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1 Some Very Useful Theorems Yuval Advanced Complex Analysis Mathcamp 2017 1 Some very useful theorems The ultimate goal of this class is to understand the geometric properties of analytic functions, but before we can do that, there are quite a few theorems that we need to prove first. All of these theorems are suprising, beautiful, and insanely useful. 1.1 Isolated roots and unique continuations Definition. A function f : C ! C is said to have isolated roots if its roots (the places where it evaluates to 0) are \spaced out". Formally, this means that if z0 2 C is a root, namely f(z0) = 0, then there is some " > 0 such that if 0 < jz − z0j < ", then f(z) 6= 0. In other words, the roots of f are not arbitrarily close to one another. Having isolated roots is a very special property. For instance, the function g : C ! C defined by g(x + iy) = x − y is an extremely nice function, but it does not have isolated roots: the entire line y = x is composed of roots of g, so every root has other roots arbitrarily close to it. Nevertheless, we have the following result: Theorem. If f : C ! C is analytic and not the zero function, then f has isolated roots. Proof. Suppose not. Then we have a root z0 that has other roots arbitrarily close to it. This means that we can find a sequence of points z1; z2;::: 2 C with f(zk) = 0 and limk!1 zk = z0. We expand f as a Taylor series centered at z0, by writing 1 X n f(z) = an(z − z0) n=0 for some collection of complex numbers an. Since f is not the zero function, one of the an must be nonzero; let m be the smallest integer such that am 6= 0. Then the above Taylor series expansion becomes 1 1 1 ! X X X an+m f(z) = a (z − z )n = a (z − z )n = a (z − z )m 1 + (z − z )n n 0 n 0 m 0 a 0 n=0 n=m n=1 m So if we define 1 X an+m g(z) = zn a n=1 m then we have written m f(z) = am(z − z0) (1 + g(z − z0)) Moreover, from the definition of g, we see that lim g(z − z0) = 0 z!z0 This, along with the fact that zk ! z0, implies that for some k large enough, we have that g(zk − z0) 6= −1. Therefore, plugging in such a k to our formula for f, we get that m 0 = f(zk) = am(zk − z0) (1 + g(zk − z0)) 6= 0 since every term on the right-hand side is nonzero. This is a contradiction, as desired. Corollary. Suppose f1; f2 : C ! C are analytic functions, and suppose they are equal on some non-discrete set (e.g. a line segment, or a curve, or a disk). Then they are equal everywhere. Proof. Consider the function f = f1 − f2. It is analytic, and we wish to prove that it's the zero function (since that implies that f1 = f2). For this, observe that f(z) = 0 whenever f1(z) = f2(z), by definition. Additionally, since we assumed that f1; f2 agreed on some non-discrete set, this implies that the roots of f cannot be isolated. So by the previous theorem, we must have that f = 0, as desired. 1 Yuval Advanced Complex Analysis Mathcamp 2017 This result is quite powerful, and in my opinion, very surprising. It tells us that the behavior of an analytic function on some tiny segment actually determines its behavior on the whole complex plane. It also makes the Cauchy Integral Formula seem less impressive|that theorem allows us to determine what an analytic function does inside a region based on what it does on the boundary, but this result is much more general than that. One corollary of this result is a new proof of a theorem that we already know, namely that the conjugation function f(z) = z is not analytic. Indeed, this function agrees with the identity function g(z) = z on the non-discrete set R ⊂ C. Since g is analytic, if f were analytic, then they'd have to agree everywhere, and they don't; thus, f is not analytic. 1.2 The Argument Principle Our next result is a very useful consequence of the Residue Theorem, which tells us how to count roots and poles of meromorphic functions. Theorem (The Argument Principle). Let f : C ! C be meromorphic, and let C be a simple closed curve. Suppose that f is never zero and has no poles on C. Then 1 Z f 0(z) dz = R − P 2πi C f(z) where R is the number of roots of f inside C, and P is the number of poles of f inside C, and both are counted with multiplicity. Proof. First, we observe a very useful property of the quantity f 0=f, which is called the logarithmic derivative of f (for the reason why, see the homework!). Namely, if f1; f2 are two functions, then by the product rule, (f f )0 f 0 f + f f 0 f 0 f 0 1 2 = 1 2 1 2 = 1 + 2 f1f2 f1f2 f1 f2 In other words, the logarithmic derivative of a product of functions is the sum of the logarithmic derivatives of the individual functions. Now, recall the Residue Theorem. It tells us that 1 Z f 0(z) X f 0 dz = res 2πi f(z) z f C poles z of f 0=f in C So where does f 0=f have poles? Whenever f is zero, we're dividing by zero, so we certainly expect a pole there. Additionally, wherever f has a pole, then f 0 will also have a pole there, so we expect a pole of f 0=f. Finally, no other point will be a pole: at any other point, we will be dividing an analytic function (namely f 0) by a nonzero analytic function (namely f), so we will get something analytic as the quotient. So now we just need to understand the residues at these two types of points, namely roots of f and poles of f. Let's start with the roots: suppose z0 is a root of f with multiplicity m. Recall that this means that we can write m f(z) = (z − z0) g(z) for some analytic function g with g(z0) 6= 0. Then using our additivity formula, we see that 0 m 0 0 0 f (z) [(z − z0) ] g (z) m g (z) = m + = + f(z) (z − z0) g(z) z − z0 g(z) 0 Since g is analytic and nonzero near z0, we see that g =g is also analytic there. So that term contributes nothing to the residue. So the only term that matters is m=(z − z0), whose residue at z0 is precisely m. This means that at a root of multiplicity m, f 0=f has a simple pole with residue precisely m. 2 Yuval Advanced Complex Analysis Mathcamp 2017 On the other hand, if z1 is a pole of f of order n, then we can write −n f(z) = (z − z1) h(z) for some analytic function h with h(z1) 6= 0. Then we get that 0 −n 0 0 0 f (z) [(z − z1) ] h (z) −n h (z) = −n + = + f(z) (z − z1) h(z) z − z1 h(z) 0 which again implies that f =f has a simple pole of residue −n at z1. So when we add up all residue of f 0=f inside C, we find that this sum is precisely R − P , as desired. This theorem is extremely useful in lots of contexts. For one cool application, recall that the Riemann Hypothesis says that a certain function (the Riemann ζ function) has its (\non-trivial") roots on the line 1 13 f 2 + it j t 2 Rg. This has been checked for the 10 smallest roots, and they indeed all lie on this line. But how do people actually check this? It turns out that the way they do so is by numerically integrating ζ0/ζ over a big rectangle near this critical line, which tells them how many roots are in this rectangle by the Argument Principle. Using the intermediate value theorem on a related function defined on the critical line, they can also count how many roots are actually on the line, and then by comparing this number, they can confirm that all the roots are on the line. 1.3 Rouch´e'sTheorem Rouch´e'sTheorem is another very useful tool, which tells us that when we perturb a function a little bit, then the number of roots in a region is unchanged. Theorem (Rouch´e). Suppose that f; g are analytic functions, C is a simple closed curve, and for every point z on the curve C, we have that jf(z)j > jg(z)j Then f and f + g have the same number of roots inside C. Proof. We're going to introduce a \time" parameter t 2 [0; 1], and define ft(z) = f(z) + tg(z) In other words, at time 0, we have that f0 is just f, whereas at time 1, f1 = f + g. So we can imagine ft as being a sort of evolution over time of our function as it goes from f to f + g. Since jfj > jgj on C, we necessarily have that ft has no roots on the curve C.
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