Question

If $z=x-iy$ and $z^{1/3}=p+iq$, then $\frac{\frac{x}{p}+\frac{y}{q}}{p^2+q^2}$ equals

• A. $2$
• B. $-1$
• C. $1$
• D. $-2$

Answer 1

The correct option is D $-2$

$z^{1/3}=p+iq$
$\Rightarrow x-iy=(p+iq{)}^3$
$\Rightarrow x-iy=p^3-3p{q}^2+i(3p^{2}q-q^3)$
$\Rightarrow x=p^3-3p{q}^2$ and $y=-(3p^{2}q-q^3)$
$\Rightarrow \frac{x}{p}=p^2-p{q}^2$ and $\frac{y}{q}=-(3p^2-q^2)$ .... (*)
Adding the equations of (*) we get
$\frac{x}{p}+\frac{y}{q}=-2(p^2+q^2)$