Question

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

Answer 1

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

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

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

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

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

Comparison
on equating real and imaginary parts, we get

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

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

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

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

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

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