2015/020) If $(x+yi)^3 = u + vi$ Prove that $\dfrac{u}{x} + \dfrac{v}{y} = 4(x^2 - y^2)$

$(x+yi)^3 = u + vi$
expand LHS

$x^3 + 3ix^2 y - 3 xy^2 - iy^3 = (x^3 - 3y^2) + i(3x^2y - y^3)$
= $x(x^2- 3y^2) + iy(3x^2 - y^2)$

equate real part on both sides and imaginary parts

to get $u = x(x^2-3y^2)$
$v = y(3x^2-y^2)$

so $\dfrac{u}{x} + \dfrac{v}{y}= (x^2-3y^2) + (3x^2-y^2) = 4x^2 - 4y^2$