Question

If $z=x+iy,z^{1/3}=a-ib$ and $\frac{x}{a}-\frac{y}{b}=k(a^2-b^2),$ then $k=$

Answer 1

$(x+iy{)}^{1/3}=a-ib$
$\Rightarrow x+iy=(a-ib{)}^3\Rightarrow x+iy=(a^3-3a{b}^2)+i(b^3-3a^{2}b)$
$\Rightarrow x=a^3-3a{b}^2,y=b^3-3a^{2}b$

$\therefore \frac{x}{a}-\frac{y}{b}=a^2-3b^2-b^2+3a^2=4(a^2-b^2)$

Hence $k=4$