Question

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

• A. $1$
• B. $-1$
• C. $2$
• D. $-2$

Answer 1

Answer: (D)

$z^{1/3}=p+iq$
$\therefore x-iy={\left(p+iq\right)}^3$
$=p^3+3p^2\left(iq\right)+3p{\left(iq\right)}^2+{\left(iq\right)}^3$
$\Rightarrow x=p^3-3p{q}^2$ and $-y=3p^{2}q-q^3$
$\Rightarrow \frac{x}{p}=p^2-3q^2$ and $\frac{-y}{q}=3p^2-q^2$
$\Rightarrow \frac{x}{p}+\frac{y}{q}=-2\left(p^2+q^2\right)$
$\therefore \frac{\left(\frac{x}{p}+\frac{y}{q}\right)}{\left(p^2+q^2\right)}=-2$