Quick Review on Complex Analysis (ep01-06)

19 Nov, 2023

This is a quick review on what is taught in the lectures by Richard E. Borcherds.

Differentiablity

If we write a complex function $w (z) = u (x + i y) + i v (x + i y)$ into the vector form in Gaussian plane, we would have:

w (x, y) = [\begin{matrix} u (x, y) \\ v (x, y) \end{matrix}]

And thus if $w$ is differentiable $d z = A d (x + i y)$ :

d w (x + i y) = [\begin{matrix} R e A - I m A \\ I m A R e A \end{matrix}] [\begin{matrix} d x \\ d y \end{matrix}]

By the differentiability of functions $ℝ^{2} \to ℝ^{2}$ we understand if the reaction to small disturbance is almost linear:

d \vec{f} = [\begin{matrix} \frac{\partial u}{\partial x} \frac{\partial u}{\partial y} \\ \frac{\partial v}{\partial x} \frac{\partial v}{\partial y} \end{matrix}] [\begin{matrix} d x \\ d y \end{matrix}]

Comparing the two expression, it is obvious to see:

\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}, \frac{\partial v}{\partial x} = - \frac{\partial u}{\partial y} .

This is known as the Cauchy-Riemann equation. Differentiable complex functions are referred to as holomorphic.

A direct result from Cauchy-Riemann equation would be the real part function $u$ should satisfy:

\frac{\partial^{2} u}{\partial x^{2}} = \frac{\partial}{\partial x} (\frac{\partial u}{\partial x}) = \frac{\partial}{\partial x} (\frac{\partial v}{\partial y}) = \frac{\partial^{2} u}{\partial x \partial y} = \frac{\partial}{\partial y} (\frac{\partial u}{\partial y}) = - \frac{\partial^{2} u}{\partial y^{2}}

That is, $u$ satisfies the Laplace equation:

\nabla^{2} u = 0

$u$ is a harmonic function.

We can prove that for any given harmonic function $u$ , $v$ is well-defined on a simply-connected set such that $w = u + i v$ is holomorphic. Actually, denote $\frac{\partial u}{\partial x} = f, \frac{\partial u}{\partial y} = g$ , the solution of differential equation:

\frac{\partial v}{\partial x} = - g, \frac{\partial v}{\partial y} = f

is unique upto some constant. A quick proof would be simply integrate function $f, g$ along path $(0, 0) \to (x, 0) \to (x, y)$ . A critical observation is that by Cauchy-Riemann, the integral is actually path-irrelevant.