Analytical Functions of a Complex Variable

Solutions to

$${\partial^{2}F\over \partial x^{2}} + {\partial^{2}F\over \partial y^{2}} = 0,$$

can be found by setting

$$F = F(z), \qquad z = x + iy.$$

Here $F(z)$ is an analytic function of the complex variable $z$. Splitting the function $F(z)$ into real and imaginary parts we obtain

$$F(z) = u(x,y) + i v(x,y),$$

and both $u(x,y)$ and $v(x,y)$ satisfy Laplace's equation.

Example 5. .2Consider solutions to Laplace's equation with

$$F(x,0) = {1\over 1 + x^{2}}, \qquad F(x,y) \rightarrow 0 \hbox{ as } x,y \rightarrow \infty.$$

In this case we may try

$$F(z) = {1\over 1 + z^{2}} = {1\over 1 + (x+iy)^{2}}.$$

Expanding the square and expressing in terms of real and imaginary parts we obtain

$${1\over 1 + x^{2} - y^{2} + i2xy} = {1 + x^{2} - y^{2} - i2xy\over (1 + x^{2} - y^{2})^{2} + 4 x^{2}y^{2}}.$$

Thus, we have

$$u = {1 + x^{2} - y^{2}\over (1 + x^{2} - y^{2})^{2} + 4x^{2}y^{2}}$$

and

$$v = {-2xy\over (1 + x^{2} - y^{2})^{2} + 4x^{2}y^{2}}.$$

Using the real part for the potential, $F$, the solution to Laplace's equation is

$$F(x,y) = {1 + x^{2} - y^{2}\over (1 + x^{2} - y^{2})^{2} + 4x^{2}y^{2}}.$$

It is straightforward to show that both Laplace's equation and the boundary conditions are satisfied.