Quick Review on Complex Analysis (ep07-13)

22 Nov, 2023

This is a summary of the online class I recently watched. I also notice that it is available on Bilibili.

Integrations

It is noteworthy that the integration on $ℂ$ seems path-relevant, however, it is not if the function is holomorphic. Cauchy's Integration Theorem states that:

If $w (z)$ holomorphic on a simply-connected set $D$ then

$$ \oint\limits_{\partial D}w(z)\mathrm{d}z=0 $$

This means the integration is terminal-relevant only, for $\int_{C_{1}} w (z) d z \equiv \int_{C_{2}} w (z) d z$ if path $C_{1}$ and $C_{2}$ are homotopic.

Notice that if $w (z) = u + i v$ then the integral can be written as a line integral:

\oint_{\partial D} w (z) z = \oint_{\partial D} u + i v d x + i y = \oint_{\partial D} u d x - v d y + i \oint_{\partial D} v d x + u d y,

Apply Green's theorem to both integration:

\oint_{\partial D} w (z) d z = \iint_{D} (- \frac{\partial u}{\partial y} - \frac{\partial v}{\partial x}) d x d y + i \iint_{D} (\frac{\partial v}{\partial y} + \frac{\partial u}{\partial x}) d x d y = 0

The last equation is the result of Cauchy-Riemann equation.

A direct result of Cauchy's integration thm is the following thm

Given $f (z)$ a holomorphic function on $D$ , $C$ is any closed path in $D$ , then:

$$ f(\omega)=\frac{1}{2\pi i}\oint\limits_C\frac{f(z)}{z-\omega}\mathrm{d}z $$

Or, by shift the function a bit

f (0) = \frac{1}{2 π i} \oint_{C} \frac{f (z)}{z} d z

The proof goes as following. Decompose $f$ into a constant $f (0) = C$ and a holomorphic function $g (z)$ which $g (0) = 0$ . Now that by selecting a specific path (due to Cauchy integral)

\oint \frac{C}{z} d z = C \int_{0}^{2 π} \frac{1}{R e^{i θ}} i R e^{i θ} d θ = C 2 π i

And let the radius be small enough so $g (z) < ϵ$

\oint \frac{g (z)}{z} d z \leq 2 π R \frac{ϵ}{R} = 2 π ϵ \to 0

Combine the two: $\oint \frac{f (z)}{z} d z = 2 π i f (0)$ hence the result.

Notice that $\partial D$ is closed, hence finite, $f$ being holomorphic, hence bounded

\frac{d^{n}}{d ω^{n}} f (ω} = \frac{1}{2 π i} \oint f (z) \frac{d^{n}}{d ω^{n}} \frac{1}{z - ω} d z = \frac{n!}{2 π i} \oint \frac{f (z)}{(z - ω)^{n + 1}} d z

That is, if $f$ is holomorphic, then it is infinitely differentiable, it is analytic.

Laurant Series and Residual

This part is missing from the lecture, but I made it up by consulting other resources nonetheless.

Notice first that if $\frac{1}{z} \sum_{k} {(\frac{ω}{z})}^{k}$ uniformly converges to $\frac{1}{z - ω}$ in a compact set, then the Cauchy integral formula can be written in to a sum of polynomials.

However, notice that the convergence region obviously will not contain $ω$ itself, we have to consider the annulus around $ω$ thus gives

f (z_{0}) = \frac{1}{2 π i} \oint_{r_{1}} \frac{f (z)}{z - ω} d z - \frac{1}{2 π i} \oint_{r_{2}} \frac{f (z)}{z - ω} d z

expand the expression, the series is known as Laurant series.

f (z_{0}) = \sum_{n = 1} b_{n} (z_{0} - ω)^{- n} + \sum_{n = 0} (z_{0} - ω}^{n}

where $r_{2} \leq | z_{0} - ω | \leq r_{1}$ and

b_{n} = \frac{1}{2 π i} \oint_{r_{2}} f (z) (z - ω)^{n - 1} d z, a_{n} = \frac{1}{2 π i} \oint_{r_{1}} \frac{f (z)}{(z - ω)^{n + 1}} d z

The coefficients of Laurant series show the type of poles of $ω$ , denote the number of non-zero coefficients in $b_{n}$ as $d$ , the order of a pole.

if $d = 0$ then the pole is removable by set $f (ω) = {lim}_{z \to ω} f (z)$
if $d = 1$ then it is a simple pole
if $d \to \infty$ then it is an essential singularity.

We also define residual of $f (z)$ at $ω$ as following:

R e s (f, ω) = b_{1}

This would be used in the following episode to get residual thm.