This seminar by ness took place on 26th October 2008 20:00 UTC, in #mathematics channel of irc.freenode.net.

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Any explanation or errors regarding this seminar should be recorded in the discussion page.

## Seminar Edit

[16:01] <ness> ok then [16:02] <ness> so in the last seminar we covered differentiation of functions C -> C (where C denotes the complex numbers) [16:02] <ness> generalizing the notion of various limits in doing so [16:02] <ness> there are mainly two things you should have taken from that seminar: [16:03] <ness> 1. many things that you know about real analysis continue to hold in the complex case. These include the familar rules of differentiation etc [16:04] <ness> 2. a complex function to be differentiable imposes a certain rigidity. This is expressed most explicitly in the Cauchy-Riemann equations (CR) [16:04] <ness> We will now consider consider integration [16:05] <ness> Recall that in R, an integral is defined by the function we integrate and the interval of integration, where we have (for sufficiently nice functions) an intuitive geometric interpretation as an (oriented) area [16:06] <ness> now the straigtforward generasization of a real interval is a curve in C, where I don't want to define in detail what a curve is (because the details are somewhat messy) [16:06] <ness> but one thing is clear: if we have a sufficiently nice function gamma:[a;b] -> C then its image is a curve [16:06] <ness> (where a and b are real) [16:07] <ness> there is one problem in generalizing the integral to arbitrary curves, and that is that there is no directly intuitive equivalent to "area" [16:08] <ness> however, we proceed in formal analogy to the real case, and just see what we get (it will be different from the contour integral in R^2) [16:09] <ness> consider figure 3 on page two

[16:09] <ness> I will try to convey the geometric idea of what we mean by an integral of f(z) "along the curve Gamma" [16:09] <ness> the (smooth) arc in the figure is supposed to represent a curve [16:10] <ness> and the three line segments represent a polygonal approximation to that curve [16:11] <ness> I call them d_1, d_2 and d_3, where d_i is to be taken as the *complex* vector that represents the line segment [16:12] <ness> now arbitrarily on the line segment represented by d_i we chose a complex number c_i (so c_i = d_1 + d_2 + ... +d_{i-1} + r*d_i, where r is in [0, 1]) [16:12] <ness> and then we define as an approximation of the integral (i.e. a riemann sum) d_1 * f(c_1) + d_2 o f(c_2) + d_3 * f(c_3) [16:13] <ness> ups, the o should of course be a * [16:13] <ness> now it is clear that there are infinitely many ways of approximating the curve, and infinitely many ways to choose the c_i [16:14] <ness> but for any choice of c_i and d_i we define I_{alpha} = sum_{i} f(c_i) * d_i [16:15] <ness> (where alpha is supposed to label the choice of c_i and d_i) [16:16] <ness> and then if, regardless of how we choose the d_i, so long as they approximate the curve linearly, and regardless of how we choose the c_i, so long as they lie on the d_i, the I_{alpha} converges to the same value if we use increasingly better approximations (increasingly many n), then we say that the common limit is the integral of f along Gamma [16:17] <ness> now this is a rather messy and long wordy description, but we will see in a minute that there is in fact an easy expression for that in terms of things we know [16:18] <ness> but before proceeding along this line, let me demonstrate that we can actually use this geometric definition to evaluate integrals [16:18] <ness> suppose we want to calculate the integral of f(z) = 1/z along the unit circle (oriented counterclockwise), and suppose we know that the integral exists [16:19] <ness> now consider figure 4

