Question

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

• A. $-2$
• B. $-1$
• C. $2$
• D. $1$

Answer 1

Answer: (A)

$-2$

We have $z^{\frac{1}{3}}=p+iq$

[Cubing both sides]
$z=(p+iq{)}^3=p^3-q^3+3p\cdot iq\cdot (p+iq)$

$z=x-iy$

Comparing the real and imaginary parts,

$\Rightarrow x=p^3-3p{q}^2$

$\Rightarrow y=q^3-3p^{2}q$
So, $\frac{x}{p}=p^2-3q^2$

$\frac{y}{q}=q^2-3p^2$

$\therefore \frac{\frac{x}{p}+\frac{y}{q}}{(p^2+q^2)}=\frac{-2(p^2+q^2)}{(p^2+q^2)}=-2$

Hence, option A.