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Find the imaginary part whose real part is $u = x^3 - 3xy^2 + 3x^2 - 3y^2 + 1$
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Given: real part $u = x^3 - 3xy^2 + 3x^2 - 3y^2 + 1$

Let $\phi_{1}(x, y) = \frac{\partial u}{\partial x}, \phi_{2}(x,y) = \frac{\partial u}{\partial y}$

i.e.

$u_{x} = 3x^2 - 3y^2 + 6x = \phi_{1}(x, y)$ and $u_{y} = -6xy - 6y = \phi_{2}(x, y)$

By Milne Thompson

$f^{1}(z) = \phi_{1}(z, 0) - i\phi_{2}(z, 0)$

i.e. substitute, x = z, y = 0 in $\phi_{1}$ & $\phi_{2}$

$\therefore \phi_{1}(z,0) = 3z^2 + 6z$

$\phi_{2}(z, 0) = 0$

$\therefore f^{1}(z) = 3z^2 + 6z - i(0) = 3z(z + 2)$

Integrating the above equation

$f(z) = \int 3z^2 + 6zdz \\ = 3\frac{z^3}{3} + \frac{6z^2}{2} + c \\ = z^3 + 3z^2 + c$

Now, Z = x + iy

$f(z) = (x + iy)^3 + (x + iy)^2 + c \\ f(z) = x^3 + 3x^2(iy) + 3x(iy)^2 + (iy)^3 + 3(x^2 + 2xyi + (iy)^2) \\ \therefore u + iv = (x^3 + 3x^2 - 3xy^2 - 3y^2) + i(3x^2 y - y^3 + 6xy) \\ \therefore u = (x^3 + 3x^2 - 3xy^2 - 3y^2)$ & $V = (3x^2 y - y^3 + 6xy)....$(Imaginary part)

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