Mathematics - Complex Number Question with Solution | TestHub

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