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  • If (x+iy)^3=u+iv, then show that u/x+v/y=4(x^2-y^2)

Q : 16       If  \small (x+iy)^3=u+iv, then show that   \small \frac{u}{x}+\frac{v}{y}=4 (x^2-y^2).

Answers (1)

best_answer

it is given that
\small (x+iy)^3=u+iv 
Now, expand the Left-hand side
x^3+(iy)^3+3.(x)^2.iy+3.x.(iy)^2= u + iv
x^3+i^3y^3+3x^2iy+3xi^2y^2= u + iv
x^3-iy^3+3x^2iy-3xy^2= u + iv                                            (\because i^3 = -i \ \ and \ \ i^2 = -1)
x^3-3xy^2+i(3x^2y-y^3)= u + iv
On comparing real and imaginary part. we will get,
u = x^3-3xy^2 \ \ \ and \ \ \ v = 3x^2y-y^3
Now,
\frac{u}{x}+\frac{v}{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)
Hence proved

Posted by

Gautam harsolia

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