Question

If, \( u=y^3-3x^2y \) be a harmonic function then its corresponding analytic function \( f(z) \), where \( z=x+iy \), is:

• A. \( f(z) = z^2 + C \); where C is an arbitrary constant
• B. \( f(z) = i(z^2+C) \); where C is an arbitrary constant
• C. \( f(z) = z^3 + C \); where C is an arbitrary constant
• D. \( f(z) = i(z^3+C) \); where C is an arbitrary constant

Hint:

The Milne-Thomson method is an extremely fast way to reconstruct an analytic function from its real or imaginary part without explicitly finding the harmonic conjugate first. The formulas to remember are: - If \(u\) is given: \( f'(z) = u_x(z, 0) - i u_y(z, 0) \) - If \(v\) is given: \( f'(z) = v_y(z, 0) + i v_x(z, 0) \) Then integrate to find \(f(z)\).

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
An analytic function \( f(z) = u(x,y) + iv(x,y) \) has a real part \(u\) and imaginary part \(v\) that are harmonic conjugates, meaning they satisfy the Cauchy-Riemann equations: \( u_x = v_y \) and \( u_y = -v_x \). Given one part (e.g., \(u\)), we can find the other part (\(v\)) and construct the analytic function \(f(z)\) using the Milne-Thomson method.

Step 2: Key Formula or Approach (Milne-Thomson Method):
If the real part \(u(x,y)\) is given, the derivative of the analytic function can be expressed as: \[ f'(z) = u_x(x,y) - i u_y(x,y) \] In the Milne-Thomson method, we replace \(x\) with \(z\) and \(y\) with \(0\) in this expression to get \(f'(z)\) directly in terms of \(z\). \[ f'(z) = u_x(z, 0) - i u_y(z, 0) \] Then, we integrate \(f'(z)\) to find \(f(z)\).

Step 3: Detailed Explanation:
Given \( u(x,y) = y^3 - 3x^2y \). First, find the partial derivatives: \[ u_x = \frac{\partial u}{\partial x} = -6xy \] \[ u_y = \frac{\partial u}{\partial y} = 3y^2 - 3x^2 \] Now, use 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 these into the formula for \(f'(z)\): \[ f'(z) = u_x(z,0) - i u_y(z,0) = 0 - i(-3z^2) = 3iz^2 \] Now, integrate \(f'(z)\) with respect to \(z\) to find \(f(z)\): \[ f(z) = \int 3iz^2 dz = 3i \int z^2 dz = 3i \frac{z^3}{3} + C' = iz^3 + C' \] Let's see if this matches any options. It 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: Final Answer:
The corresponding analytic function is \( f(z) = i(z^3 + C) \).