Wednesday, 20 August 2014

I Refuse to Name This "Complex Analysis...Made Simple": 2

Complex differentiability, what's up with that, ammirite?

So, it turns out that if a complex function \(f: U\rightarrow \mathbb{C}\) is complex differentiable for an open plane region \(U\subset\mathbb{C}\), then \(f\) can be written as a power series around each point in \(U\). What is UP with that, ammirite??

Plus, once a function is complex differentiable ONCE, it's complex differentiable INFINITELY MANY TIMES. WHAT IS UP WITH THAT, AMMIRITE???

Wow...tough crowd, *awkward turtle time*. =P

Now, in order to show that a function is complex differentiable, you just need to write down its derivative (via first principles) and show that the limit exists. And here's where the complex case really differentiates itself from the real case. *insert smug smile for that pun*

Well, for one thing, we know that differentiability implies continuity. So, let's backtrack for a second and look at the continuity of a function. In the real case, to check continuity of a function \(g:\mathbb{R}\rightarrow\mathbb{R}\) at a point \(x_0\in\mathbb{R}\), we just need to check that the limit of \(g\) as we approach \(x_0\) from the left hand side of \(x_0\) agrees with what we get as we approach from the right hand side of \(x_0\). For complex functions from a planar domain, it's trickier. We've obviously got to check that the limit agrees in every which way, and there are uncountably infinitely many directions in which you can approach some point \(z_0=x_0+iy_0\in U=\mathrm{Dom}(f)\). And in fact, it's not even enough to just check the "linear directions" can be seen from the following example: \begin{align}f(x+yi)=\frac{x^2y}{x^4+y^2}.\end{align} Taking the limit as we approach \(0\) from any linear direction \(\alpha\in\mathbb{C}-\{0\}\), we get \(0\). Concretely written out: \begin{align}\lim_{t\to 0^+} f(t\alpha)=0.\end{align} However, taking the limit along a non-linear curve, we see that \begin{align}\lim_{t\to 0^+} f(t+t^2i)=\frac{1}{2}.\end{align} Thus, we see that this particular function \(f\) is not continuous at \(0\).

In order to check continuity in general, we really have to check every direction. This is done by going through an \(\epsilon-\delta\) convergence proof.

Um...ookay, maybe I'll talk more about complex differentiability next time. Or possibly when you guys start to look at them sweet sweet power series results!

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