[16:19] <ness> we approximate the unit circle by tangents [16:19] <ness> we will choose all of the line segments to have a common length d [16:20] <ness> we first need to find an expression of d_i [16:21] <ness> let's say that d_i is the tangent segment of length d at z_i, where z_i is somewhere on the unit circle [16:22] <ness> then, by definition, d_i is orthogonal to z_i, i.e. d_i is proportional to i*z_i (recall that multiplication by i means rotation by 90 degrees). But the length of z_i is unity, so d_i = i*z_i*d [16:23] <ness> next we choose c_i = z_i, then f(c_i) = f(z_i) = 1/z_i = conj(z_i), as |z_i|=1, where conj(z) denotes the complex conjugate of z. [16:24] <ness> thus in our riemann sum approximating the integral, the ith term will equal d_i*f(c_i) = d*z_i*i*conj(z_i) = d*i*|z_i| = d*i, as we chose z_i on the unit circle [16:25] <ness> so we can pull a factor of i out of the sum, and are left with sum_{i} d. But in the limit as we get better and better approximations, this sum must clearly equal the length of our curve [16:25] <ness> so we have shown that the integral of 1/z along the unit circle = 2pi*i [16:26] <ness> this is a very fundamental integral that occurs numerous times in complex analysis [16:26] <ness> but let us now return to a more formal definition of the contour integral [16:27] <ness> you can convince yourself that, if we parametrize a curve Gamma as gamma:[a;b] -> C, the wordy description from above reduces to integral_a^b f(gamma(t)) * gamma'(t) dt [16:28] <ness> in fact, if you dislike the geometric description, we can take this as the essential definition of the contour integral [16:29] <ness> for one thing, it can be seen easily that this definiton is well-behaved in the sense that the integral does not depend on the parametrization of Gamma. this can be shown most easily by using the substitution rule for functions R->C which can proved easily [16:30] <ness> but let us instead redo the above integral using the parametric evaluation [16:31] <ness> we can of course parametrize the unit circle by gamma(t) = e^{it}, for t in [0; 2pi]. Then gamma'(t) = i*e^{it}, so I = integral_0^{2pi} 1/e^{it} * i * e^{it} dt = 2pi*i [16:32] <ness> we have indeed proved more, for in the geometric evaluation we have assumed that the integral exists, which we didn't need here [16:32] <ness> so much for the definition [16:32] <ness> I have spent quite a lot of time on this because I think it is important to have a geometric view of the things [16:33] <ness> let us look at some basic properties of the integral defined this way [16:33] <ness> all of them can be shown easily using the parametric evaluation [16:34] <ness> first, the integral is linear in the integrand: integral_{Gamma} [a*f(z) + b*g(z)] dz = a*integral_{Gamma} f(z) dz + b*integral_{Gamma} g(z) dz [16:34] <ness> but furthermore, the integral is also "linear in the contour" [16:35] <ness> for if we denote by Gamma = Gamma_1 + Gamma_2 the contour (path) that first traverses Gamma_1, then Gamma_2, and by -Gamma the path that traverses in the opposite direction, the following holds [16:36] <ness> integral_{Gamma_1 + Gamma_2} f(z) dz = integral_{Gamma_1} f(z) dz + integral_{Gamma_2} f(z) dz [16:36] <ness> and integral_{-Gamma} f(z) dz = -integral_{Gamma} f(z) dz [16:36] <ness> all of these are trivial [16:37] <ness> now let's have a look at some of the more interesting properties [16:37] <ness> I stressed in the beginning that a function being complex differentiable imposed a large certain amount of rigidity, and we can also see this in integration, for the following holds: [16:38] <ness> if f(z) is complex differentiable inside Gamma, then the loop integral of f around Gamma vanishes [16:39] <ness> this is were the aforementioned difficulties with the definition of a curve surface again (what is inside and what outside), and I won't go into details here. the keywords are jordan curve, piecewise smooth curve, and so on [16:40] <ness> I won't prove this theorem either, here is a rough sketch: you can write integral_{Gamma} f(z) dz = integral_{Gamma} [u(x,y) + iv(x,y)] (dx + idy), then use Green's theorem to convert that into a surface integral, and simplify using the CR [16:41] <ness> it was of course obvious from the start that the CR played an essential role here [16:41] <ness> you might think (if you haven't seen comparable things before) that vanishing loop integrals are not very interesting, but something very remarkable follows from this: [16:42] <ness> if f(z) is complex differentiable inside an open, simply connected set G, then integrals along contours inside that set only depend on the beginning and end points, i.e. they are path independent [16:43] <ness> consider figure 5

[16:43] <ness> let's assume that our function is differentiable on the entire sheet. Then what I just said means that the integral along A is the same as the integral along B [16:44] <ness> from that figure, we can also easily see the relation to vanishing loop integrals: [16:44] <ness> from the theorem, we know that (symbolically) integral_{A+B} = 0 = integral_{A+C}. [16:45] <ness> But from the basic properties, integral_{A+B} = integral_A + integral_B, so we have integral_A + integral_B = integral_A + integral_C, which of course implies integral_A = integral_B (make sure you see that geometrically) [16:45] <ness> or [16:45] <ness> er [16:46] <ness> the last statement should of course read integral_B = integral_C [16:47] <ness> so that means that we can continuosly deform a contour of integration in any way we like, so long as we stay in the region of holomorphy [16:47] <ness> (holomorphic is another term for complex differentiable) [16:48] <ness> so the integral from above immediatly tells us that the integral around *any* counterclockwise closed loop of f(z) = 1/z equals 2pi*i [16:49] <ness> here we of course assume that the loop goes aronud the origin, but doesn't "wind" around it more that once either [16:49] <ness> we also know that the loop integral around any other loop (not enclosing the origin) must be zero [16:50] <ness> there are a number of other interesting results on contour integration with closed loops, treating cases where there are singularities inside the contour, i.e. where the former theorem does not apply [16:51] <ness> we can begin with a general idea: [16:51] <ness> suppose we have a complicated loop, as depicted in the upper left image of figure 6

