Question

find the imaginary part of the analytic function whose real part is x^3 - 3xy^2 +3x^2 - 3y^2​

Answer 1

Answer:

find the imaginary part of the analytic function whose real part is x^3 - 3xy^2 +3x^2 - 3y^2

Answer 2

Answer:

The imaginary part = i(3x²y - y³ + 6xy)

Step-by-step explanation:

The real part, U = x³ - 3xy² + 3x² - 3y²

• Step-1: Ф1(x,y) = du/dx = $\frac{d}{dx}$ (x³ - 3xy² + 3x² - 3y²) = 3x² - 3y² + 6x

                    Ф2(x,y) = du/dy = $\frac{d}{dy}$ (x³ - 3xy² + 3x² - 3y²) = - 6xy - 6y

• Step-2: Putting x = z and y = 0 we get,

                    Ф1(z,0) =  3z² + 6z

                    Ф2(z,0) =  0

• Step-3: According to Milne Thompson, f'(z) = Ф1(z,0) - iФ2(z,0)

                                                                          f'(z) = 3z² + 6z

       By integrating both sides , f(z) = ∫(3z² + 6z)dz + c

                                                    f(z) = z³ + 3z² + c (c=integration constant)

• Step-4: Putting z = x + iy in f(z) we get,

                      f(x,y) = (x + iy)³ + 3(x + iy)² + c

                               = (x³ + i3x²y - 3xy² - iy³) + 3(x² - y² + i2xy) +c

                               = (x³- 3xy² + 3x² - 3y²) + i (3x²y- y³ + 6xy)

• Conclusion: The real part = x³- 3xy² + 3x² - 3y²

                              The imaginary part = 3x²y- y³ + 6xy