Question

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

Answer 1

$(x+iy)={(a+ib)}^3$

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

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

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

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

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

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

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