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)

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