written 7.7 years ago by |
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)