Step 1: Analytic Function Fundamentals:
An analytic function \( f(z) = u(x,y) + iv(x,y) \) comprises a real part \(u\) and an imaginary part \(v\). These parts are harmonic conjugates, satisfying the Cauchy-Riemann equations: \( u_x = v_y \) and \( u_y = -v_x \). Given one part (e.g., \(u\)), we can determine the other (\(v\)) and construct \(f(z)\) using the Milne-Thomson method.
Step 2: Milne-Thomson Method:
If \(u(x,y)\) is provided, the derivative of the analytic function is:
\[ f'(z) = u_x(x,y) - i u_y(x,y) \]In the Milne-Thomson method, substitute \(x\) with \(z\) and \(y\) with \(0\):
\[ f'(z) = u_x(z, 0) - i u_y(z, 0) \]Then, integrate \(f'(z)\) to find \(f(z)\).
Step 3: Detailed Example:
Given \( u(x,y) = y^3 - 3x^2y \).
First, calculate the partial derivatives:
\[ u_x = \frac{\partial u}{\partial x} = -6xy \]\[ u_y = \frac{\partial u}{\partial y} = 3y^2 - 3x^2 \]Apply the Milne-Thomson method. Replace \(x\) with \(z\) and \(y\) with \(0\):
\[ u_x(z,0) = -6(z)(0) = 0 \]\[ u_y(z,0) = 3(0)^2 - 3z^2 = -3z^2 \]Substitute into the formula for \(f'(z)\):
\[ f'(z) = u_x(z,0) - i u_y(z,0) = 0 - i(-3z^2) = 3iz^2 \]Integrate \(f'(z)\) to find \(f(z)\):
\[ f(z) = \int 3iz^2 dz = 3i \int z^2 dz = 3i \frac{z^3}{3} + C' = iz^3 + C' \]This matches the form of option (D).
Let's check the result. If \( f(z) = iz^3 + C' \), let \( C' = C_1 + iC_2 \).
\[ f(z) = i(x+iy)^3 + C_1+iC_2 = i(x^3 + 3x^2(iy) + 3x(iy)^2 + (iy)^3) + C_1+iC_2 \]\[ = i(x^3 + 3ix^2y - 3xy^2 - iy^3) + C_1+iC_2 = ix^3 - 3x^2y - 3ixy^2 + y^3 + C_1+iC_2 \]\[ = (y^3 - 3x^2y + C_1) + i(x^3 - 3xy^2 + C_2) \]The real part \( u(x,y) = y^3-3x^2y + C_1 \). This matches the given \(u\) (up to a constant which can be absorbed into C).
The solution \( f(z) = iz^3 + C' \) can be written as \( f(z) = i(z^3 - iC') = i(z^3 + C) \) where \(C = -iC'\) is another arbitrary complex constant.
Step 4: Solution:
The analytic function is \( f(z) = i(z^3 + C) \).