac{{\displaystyle \frac{p^3-3p{q}^2}{p}}+{\displaystyle \frac{q^3-3p^{2}q}{q}}}{p^2+q^2}$

$$\begin{gathered} \Rightarrow \frac{{\displaystyle \frac{x}{p}}+{\displaystyle \frac{y}{q}}}{p^2+q^2}=\frac{p^2-3q^2+q^2-3p^2}{p^2+q^2} \\ \Rightarrow \frac{{\displaystyle \frac{x}{p}+\frac{y}{q}}}{p^2+q^2}=\frac{-2p^2-2q^2}{p^2+q^2} \\ \Rightarrow \frac{{\displaystyle \frac{x}{p}+\frac{y}{q}}}{p^2+q^2}=\frac{-2\left(p^2+q^2\right)}{p^2+q^2} \\ \Rightarrow \frac{{\displaystyle \frac{x}{p}+\frac{y}{q}}}{p^2+q^2}=-2 \end{gathered}$$

Therefore, $\frac{{\displaystyle \frac{x}{p}}+{\displaystyle \frac{y}{q}}}{p^2+q^2}$ is equal to $-2$.

Hence, (D) is the correct option.