Question

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

Answer 1

Given, $(x+iy{)}^{\frac{1}{3}}=a+bi$

cube both sides,

$x+iy=(a^3-3a{b}^2)+(3a^{2}b-b^3)i$

comparing real and imaginary parts,

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

Adding both ,we get,

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