Question

If $(x+iy{)}^3=u+iv$, then prove that $\frac{u}{x}+\frac{v}{y}=4(x^2-y^2)$.

Answer 1

Given, $(x+iy{)}^3=u+iv$
$x^3+(iy{)}^3+3.x.iy(x+iy)=u+iv$
$x^3+i^3{y}^3+3x^{2}yi+3x{y}^2{i}^2=u+iv$
$x^3-i{y}^3+3x^{2}yi-3x{y}^2=u+iv$
$(x^3-3x{y}^2)=i(3x^{2}y-y^3)=u+iv$
On equating real and imaginary parts, we get
$u=x^3-3x{y}^2,v=3x^{2}y-y^3$
$\therefore \frac{u}{x}+\frac{v}{y}=\frac{x^3-3x{y}^2}{x}+\frac{3x^{2}y-y^3}{y}$

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

$=x^2-3y^2+3x^2-y^2=4x^2-4y^2=4(x^2-y^2)$