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.