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
, 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
for real
. 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.
![\[ \lim_{\Delta z \to 0} \frac{f(z + \Delta z) - f(z)}{\Delta z} \]](//latex.artofproblemsolving.com/4/4/d/44da5f4b3f4ee32e9eb9fc46802a7b05656991af.png)
Mathematicians use
to represent complex numbers. You see,
stands for... uh, never mind.
If
is a function and
is a complex value so that the complex limit above exists, then we say that
is complex differentiable at
. Except nobody says that, because it's boring. All the cool people say that
is holomorphic at
. 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
![\[ \int_{-1}^1 \frac{1}{z} dz \]](//latex.artofproblemsolving.com/a/a/a/aaa5116b3190841bf5eaf66f2df838ec2a2f9a7f.png)
which is
if you go down,
if you go up, and
if you loop around zero exactly
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
is complex differentiable holomorphic inside a closed set that basically doesn't have any holes, and
is any closed curve in the closed set, then ![\[ \int_C f(z) dz = 0. \]](//latex.artofproblemsolving.com/c/c/1/cc11466da76875b7cbe01ee8d23a0a22d4702987.png)
To prove this, we first prove something simpler:
Goursat's theorem: If
is complex differentiable holomorphic on a closed triangle and
is the triangle's boundary, then ![\[ \int_C f(z) dz = 0. \]](//latex.artofproblemsolving.com/c/c/1/cc11466da76875b7cbe01ee8d23a0a22d4702987.png)
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
, the one with the largest absolute value of boundary integral. Then we can bound the original integral:
![\[ \left|\int_C f(z) dz\right| \leq 4\left|\int_T f(z) dz\right|. \]](//latex.artofproblemsolving.com/8/f/1/8f1017ba8aec1b55b7b190a9aedac281bfe966df.png)
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
times, our evil hypothetical triangle has diameter
and perimeter
times that of the original, and its boundary integral bounds the original integral within
times of it.
The
seems like it'll ruin everything, but the
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
in all of them.
Grounds-up justification
Now that we have that point
, we examine
's behavior near
. Holomorphicity really means that you can approximate
with a linear function near
. Explicitly, write
![\[f(p + \delta) = m\delta + \psi(\delta) + b\]](//latex.artofproblemsolving.com/a/d/3/ad33757080f1e78fa54b14ff51a2bf5c5af859c0.png)
where
and
are constants not depending on
. So
is an auxiliary function measuring
's departure from linearity. Then complex diff--- yes, fine, holomorphicity really means that
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
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
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
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
from cutting triangles. That is, by picking
big enough, we can make
arbitrarily small in
![\[ \sup_{z \in T} \psi(z) \leq \epsilon \operatorname{diam} T = \epsilon 2^{-n} \operatorname{diam} C. \]](//latex.artofproblemsolving.com/3/e/b/3ebc9abda060882910a557b521f87d66e911ea9a.png)
And then if you plug everything back you realize you can bound
by
times a constant, so take
to infinity and
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 definingan 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!

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
![$\mathbb{R}[x] / \langle x^2 + 1 \rangle$](http://latex.artofproblemsolving.com/8/8/b/88bd026b4f8c42d886e39e22b3f2ae45b284498b.png)


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.
![\[ \lim_{\Delta z \to 0} \frac{f(z + \Delta z) - f(z)}{\Delta z} \]](http://latex.artofproblemsolving.com/4/4/d/44da5f4b3f4ee32e9eb9fc46802a7b05656991af.png)
Mathematicians use


If






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
![\[ \int_{-1}^1 \frac{1}{z} dz \]](http://latex.artofproblemsolving.com/a/a/a/aaa5116b3190841bf5eaf66f2df838ec2a2f9a7f.png)
which is




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


![\[ \int_C f(z) dz = 0. \]](http://latex.artofproblemsolving.com/c/c/1/cc11466da76875b7cbe01ee8d23a0a22d4702987.png)
To prove this, we first prove something simpler:
Goursat's theorem: If


![\[ \int_C f(z) dz = 0. \]](http://latex.artofproblemsolving.com/c/c/1/cc11466da76875b7cbe01ee8d23a0a22d4702987.png)
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

![\[ \left|\int_C f(z) dz\right| \leq 4\left|\int_T f(z) dz\right|. \]](http://latex.artofproblemsolving.com/8/f/1/8f1017ba8aec1b55b7b190a9aedac281bfe966df.png)
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




