Question

If ${\left(x-iy\right)}^{\frac{1}{3}}=a-ib$, then show that $\frac{x}{a}+\frac{y}{b}=4\left(a^2-b^2\right)$.

Answer 1

${\left(x-iy\right)}^{\frac{1}{3}}=\left(a-ib\right)$

Cubing both side

$x-iy={\left(a-ib\right)}^3$

$=a^3-{\left(ib\right)}^3-3\times a\times ib\left(a-ib\right)$

$=a^3+i{b}^3-3a^{2}ib-3a{b}^2$

$=\left(a^3-3a{b}^2\right)+i\left(b^3-3a^{2}b\right)$

Comparing real & imaginary part

$x=a^3-3a{b}^2\Rightarrow \frac{x}{a}=a^2-3b^2$

$y=-\left(b^3-3a^{2}b\right)$

$\Rightarrow \frac{y}{b}=3a^2-b^2$

We get

$\frac{x}{a}+\frac{y}{b}=a^2-3b^2+3a^2-b^2$

$=4\left(a^2-b^2\right)$