Complex analysis 101: hmm, a "Cauchy" tag might be a good idea

by math_explorer, Aug 21, 2013, 7:53 AM

"Complex analysis" basically means analysis, except you use complex numbers. Analysis is basically calculus, except fancier. Duh. Actually I read Wikipedia and searched a little and I don't really get where "calculus" and "analysis" diverge. It's like if people keep saying "calculus" then it doesn't sound smart anymore so you need a new fancier name.

So what happens in complex analysis?

First, of course, we define complex numbers. You can think of them as ordered pairs of reals with wonky multiplication, or as the quotient group $\mathbb{R}[x] / \langle x^2 + 1 \rangle$, or whatever. Go look up that stuff yourself. We need to get a little complex book-keeping out of the way: we have the absolute value $|x + iy| = \sqrt{x^2 + y^2}$ for real $x, y$. That's all I can think of.

Okay let's extend calculus to complex numbers.

First, we have the limit, which is defined exactly the same way as in boring single-variable calculus. You can take a shortcut by delegating to the real limit, but it boils down to epsilon-delta or epsilon-big-N. You need to use the absolute value, though.

Next, we have the derivative, which is defined exactly the same way as in boring single-variable calculus.

\[ \lim_{\Delta z \to 0} \frac{f(z + \Delta z) - f(z)}{\Delta z} \]

Mathematicians use $z$ to represent complex numbers. You see, $z$ stands for... uh, never mind.

If $f$ is a function and $z$ is a complex value so that the complex limit above exists, then we say that $f$ is complex differentiable at $z$. Except nobody says that, because it's boring. All the cool people say that $f$ is holomorphic at $z$. Or regular, or analytic, but don't use that until later because we haven't proven what "analytic" means yet.

PROTIP: Want to be cool? Just add lots of prefixes to -morphic and -morphism. Yaaaayyyy!

Finally, we have the integral, which is... uh.

Well. When we integrate from one number to another in boring single-variable calculus, we're stuck on a line so there's only one way to get from the first number to the second. Even if you imagine going back and forth, everything extraneous just cancels out quickly. But in the complex plane, there are lots of ways. You can go in a line, or go in a circle, or take a detour to the xkcd number to trace the Pikachu curve before coming back. So that's kind of annoying.

And you can't really work around this either, because some complex integrals do depend on which curve you take. The obvious example is

\[ \int_{-1}^1 \frac{1}{z} dz \]

which is $\pi i$ if you go down, $-\pi i$ if you go up, and $1337\pi i$ if you loop around zero exactly $668$ times counterclockwise.

Oh well, we'll just have to specify the curve in an integral. And then we have to do a lot of work defining a curve, and proving that different parametrizations of the same curve don't make anything blow up, and so on. This is boring; I'm skipping it.

Then we realize that for a lot of functions, we don't actually need that.

Cauchy's theorem: If $f$ is complex differentiable holomorphic inside a closed set that basically doesn't have any holes, and $C$ is any closed curve in the closed set, then \[ \int_C f(z) dz = 0. \]

To prove this, we first prove something simpler:

Goursat's theorem: If $f$ is complex differentiable holomorphic on a closed triangle and $C$ is the triangle's boundary, then \[ \int_C f(z) dz = 0. \]

Pf. We induct on the size of the triangles. Wait, we can't do that.

We cut the big triangle into four smaller triangles along the midpoints of the sides, and consider the four integrals around their boundaries. An easy consequence of the integral is that if you integrate once across a curve this way and integrate back, the two results cancel out; so the sum of the integrals around the small triangles is the integral around the big triangle.

Well, we want the integral to be zero, so we want to disprove any hypothetical integrals that aren't zero. Let's focus on the "worst" triangle $T$, the one with the largest absolute value of boundary integral. Then we can bound the original integral:
\[ \left|\int_C f(z) dz\right| \leq 4\left|\int_T f(z) dz\right|. \]

That's not very useful if we do it just once. So we repeat! Keep cutting the worst triangle into four, and pick the worst of those four, and go on forever. After we cut $n$ times, our evil hypothetical triangle has diameter $2^{-n}$ and perimeter $2^{-n}$ times that of the original, and its boundary integral bounds the original integral within $4^n$ times of it.

The $4^n$ seems like it'll ruin everything, but the $2^{-n}$s might be able to cancel it out...

Well, complex differentiability holomorphicity isn't very useful unless we have a point to use it on. And luckily, the triangles we're picking out actually do give us a point; since they keep shrinking and always include the next one, they converge to something, which is an emphatic and mostly incorrect way of saying that there is exactly one point $p$ in all of them.

Grounds-up justification

Now that we have that point $p$, we examine $f$'s behavior near $p$. Holomorphicity really means that you can approximate $f$ with a linear function near $p$. Explicitly, write

\[f(p + \delta) = m\delta + \psi(\delta) + b\]

where $m = f'(p)$ and $b = f(p)$ are constants not depending on $\delta$. So $\psi$ is an auxiliary function measuring $f$'s departure from linearity. Then complex diff--- yes, fine, holomorphicity really means that \[ \lim_{\delta \to 0} \frac{\psi(\delta)}{\delta} = 0, \] that the departure from linearity is less than linear --- which is kind of obviously the point if you think about it. I think that's an important interpretation of what differentiability means in the real case, too; guess I never learned calculus properly.

So, what happens to that integral? Well, you can actually integrate the part like $m\delta + b$ very easily, since such functions have primitives, which is the cool, complex-analysis way of saying "antiderivatives". They're just polynomials, and as a result, integrals with the equal starting and ending point just vanish, just like in single-variable calculus.

As for the $\psi$ part: the whole less-than-linear part means that, by shrinking the diameter enough from cutting triangles, we can make the supremum absolute value of $\psi$ on the integral as small as we want relative to the diameter. Just "as small as we want" isn't strong enough, since we have to counteract the multiple $4^n$ from cutting triangles. That is, by picking $n$ big enough, we can make $\epsilon$ arbitrarily small in

\[ \sup_{z \in T} \psi(z) \leq \epsilon \operatorname{diam} T = \epsilon 2^{-n} \operatorname{diam} C. \]

And then if you plug everything back you realize you can bound \[ \left|\int_C f(z) dz\right| \] by $\epsilon$ times a constant, so take $n$ to infinity and $\epsilon$ goes to 0 and you're done!

Okay, we proved it for triangles. Then it turns out that Cauchy's theorem is pretty much just defining an antiderivative a primitive for a holomorphic function as the value you get by integrating to each point from a fixed point along some simple pattern of horizontal and vertical lines, and handwaving that this is really a primitive by indicating how the stuff you get when you holomorphize differentiate it can be shown, by cutting up into lots of canceling paths and Goursat triangles, to equal the original function.

The region can't have holes because otherwise you won't always be able to keep the simple patterns of horizontal and vertical lines varying continuously when defining the would-be primitive, which is necessary to reduce them to tiny triangles by canceling.

In reality, you can "easily" prove that the function you defined really is a primitive for most given valid regions, like a circle. The annoying thing is that it's not easy to formalize the idea of a valid region and prove that there's always a path from that fixed point that satisfies everything you want it to, and there are so many regions that we might want to use that doing all this work is just too infeasible (read: boring). So, handwave!

:bye:

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calculus is for calculating

analysis is for analysating

i think it's pretty clear :O

by pythag011, Aug 22, 2013, 4:48 AM

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