The


Well,

Grounds-up justification
Here I'm providing a proof that goes all the way back to the least-upper-bound-property and basic topology. Mostly because I realized I couldn't remember all the steps of this stuff.
Let's call that property the "shrinking convergence" property --- that if you have an infinite sequence of sets of some type where each one contains the next and they get infinitely small, there exists one point that they all contain. (Note that the uniqueness of such a point, if it does exist, is easy: if there were two points, they'd have positive distance, and sets could not shrink to be infinitely small while containing both. (So we need to rigorously define infinitely small: their diameters converge to zero. (But what's the diameter? The supremum of the distance between any two points, right.))) Sounds fun and not at all mathematical, right?
Again, topologically, a compact set is a set for which, in any family of open sets whose union contains the whole set (an open subcover), there exists a finite subfamily of open sets whose union still contains the whole set (a finite subcover). Compact sets are bounded: consider the open subcover containing a neighborhood of radius 1 around every single point of the compact set. Compact sets are closed: if not, then there's a limit point
that's not in the compact set, and you can consider the open subcover containing a neighborhood that barely doesn't contain
around every single point of the compact set. And compactness is "heritable" (not actual jargon) in the sense that a compact set in one metric space is still compact in a bigger one, and a closed subset of a compact set is compact: union every covering open set with the complement of the to-be-proved-compact subset, which is open, and find a finite subcovering of that.
In metric spaces
(and isomorphic spaces, like
in particular which is isomorphic to
), these two conditions are also sufficient for compactness.
Now, compactness implies the shrinking convergence property. Again, we use the complement trick. If we have a sequence of shrinking compact sets where no point is included in all of them, then the complements of all the compact sets cover the biggest compact set. But a finite subcover of that would necessarily not include the smallest complemented compact set.
But weirdly, we prove the shrinking convergence property first, and compactness later, for
-cells, which are
-dimensional intervals/rectangles aligned to the axes. You know what I mean. Or if you don't, since we really only need
you can just pretend
-cells are rectangles. Shrinking convergence for
-cells is straightforward, actually: consider each dimension separately and take a least-upper-bound of the lower bounds, which is not greater than each of the upper bounds, since the upper bounds are also upper bounds of the lower bounds. [insert yo dawg image macro here]
To get compactness of
-cells, we employ an eerily familiar divide-and-conquer method: chop the
-cells up into
smaller
-cells along the midpoints of the edges. If there's an open subcover without a finite subcover for the big
-cell, there's an open subcover without a finite subcover for a small
-cell, and we repeat this. Then by shrinking convergence we get a point in all those allegedly uncoverable
-cells; but that point is covered by some open set, which has to cover one of those
-cells in the shrinking sequence, however small, since it's open. Then we have a contradiction.
Then compactness of bounded sets follows from the "heritability" we mentioned. Whew!
--- end tortuous hide tag ---
Let's call that property the "shrinking convergence" property --- that if you have an infinite sequence of sets of some type where each one contains the next and they get infinitely small, there exists one point that they all contain. (Note that the uniqueness of such a point, if it does exist, is easy: if there were two points, they'd have positive distance, and sets could not shrink to be infinitely small while containing both. (So we need to rigorously define infinitely small: their diameters converge to zero. (But what's the diameter? The supremum of the distance between any two points, right.))) Sounds fun and not at all mathematical, right?
Again, topologically, a compact set is a set for which, in any family of open sets whose union contains the whole set (an open subcover), there exists a finite subfamily of open sets whose union still contains the whole set (a finite subcover). Compact sets are bounded: consider the open subcover containing a neighborhood of radius 1 around every single point of the compact set. Compact sets are closed: if not, then there's a limit point


In metric spaces



Now, compactness implies the shrinking convergence property. Again, we use the complement trick. If we have a sequence of shrinking compact sets where no point is included in all of them, then the complements of all the compact sets cover the biggest compact set. But a finite subcover of that would necessarily not include the smallest complemented compact set.
But weirdly, we prove the shrinking convergence property first, and compactness later, for





To get compactness of








Then compactness of bounded sets follows from the "heritability" we mentioned. Whew!
--- end tortuous hide tag ---
Now that we have that point





![\[f(p + \delta) = m\delta + \psi(\delta) + b\]](http://latex.artofproblemsolving.com/a/d/3/ad33757080f1e78fa54b14ff51a2bf5c5af859c0.png)
where





![\[ \lim_{\delta \to 0} \frac{\psi(\delta)}{\delta} = 0, \]](http://latex.artofproblemsolving.com/2/0/c/20cbbcf2befeac3f70f825c8992ebe072f7796ca.png)
So, what happens to that integral? Well, you can actually integrate the part like

As for the





![\[ \sup_{z \in T} \psi(z) \leq \epsilon \operatorname{diam} T = \epsilon 2^{-n} \operatorname{diam} C. \]](http://latex.artofproblemsolving.com/3/e/b/3ebc9abda060882910a557b521f87d66e911ea9a.png)
And then if you plug everything back you realize you can bound
![\[ \left|\int_C f(z) dz\right| \]](http://latex.artofproblemsolving.com/3/e/c/3ecc2868810d1acd886c6174bda785f8162825ae.png)



Okay, we proved it for triangles. Then it turns out that Cauchy's theorem is pretty much just defining
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!
