Question

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

Answer 1

${(x+iy)}^3=u+iv$

$\Rightarrow x^3+{\left(iy\right)}^3+3.x.iy(x+iy)=u+iv$ $[\because (a+b{)}^3=a^3+3ab(a+b)+b^3]$

$\Rightarrow x^3+i^3{y}^3+3x^{2}yi+3x{y}^2{i}^2=u+iv$

$\Rightarrow x^3-{iy}^3+3x^{2}yi-3x{y}^2=u+iv$

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

On equating real and imaginary parts, we obtain

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

$\therefore \frac{u}{x}+\frac{v}{y}=\frac{x^3-3x{y}^2}{x}+\frac{3x^{2}y-y^3}{y}$

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

$=x^2-{3y}^2+{3x}^2-y^2$

$={4x}^2-{4y}^2$

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

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

Hence, it proved.