Question

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

• A. $-2$
• B. $-1$
• C. $1$
• D. $2$
• E. $0$

Answer 1

Answer: (A)

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