Thursday, 11 September 2014

I Refuse to Name This "Complex Analysis...Made Simple": 3

Hey guys! Let us begin somewhat worryingly: with an apology. the dark corridoors of my decrepit heart, a solitary ray of light lingers on. It shines upon my macabre soulscape, and whispers forth a faint, but grave exhortation: "...the truth! By God they deserve the truth!..."

That's right - you've been lied to! AND I AM SO SORRY. Specifically, you've been taught that a complex function \(f: U\rightarrow\mathbb{C}\) is a called analytic on \(U\) if it's complex-differentiable at every point on \(U\). BUT THIS IS A LIE! This is in fact the definition for holomorphicity. So, then, what the heck is an analytic function? Well, a function \(f:U\rightarrow\mathbb{C}\) is analytic if it can be locally written expressed as a convergent power series at every point in \(U\). So why the mixed up in nomenclature? To be honest, I have no idea - but I will say this: it turns out that the two criteria are equivalent - a function is holomorphic if and only if it's analytic!

*Phew* glad I got THAT off my chest. =P

There're a few cool consequences of this equivalence between holomorphic and analytic functions. Firstly, power series are infinitely complex-differentiable, so holomorphic functions are infinitely differentiable! This is kinda freaky, coz the definition of holomorphic only requires that \(f\) be complex-differentiable - it doesn't even say that the derivative should be continuous...and yet...we are, basking in its smug infinite differentiability.

Secondly, power series are basically just like beefed up polynomials, right? And we should know from highschool algebra that any two order \(n\) polynomials which agree on \(n+1\) points have to agree everywhere. Well, that observation also beefs up for analytic functions: any two analytic/holomorphic functions which agree on a countable collection of points with an accumulation (limit) point agree everywhere!

We can also use this characterisation to help us get a better feel for some of the results or phenomena that we see in complex analysis. You've seen from your tute questions that it is possible to have a vanishing contour integral of a function \(f\) around some \(\Gamma\) bounding a simply connected domain \(U\) so that \(f\) has a pole inside of \(U\). Take for example the following integral: \begin{align}\oint_{|z|=1}\frac{\mathrm{d}z}{z^2}=\int_0^{2\pi}\frac{i\exp(i\theta)}{\exp(2i\theta)}\mathrm{d}\theta=\int_0^{2\pi}i\exp(-i\theta)\mathrm{d}\theta=0.\end{align} So, is there a nice criterion for when such integrals will vanish? Well, let's assume (to make things easy) that the pole for \(f\) is positioned at \(0\) and that the pole is of order \(n\). This means that \(z^n f(z)\) is holomorphic at \(0\) and hence has a power series expansion at this point: \begin{align} z^nf(z)&=a_0+a_1z+a_2z^2+\ldots+a_{n-1}z^{n-1}+a_nz^n+\ldots\\ f(z)&=a_0z^{-n}+a_1z^{1-n}+a_2z^{2-n}+\ldots+a_{n-1}z^{-1}+a_n+\ldots\end{align} Now, doing a contour integral around the unit circle \(|z|=1\), we expect everything to vanish except possibly \(\oint_{|z|=1}a_{n-1}z^{-1}\mathrm{d}z\). This term contributes \(2\pi i a_{n-1}\) to the integral, and so a contour integral around a pole of a function vanishes precisely when the "power-series" has no \(z^{-1}\) term. In fact, the coefficient of the \(z^{-1}\) term is so important that we've given it a name: it's called the residue.

Soooooooooo...complex simple, right? =)

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