Q You mentioned that in complex analysis we study analytic functions, or, in another name, holomorphic functions. Pray tell me, what are they?

Transcription

1 COMPLEX ANALYSIS PART 2: ANALYTIC FUNCTIONS Q You mentioned that in complex analysis we study analytic functions, or, in another name, holomorphic functions. Pray tell me, what are they? A There are many ways to define analytic functions. To give you a quick idea, we can describe analytic functions as a generalization of something familiar: polynomials. Q How can that be? A Well, could you give me a general expression for polynomials? Q Sure: p(z) = c 0 + c 1 z + + c N z N, where c n are constants, allowed to be complex. A Fine. The way you put p(z) here can be regarded as a finite power series expansion at z = 0. An expansion of p(z) at an arbitrary point z = z 0 is something like... Q I know: p(z) = a 0 + a 1 (z z 0 ) + + a N (z z 0 ) N, or using the summation convention. p(z) = N n=0 a n(z z 0 ) n, A A natural generalization of this is a power series expansion n=0 a n(z z 0 ) n. We say that a complex function f(z) is analytic at a point z = z 0 if it has a power series expansion at z = z 0, say f(z) = a n(z z 0 ) n (1) n=0 which is valid for z in a neighborhood of z 0. Q I want to know the precise meaning of what you have said about (1). A An easy way to make this precise is by saying this: there exists some r > 0 and some sequence {ε N } of positive numbers decreasing to zero such that f(z) N a n(z z 0 ) n n=0 < ε N for all N and for all z satisfying z z 0 r; in other words, the series in (1) converges uniformly to f on the closed disk D(z 0 ; r) {z C z z 0 r }. 1

2 Q So, locally, f can be uniformly approximated by polynomials at z 0. A Yes. Actually this property turns out to be equivalent to f being analytic at z 0. Q An analytic function, I presume, is a complex function analytic at each point of its domain, which is usually assumed to be an open set. A That s correct. Q So we need to spend some time on tests of convergence of power series, which we have learned in advanced calculus. A Not so. We rarely need these tests. Q Why? A The convergence of a series is often guaranteed by the theory and hence there is no need to worry about them. This shows the strength of complex analysis. Q I am glad to hear that. Testing the convergence of a series is often very annoying. A I agree, an unpleasant job. Happily we can avoid it in complex analysis. Now we consider another way to describe analytic functions. Take a complex-valued function defined on an open region in the complex plane C, say f. The variable of f is designated by the usual symbol z so that we may write f(z) for f if we wish. The real and the imaginary parts of z are, as usual, denoted by x and y respectively: z = x+iy. Since f(z) = f(x+iy), we may regard f as a function of two real variables x and y. Thus the partial derivatives x f can be expressed in terms of them: and make sense. The differential df of df = dx + dy. (2) x However, often we do not use this identity in actual computation. For example, if f(z) = z 2, then df = 2z dz. Another example: if f(z) = z 2, then f(z) = z z hence df = z d z + z dz. Q Wait a minute! Writing z = x + iy df = 2x dx + 2y dy. Are you sure your answer is the same? and as usual, we have z 2 = x 2 + y 2, which gives A Yes. Let us check. From z = x + iy we have z = x iy. Take differential: dz = dx + idy, d z = dx idy (3) 2

3 Thus z d z + z dz = (x + iy)(dx idy) + (x iy)(dx + idy) =.... Q Now I can tell this is the same answer after simplification. A In (1), the differential df of f is expressed in terms of dx and dy. Hence, by using (3), df can also be expressed in terms of dz and d z. For a complex function f, expressing df in terms of dz and d z turns out to be more desirable. Q Let me follow your suggestion and have a try: (3) gives dx = (dz + d z)/2 and dy = (dz d z)/2i. So df = dx + dy = x x 1 ( (dz + d z) + 2 i ) (dz d z) 2 = 1 ( 2 x i ) dz + 1 ( 2 x + i ) dz. This does not look very pleasing. A It is fine. When you get used to it, it will appeal to you. Now we define / z and / z in such a way that the identity (4) df = dz + d z (5) z z holds. Comparing (5) with the previous identity (4), naturally we define: z = 1 ( 2 x i ), z = 1 ( 2 x + i ). (6) The second identity gives the definition of the - operator / z. Q How do you pronounce? A Dee bar. / z is called the dee bar operator. Q Is it important? A Yes, very important! It is the simplest first order elliptic differential operator. Q Then why some textbooks in complex analysis do not mention it? A Well, because many textbooks in math are not written by good mathematicians. Consult books in complex analysis written by good mathematicians, such as Henri Cartan, Walter Rudin, Raghavan Narashimhan, John Conway, Lars Hörmander, etc. You ll see it in any of these books. 3

4 Q Then why don t we pick a book written by a good mathematician as our textbook? A This is not a bad idea for a serious student. But the sophisticated style of a good mathematician is too demanding for our majority of students today. An instructor does not want to take a risk to choose good but demanding books and to end up with bad evaluation. O.K. We should get back to our mathematical discussion. Q Where are we now? A We have arrived at (5) and (6). From (5) we see that, if holds, then the complex derivative df/dz exists, with z = 0 (7) df dz = z. (8) (7) is called the Cauchy-Riemann equation, or CR-equation for short. Q This does not look like the Cauchy-Riemann equations to me. A I know what you mean. What you have in mind is the real version of my CR-equation here. Let u be the real part of f and v be the imaginary part so that Now ( u x + i = x + i v ) x ( u = x v ) ( u + i + v ) x 2 z = ( u + i + i v ) f = u + iv. Setting the real and the imaginary parts of the right hand side to zero, we ll get your real version of Cauchy-Riemann equations. Q What happens if the CR-equation (7) fails? A Then the complex derivative df/dz does not exist. Let us rewrite (5) formally as df dz = z + d z z dz (9) The symbol d z/dz does not make sense. For example, at z = 0 the complex derivative d z/dz by definition is the limit lim z 0 z/z, which does not exist. So, only when (7) holds, we can get rid of this expression from (9). 4

