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)

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