Milne-Thomson method for finding a holomorphic function

In mathematics, the Milne-Thomson method is a method for finding a holomorphic function whose real or imaginary part is given. It is named after Louis Melville Milne-Thomson.

Introduction
Let $$z = x + iy$$ and $$ \bar {z}\ = x - iy$$ where $$x$$ and $$y$$ are real.

Let $$f(z) = u(x,y) + iv(x,y)$$ be any holomorphic function.

Example 1: $$z^4 = (x^4-6x^2y^2+y^4) +i(4x^3y-4xy^3)$$

Example 2: $$\exp(iz)=\cos(x)\exp(-y)+i\sin(x)\exp(-y)$$

In his article, Milne-Thomson considers the problem of finding $$f(z)$$ when 1. $$u(x,y)$$ and $$v(x,y)$$ are given, 2. $$u(x,y)$$ is given and $$f(z)$$ is real on the real axis, 3. only $$u(x,y)$$ is given, 4. only $$v(x,y)$$ is given. He is really interested in problems 3 and 4, but the answers to the easier problems 1 and 2 are needed for proving the answers to problems 3 and 4.

1st problem
Problem: $$u(x,y)$$ and $$v(x,y)$$ are known; what is $$f(z)$$?

Answer: $$f(z)=u(z,0)+iv(z,0)$$

In words: the holomorphic function $$f(z)$$ can be obtained by putting $$x = z$$ and $$y = 0$$ in $$u(x,y)+iv(x,y) $$.

Example 1: with $$u(x,y)=x^4-6x^2y^2+y^4$$ and $$v(x,y)=4x^3y-4xy^3$$ we obtain $$f(z)=z^4$$.

Example 2: with $$u(x,y)=\cos(x)\exp(-y)$$ and $$v(x,y)=\sin(x)\exp(-y)$$ we obtain $$f(z)=\cos(z)+i\sin(z)=\exp(iz)$$.

Proof:

From the first pair of definitions $$x = \frac{z + \bar {z}}{2}$$ and $$y = \frac{z - \bar {z}}{2i}$$.

Therefore $$f(z)= u \left( \frac{z + \bar {z}}{2}\, \frac{z - \bar {z}}{2i}\right) + iv\left( \frac{z + \bar {z}}{2}\ , \frac{z - \bar {z}}{2i}\right) $$.

This is an identity even when $$x$$ and $$y$$ are not real, i.e. the two variables $$z$$ and $$ \bar {z}\ $$ may be considered independent. Putting $$\bar {z} =z $$ we get $$f(z) = u(z,0) + iv(z,0) $$.

2nd problem
Problem: $$u(x,y)$$ is known, $$v(x,y)$$ is unknown, $$f(x+i0)$$ is real; what is $$f(z)$$?

Answer: $$f(z)=u(z,0)$$.

Only example 1 applies here: with $$u(x,y)=x^4-6x^2y^2+y^4$$ we obtain $$f(z)=z^4$$.

Proof: "$$f(x+i0)$$ is real" means $$v(x,0)=0$$. In this case the answer to problem 1 becomes $$f(z)=u(z,0)$$.

3rd problem
Problem: $$u(x,y)$$ is known, $$v(x,y)$$ is unknown; what is $$f(z)$$?

Answer: $$f(z)=u(z,0)-i \int u_y(z,0) dz$$ (where $$u_y(x,y)$$ is the partial derivative of $$u(x,y)$$ with respect to $$y$$).

Example 1: with $$u(x,y)=x^4-6x^2y^2+y^4$$ and $$u_y(x,y)=-12x^2y+4y^3$$ we obtain $$f(z)=z^4+iC $$ with real but undetermined $$C$$.

Example 2: with $$u(x,y)=\cos(x)\exp(-y)$$ and $$u_y(x,y)=-\cos(x)\exp(-y)$$ we obtain $$f(z)=\cos(z)+i\int\cos(z)dz=\cos(z)+i(\sin(z)+C)=\exp(iz)+iC$$.

Proof: This follows from $$f(z)=u(z,0)+i \int v_x(z,0) dz$$ and the 2nd Cauchy-Riemann equation $$u_y(x,y)=-v_x(x,y)$$.

4th problem
Problem: $$u(x,y)$$ is unknown, $$v(x,y)$$ is known; what is $$f(z)$$?

Answer: $$f(z)=\int v_y(z,0)dz+i v(z,0)$$.

Example 1: with $$v(x,y)=4x^3y-4xy^3$$ and $$v_y(x,y)=4x^3-12xy^2$$ we obtain $$f(z)=\int 4z^3dz+i0=z^4+C$$ with real but undetermined $$C$$.

Example 2: with $$v(x,y)=\sin(x)\exp(-y)$$ and $$v_y(x,y)=-\sin(x)\exp(-y)$$ we obtain $$f(z)=-\int\sin(z)dz+i\sin(z)=\cos(z)+C+i\sin(z)=\exp(iz)+C$$.

Proof: This follows from $$f(z)=\int u_x(z,0) dz +i v(z,0)$$ and the 1st Cauchy-Riemann equation $$u_x(x,y)=v_y(x,y)$$.