Problem Statement
If $(x + iy)^3 = u + iv$, then prove that:
$\frac{u}{x} + \frac{v}{y} = 4(x^2 - y^2)$
Verified Solution & Marking Scheme
Expand (x + iy)³ Using Binomial Theorem
$(x + iy)^3 = x^3 + 3x^2(iy) + 3x(iy)^2 + (iy)^3$
$= x^3 + 3x^2 y i - 3xy^2 - i y^3$
$= (x^3 - 3xy^2) + i (3x^2 y - y^3)$
Equate Real and Imaginary Parts
Given $(x + iy)^3 = u + iv$:
$u = x^3 - 3xy^2 = x(x^2 - 3y^2)$
$v = 3x^2 y - y^3 = y(3x^2 - y^2)$
Evaluate u/x and v/y and Add
$\frac{u}{x} = x^2 - 3y^2$
$\frac{v}{y} = 3x^2 - y^2$
$\frac{u}{x} + \frac{v}{y} = (x^2 - 3y^2) + (3x^2 - y^2) = 4x^2 - 4y^2 = 4(x^2 - y^2)$.