5 Q The derivative df/dz is defined by taking certain limit of a quotient. But throughout our discussion you did not use limits. A Yes, this is because our discussion is informal. The symbols dz, df, dx etc. are formal expressions. (However it is possible to define them rigorously.) Q I am confused. You use both words informal and formal. A I admit, it is rather confusing. But in math we introduce formal expressions often for the purpose of starting an informal discussion. To make the above discussion precise, we should replace dz, df, dx etc. by increments z, f, x etc. so that dz = dx + i dy becomes z = x + i y and instead of df we use f = f(z + z) f(z). Identity (5) is converted into something with an error term: f = z + z + o( z), z z where o( z) stands for an expression r(z, z) having the property that r(z, z)/ z tends to zero as z 0. By definition, df/dz is lim z 0 f/ z, if the limit exists. The above discussion can be put into a rigorous framework under the assumption that f(z) = f(x + iy) as a function of two real variables x and y is differentiable. Q OK. I think we shouldn t get into technical details. Let us go back to the original question about analytic functions. You gave two descriptions, one via power series and the other using complex differentiation, or equivalently, CR-equations. How can we relate these two descriptions? How do we know that they are equivalent? A Good question. To answer this, we need to take a major step: use integration. Q What makes you take this step? Why don t we continue with differentiation? A Because properties of functions often deteriorate after differentiation, and get better after integration. Another reason is that integration goes along well with taking limits. For example, suppose that you have a sequence of functions f n (of a real variable) b converging uniformly to f. Then lim n a f n(x) dx = b f(x) dx is valid, but a lim n df n /dx = df/dx in general is false. Q Ah. Now I understand why in a differential equations class we convert the initial value problem dy/dx = f(x, y) with y(t 0 ) = y 0 into the corresponding integral equation y(t) = y 0 + t t 0 f(t, y(t)) dt and then apply Picard s method. 5

6 A Precisely. The idea behind that procedure is similar to what we are going to do. Q Now, how do you integrate complex functions? A Integration f(z) dz can be carried out in the same way as the first year calculus, if you know enough formulas for differentiation. For example: z 2 dz = z 3 /3 + C, e z dz = e z + C, Log z dz = 1 z + C, etc. Q But what about definite integrals? A These are replaced by so called contour integrals, which are of the form f(z) dz. C Q How do you compute C f(z) dz? A The procedure is quite straightforward. When the path C is described by the parametric equation z = z(t) (a t b), f(z) becomes f(z(t)) and dz becomes dz dt dt, or z (t) dt. Thus we have C f(z) dz = b a f(z(t))z (t) dt. For example, let us compute C dz/(z z 0), where C is the path going round a circle centered at z 0 once in the counter clockwise direction. Here f(z) = 1/(z z 0 ) and C is described by z(t) = z 0 + r cos t + i r sin t z 0 + re it (0 t 2π), where r > 0 is its radius. So dz becomes (dz/dt)dt = ire it dt. Hence C dz 2π = z z 0 0 As I have said, this is straightforward. ire it dt 2π re it = idt = 2πi. 0 Q How can we use contour integrals to explore analytic functions? A They are mainly used in the so-called Cauchy s formula. analytic functions follow more or less from this formula. The major properties of Q Can you tell me what this formula is and how to derive it? A The main step towards this formula is to establish Cauchy s theorem, which says that if f is an analytic function and D is a bounded region inside the domain of f with a nice boundary (say, is piecewise smooth), then the integral of f along the closed contour vanishes: f(z) dz = 0. I am going to explain why 6

7 this theorem is true, and leave the statement and the derivation of Cauchy s formula from this theorem for the next conversation. Here I use Green s formula to prove Cauchy s theorem. Do you remember Green s formula? Q Is it something like P dx + Q dy = D ( Q x P ) dxdy, where P and Q are functions of x and y? A Yes. OK. Let us begin. Since Green s formula involves real variables x and y, we have to put dz = d(x + iy) = dx + idy and f = u + iv in f(z) dz to get f(z) dz = (u + iv)(dx + idy) = (u dx v dy) + i(v dx + u dy). Letting P = u and Q = v in Green s formula, we get ( u dx v dy = v x u ) ( u dxdy = + v ) dxdy = 0 x D in view of the 2nd real An important way towards the solution of a Pfaffian equation is to find its group of symmetries. A 1-parameter group {Φ t } is called a symmetry group for a Pfaffian equation α = 0 if Φ t α α for all t. An important way towards the solution of a Pfaffian equation is to find its group of symmetries. A 1-parameter group {Φ t } is called a symmetry group for a Pfaffian equation α = 0 if Φ t α α for all t. CR-equation. Similarly, let P = v and Q = u to get ( u v dx + u dy = x v ) dxdy = 0. Hence we conclude f(z) dz = 0. As you can tell, CR-equations are the essential ingredient in this proof. Without them, nothing can be done. In fact, if the conclusion of Cauchy s theorem holds for any bounded region in the domain of f, then f must satisfy the CR-equations. Q Is this called Morera s theorem? A Yes, Morera s theorem is the converse of Cauchy s theorem. Good. Can you recall roughly what we have discussed today and what is tomorrow s plan? 7

8 Q Yes. We have two ways to describe analytic functions: one uses power series and the other uses CR-equations. We also have Cauchy s theorem: f(z) dz = 0 under some appropriate conditions. Tomorrow we ll use this theorem to establish Cauchy s formula, which is used to derive basic properties of analytic functions. A Good. Let us stop here for today. 8