Question

If $z=x$-iy and $z^{1 / 3}=p+i q$, 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)

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