Question

If $z=x+iy,z^{1/3}=a-ib,$ then $\frac{x}{a}-\frac{y}{b}=k\left(a^2-b^2\right)$ where $k$ is equal to

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

Answer: (D)

$z^{1/3}=a-ib\Rightarrow z={\left(a-ib\right)}^3$
$\therefore x+iy=a^3+i{b}^3-3i{a}^{2}b-3a{b}^2$.Then
$x=a^3-3a{b}^2\Rightarrow \frac{x}{a}=a^2-3b^2$
$y=b^3-3a^{2}b\Rightarrow \frac{y}{b}=b^2-3a^2$
So, $\frac{x}{a}-\frac{y}{b}=4\left(a^2-b^2\right)$