Question

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

Answer 1

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

$⇒x^3+(iy)^3+3.x.iy(x+iy)=u+iv$

$⇒x^3+i^3+3x^2yi+3xy^2i^2=u+iv$

$⇒x^3-iy^3+3x^2yi-3xy^2=u+iv$

$⇒(x^3-3xy^2)+(3x^2y-y^3)=u+iv$

On equating real and imaginary parts, we obtain

$u=x^3=3xy^2,v=3x^2y-y^3$

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

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

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

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

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

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

Hence, proved.