Question

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

• A. $1$
• B. $-1$
• C. $2$
• D. $-2$

Answer 1

Answer: (D)

$-2$

Given, $z=x-iy$ and $z^{1/3}=p+iq$
$\Rightarrow {(x-iy)}^{1/3}=p+iq$
$\Rightarrow x-iy={(p+iq)}^3$
$\Rightarrow x-iy=p^3-i{q}^3+3p^{2}qi-3p{q}^2$
$\Rightarrow x-iy=(p^3-3p{q}^2)+i(3p^{2}q-q^3)$
On comparing both sides, we get
$x=p^3-3p{q}^2$
and $-y=3p^{2}q-q^3$
$\Rightarrow \frac{x}{p}=p^2-3q^2$
and $\frac{y}{q}=q^2-3p^2$
Now, $\frac{x}{p}+\frac{y}{q}=p^2-3q^2+q^2-3p^2$
$\Rightarrow \frac{x}{p}+\frac{y}{q}=-2p^2-2q^2$
$\Rightarrow \frac{x}{p}+\frac{y}{q}=-2(p^2+q^2)$
$\Rightarrow (\frac{x}{p}+\frac{y}{q})/(p^2+q^2)=-2$