Question

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

Answer 1

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

Taking cube on both sides

$(x+iy)={(a+ib)}^3=a^3-i{b}^3+3abi(a+ib)\Rightarrow (x+iy)=a^3-i{b}^3+3a^{2}bi-3a{b}^2\Rightarrow x+iy=(a^3-3a{b}^2)+i(3a^{2}b-b^3)$

Comparing the real and imaginary parts

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

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

Hence proved.