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18.112: Functions of a Complex Variable Introduction 1 September 5, 2019

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18.112: Functions of a Complex Variable Introduction 1 September 5, 2019 18.112: Functions of a Complex Variable Lecturer: Professor Wei Zhang Notes by: Andrew Lin Fall 2019 Introduction The course syllabus can be found on Stellar, and we were also given a paper copy in class. This class will assign nine problem sets, roughly due every week except for the three weeks with exams. The two midterms will be quiz-style – they’re only going to be 80 minutes long, since that’s how much time we have for class, and they’ll be on October 3rd and November 5th. (The questions will be pretty basic for the quizzes.) The final exam will take place during finals week, date and location to be determined. Grading is 30% homework (one problem set dropped), 30% final, and 20% for each midterm. Collaboration is allowed, but we’re encouraged to solve the problems ourselves first. If we do talk with others, we should indicate all collaborators on the homework itself. As per policy, late homework won’t be accepted, but certain situations can be accommodated e.g. from S3. 1 September 5, 2019 Let’s start with some motivation for why we care about functions of a complex variable. Fact 1 (Notation) R will denote the reals, Z will denote the integers, and N will denote the natural (positive) integers throughout this class. Here, R ⊃ Z ⊃ N, and we’ll be dealing with a superset of the real numbers: C, the complex numbers. We can identify elements of C naturally as ordered pairs of real numbers: in other words, C can be thought of as a 2-dimensional vector space over R. To review, any complex number z can be written as x + yi for real numbers x and p y, where i = −1. Remark 2. Mathematics defies English here: we say "I is" instead of "I am." But why do we even care about C in the first place – are there any applications for studying this? Definition 3 Prime numbers are the set of natural numbers whose only divisors are 1 and themselves. The prime counting function π(x) = number of primes p : p < x counts the number of primes less than a real number x 2 R>0. 1 One important result in number theory is the following result: Theorem 4 (Prime number theorem) Asymptotically, x π(x) ∼ : log x That is, the limit π(x) lim = 1: x!1 x= log x Obviously, this theorem statement has nothing to do with complex numbers (technically, integers are all we really need), but it turns out that the right way to think about this theorem is to use the Riemann zeta function 1 X 1 ζ(s) = : ns n≥1 π2 (For example, we know that ζ(2) = 6 .) The important change of perspective here is that we can replace s with an arbitrary complex number, and there are a lot of interesting things that come out of this! Obviously, ζ(s) isn’t convergent everywhere, but it will turn out to be convergent for Re(s) > 1, and we’ll study this a bit more later: it turns out we can prove the prime number theorem using the complex Riemann zeta function. One good way to start studying functions of a complex variable is to restrict our attention to some “nice” functions. When we have an arbitrary continuous function f : R ! R, we generally like it a lot when we can consider the derivative f (x + h) − f (x) f 0(x) = lim : h!0 h If this quantity f 0 exists, then we call our function differentiable. Let’s see if we can copy this over to complex numbers: Definition 5 Fix a point z 2 C. Given a function f : C ! C, define the derivative f (z + h) − f (z) f 0(z) = lim ; h!0 h if it exists. We should be a bit more specific here with what we want: since h is not just a real number anymore, what does it mean for h to go to 0? To answer this, we need to think about the idea of convergence. Definition 6 A sequence of complex numbers fzng converges to z if lim jzn − zj = 0: n!1 (Note that this last statement can be interpreted as an expression in the real numbers by looking at the real and imaginary parts at the same time.) Let’s try to introduce a few more real analysis concepts into our language here, as this will make later study easier: 2 Definition 7 fzng is a Cauchy sequence if jzn − zmj ! 0 as n; m ! 1: Definition 8 A set Ω is open if every point has an open neighborhood contained in Ω. A set is closed if its complement is open. 2 Since C can be treated similarly to R , we can also use the following definition: Definition 9 A set Ω is compact if it is closed and bounded. Example 10 D(z0; r), the closed disk centered at z0 with radius r, is compact. However, D(z0; r), the open disk, is not compact, though it is bounded. The reason we care about compact sets is that every sequence whose elements are in a compact set has a convergent subsequence. Definition 11 Ω is connected if we cannot write it as a disjoint union Ω1 t Ω2 of open sets. Ω is path-connected if any two points x; y in Ω can be connected by a path that lies inside Ω. (We’ll call open connected sets regions.) n In R , being open and connected is the same as being path-connected, but this isn’t true in a general topological space. Definition 12 Let Ω be an open subset of C (which can be C itself). A function f :Ω ! C is holomorphic at z if the derivative exists at z, and f is holomorphic on Ω if f is holomorphic at every z 2 Ω. Demanding that f be holomorphic is actually pretty strong: h can approach 0 from basically any angle in the complex plane, and we need to arrive at the same answer in all directions. Example 13 The function f (z) = z n is holomorphic for all n: expanding out (z + h)n with the binomial theorem is still valid, so the power rule still holds and the derivative exists: f 0(z) = nz n−1. However, g(z) = z is not holomorphic: g(z + h) − g(z) h = ; h h and notice that h and h will have the same norm as h approaches zero, but the arguments will be different (depends on the angle of approach). Thus, g0(z) does not exist. 3 This means that we somehow want functions that involve only z, not z: how do we quantify this? Notice that 2 when we have a function f : C ! C, we can think of this as taking ordered pairs (x; y) 2 R to other ordered pairs 2 (u; v) 2 R . So let’s try to study f in this new context! Theorem 14 (Cauchy-Riemann equations) Let Ω be an open set, and let f :Ω ! C be a function where z = x +iy and f (z) = u +iv. Then f is holomorphic on Ω if and only if u and v are differentiable, satisfying @u @v @u @v = and = − : @x @y @y @x Notice that u and v are both functions of x and y, so we have four partial derivatives, and all four of them are being used here. Proof. We can get expressions in these partial derivatives by looking at specific limits: for example, f (x + iy + h) − f (x + iy) @f @u @v lim = = + i : h2R h @x @x @x h!0 Similarly, if we go along the imaginary axis, f (x + i(y + h)) − f (x + iy) 1 @f 1 @u @v lim = = + : h2R ih i @y i @y @y h!0 But if the limit exists, these two are both equal to f 0(x + iy), so we can equate real and imaginary parts: @u @v 1 @u @v = and = i : @x @y i @y @x We can state this equivalently in the following way: Proposition 15 A function f is holomorphic if and only if @f 1 @f = : @x i @y Proof. If f is holomorphic, we know that f (z + h) = f (z) + f 0(z)h + hφ(h); where φ(h) ! 0 as h ! 0. (This can also be written as df = f 0(z)dz:) We want to (intuitively) write down something that looks like @f @f df = dz + dz @z @z with the multivariable chain rule. To do this, note that df = d(u + iv) = du + idv = ux dx + uy dy + ivx dx + ivy dy; but we can also relate x and y to z and z: z + z z − z x = ; y = ; 2 2i 4 which is equivalently (in differential form) 1 1 1 1 dx = dz + dz; dy = dz − dz: 2 2 2i 2i So we can actually use the multivariable chain rule to say that @f @f @x @f @y = + ; @z @x @z @y @z @x @y and substituting in the values of @z and @z from above, we find that @f 1 @f 1 @f = + : @z 2 @x i @y Similarly, we find that @f 1 @f 1 @f = − : @z 2 @x i @y The Cauchy-Riemann equations tell us that this last quantity should be 0 (by directly writing out f = u + iv), so @f @z = 0 if and only if f is holomorphic, which yields the result we want.
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