[16:52] <ness> the function is assumed to be complex differentiable everywhere, except for the dots (we call them isolated singularities) [16:52] <ness> we can now try to continuously deform the contour of integration, without passing over the poles of course [16:52] <ness> our findings on path independence ensure that we don't change the value of the integral along the contour [16:53] <ness> I have tried to indicate one especially useful deformation in the other two images of figure 6 [16:53] <ness> you see that we can reduce the complicated curve to loop integrals around the singularities [16:56] <ness> and you can see that the value of the integral essentially only depends on the number of times that the curve "winds" around each singularity [16:56] <ness> We can define Ind [16:56] <ness> argh [16:57] <ness> We can define Ind_{Gamma} z to be "the number of times Gamma winds around z" [16:57] <ness> this of course must be an integer [16:57] <ness> we count windings in the counterclockwise orientation positive, clockwise orientation counts negative [16:58] <ness> we can further define Res_a f to be 1/{2pi*i} lim_{r -> 0} int_{C_r} f(z) dz, where C_r is a circle of radius r and center a, oriented counterclockwise [16:59] <ness> the factor of 1/{2pi*i} might look useless but it turns out to be useful ;) [17:00] <ness> then, for any Gamma inside a set G where f is holomorphic except for a finite number of isolated singularities, we know that integral_{Gamma} f(z) dz = 2pi*i sum_{a in G} Ind_{Gamma} (a) Res_a f. [17:01] <ness> The sum reduces to a finite sum over the the singularities [17:01] <ness> now time is over, so let me just say a few more words: [17:02] <ness> it turns out that for sufficiently nice functions (those which are a ratio of two holomorphic functions, also called meromorphic functions) there is a relatively easy, different way to compute Res_a f [17:03] <ness> If we formalize these ideas, we get the Residue Theorem, which is usually attributed to cauchy [17:03] <ness> Res_a f is called the residue of f at a [17:04] <ness> if you look things up (you are prepared for this now), you will see that the residue can be computed by differentiation! [17:04] <ness> so this is one really nice method to evaluate integrals [17:04] <ness> there is another theorem (perhaps more fundamental) about loop integration, called the cauchy formula [17:05] <ness> it can be used to show that complex differentiable functions are necessarily smooth and analytic [17:05] <ness> so you have a great unity in complex analysis, conformal = differentiable = smooth = analytic, and we use the term holomorphic becaus we don't like the others ;) [17:06] <ness> complex analysis finds numerous applications [17:06] <ness> and a lot more can be said about it, especially when striving for formal proofs [17:06] <ness> thanks for listening [17:07] <ness> I hop I didn't bore everyone away ^^ [17:07] <ness> are there questions? [17:08] <mark-t> I think it is somewhat typical for these to go longer than an hour, or at least not uncommon [17:08] <ness> hm, the ones I saw where about one hour [17:08] <_llll_> think it's varied a fair bit [17:08] <b0o-yeah> its entirely up to the presenter, thank you ness [17:09] <cerebrum> i did only catch parts of this, but this wasn't very rigorous. for example the "definition" of the index [17:09] <ness> sure [17:09] <mark-t> it's asking a bit much to go through all this rigorously; it would fill several weeks in a complex analysis class [17:09] <ness> I presented it in a way I would present it to someone who hasen't done any more difficult analysis than real calculus [17:10] <_llll_> plenty of books do things completely rigorously, i dont know what is usually recommeded here [17:11] <ness> if there is interest, I could give one more seminar, fleshing out calculation of residues, the cauchy formula, perhaps liouville's theorem as an example of how these things can be applied. But the important stuff I wanted to say has been said. [17:11] <mark-t> I think it's definitely worth going through laurent series and computing residues [17:12] <ness> sure [17:21] <thermoplyae> whoop, i walked away, but i agree [17:29] * ness (n=ness@xx.xxx.xxx.xxx) Quit (Remote closed the connection)