Question

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

• A. $1$
• B. $-1$
• C. $2$
• D. $-2$

Answer 1

The correct option is D $-2$

Given $z=x+iy$

$(p+iq{)}^3=x-iy$

$p^3-i{q}^3+i3q{p}^2-3p{q}^2=x-iy$ $[\because (a+b{)}^3=a^3+3a^{2}b+3a{b}^2+b^3]$

$p^3-3p{q}^2+i3q{p}^2-q^3=x-iy$

Hence

$x=p^3-3p{q}^2$ and $y=q^3-2q{p}^2$
Hence $qx=q{p}^3-3p{q}^3$ and $py=p{q}^3-3q{p}^3$
Hence $qx+py$
$=q{p}^3+p{q}^3-3(q{p}^3+p{q}^3)$
$=-2(q{p}^3+p{q}^3)$
$=-2pq(p^2+q^2)$...(i)

Hence, $\frac{\frac{x}{p}+\frac{y}{q}}{p^2+q^2}$ $=\frac{qx+py}{pq(p^2+q^2)}$
$=\frac{-2pq(p^2+q^2)}{pq(p^2+q^2)}$.......[from i]
$=-2$