Question

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

Answer 1

Let, $(x+iy{)}^3=u+vi$

or, $x^3+3x^2.\left(iy\right)+3x(iy{)}^2+(iy{)}^3=u+iv$

or, $x^3-3x{y}^2+i(3x^{2}y-y^3)=u+iv$

Comparing both sides we get,

$u=x^3-3x{y}^2$ and $v=3x^{2}y-y^3$.

Now

$(\frac{u}{x}+\frac{v}{y})$

$=x^2-3y^2+3x^2-y^2$ [Using values of $u$ and $v$]

$=4(x^2-y^2)$.