Integration in the Complex Plane (Zill & Wright Chapter 18)

Transcription

1 Integration in the omplex Plane Zill & Wright hapter 18) : omplex Variables Fall 11 ontents 1 ontour Integrals 1.1 Definition and Properties Evaluation The ML Limit irculation and Flux The auchy-goursat Theorem 5.1 Integral Around a losed Loop Independence of Path for Analytic Functions Deformation of losed ontours The Antiderivative auchy s Integral Formulas auchy s Integral Formula for Derivatives onsequences of auchy s Integral Formulas auchy s Inequality Liouville s Theorem opyright 11, John T. Whelan, and all that 1

2 Tuesday 7 September 11 Guest lecture by Dr. Tamas Wiandt. Here are some additional notes by Dr. Whelan 1 ontour Integrals 1.1 Definition and Properties ecall the definition of the definite integral xf x I fx) dx = lim x k fx k ) x k 1.1) We d like to define a similar concept, integrating a function from some point z I to another point z F. The problem is that, since z I and z F are points in the complex plane, there are different ways to get between them, and adding up the value of the function along one path will not give the same result as doing it along another path, even if they have the same endpoints. So instead we need to define a contour integral dz = fz k ) z k 1.) lim z k which in general depends on the path by which we choose to go from the initial point z I to the final point z F. To define such an integral, we parametrize the curve, i.e., write it as zt) where zt I ) = z I and zt F ) = t F. For example, if is the semi-circle in the upper half-plane, from z I = 1 to z F = 1, we can use the angular polar coördinate as the parameter, and then zt) = e it where t I = and t F = π.) Given a parametrization of the curve, we define the contour integral as dz = tf t I k k f zt) ) dz tf dt dt = f zt) ) z t) dt 1.3) t I You might worry that we could get a different result by choosing a different parametrization of the curve e.g., suppose we d chosen zt) = e it /pi in the example above), but you can show using the chain rule that the value of the integral doesn t change. To specify more concretely the value of the integral 1.3), write zt) = xt) + iyt) and fx+iy) = ux, y)+ivx, y), where xt), yt), ux, y) and vx, y) are all real functions. Then [ ][ ] dz = ux, y) + ivx, y) dx + idy = = tf t I tf t I + i [ u xt), yt) ) + iv xt), yt) )][ x t) + iy t) ] dt [ u xt), yt) ) x t) v xt), yt) ) y t) ] dt tf t I [ u xt), yt) ) y t) + v xt), yt) ) x t) ] dt 1.4)

3 Note that Zill and Wright s equation ), on page 797, which says sort of the same thing, really should be written with some parentheses, to indicate that e.g., u dx and v dy are both part of the first integral. I.e., it should read ) ) dz = u dx v dy + i v dx + u dy 1.5) I would probably take points off on a test if it were written without the parentheses! Because the contour integral is defined as the limit of a sum, it has the usual linearity properties of an integral, i.e., if α and β are complex constants, [ ] α + β gz) dz = α dz + β gz) dz 1.6) There are also contour equivalents of the properties of concatenation and reversal; just as and c a fx) dx = a b b a fx) dx = fx) dx + b a c b fx) dx, fx) dx 1.7a) 1.7b) if 1 + is the curve formed by chaining together 1 and which we can do if the final point of 1 is the initial point of ), then dz = dz + dz 1.8) and if is the curve we get by following backwards so the initial point of is the final point of and vice versa), then dz = dz 1.9) 1. Evaluation [A few examples of evaluating contour integrals.] 1.3 The ML Limit It is occasionally useful to place an upper limit on a contour integral, rather than evaluating it. A limit that can be placed on any contour integral of a continuous function along a smooth curve is the so-called ML limit: dz ML 1.1) where M is the maximum value of : and L is the length of. M for all z on 1.11) 3

4 1.4 irculation and Flux The contour integral also appears in vector calculus, where can evaluate the integral of a scalar field Ψ r) along some path rt) as tf Ψ r) ds = Ψ rt)) d r dt dt, 1.1) t I where In two dimensions, i.e., Given a vector field d r d r dt = dt d r dt. 1.13) ds = dx) + dy) 1.14) dx ) d r dt = + dt ) dy 1.15) dt F r) = F x, y) = F x x, y) ˆx + F y x, y) ŷ, 1.16) and a smooth closed curve, we re often interested in two contour integrals, known as the flux of F through and the circulation of F around. At each point on the closed curve we define a unit vector ˆt pointing along the curve in the counter-clockwise direction and another unit vector ˆn pointing perpendicular to the curve in the outward direction. Then the integrals of interest are irculation of F around = F ˆt ds 1.17a) Flux of F through = F ˆn ds 1.17b) where the arrow on the integral sign emphasizes that is a closed curve traversed counterclockwise. A little bit of geometry shows ˆt ds = ˆx dx + ŷ dy ˆn ds = ˆx dy ŷ dx 1.18a) 1.18b) so irculation = F ˆt ds = [F x dx + F y dy] 1.19a) Flux = F ˆn ds = [F x dy F y dx] 1.19b) 4

5 Now, consider our friend the Pólya vector field associated with a function = fx+iy) = ux, y) + ivx, y), defined H = ux, y) ˆx vx, y)ŷ 1.) with the minus sign on the y component. Its circulation around and flux through some closed curve are irculation of H around = [H x dx + H y dy] = [u dx v dy] 1.1a) Flux of H through = [H x dy H y dx] = [u dy + v dx] 1.1b) But if we compare this to 1.4) or 1.5), we see that these are just the real and imaginary parts of the contour integral dz! I.e., ) irculation of H = uˆx vŷ around = e dz 1.a) ) Flux of H = uˆx vŷ through = Im dz 1.b) This once again shows the convenience of working with the Pólya vector field H = uˆx vŷ; compare these statements to equations 8) and 9) in Zill and Wright, which have to make statements about the contour integral of the complex conjugate of : ) irculation of uˆx + vŷ around = e dz 1.3a) ) Flux of uˆx + vŷ through = Im dz 1.3b) Practice Problems , , , , , , , , Thursday 9 September 11 The auchy-goursat Theorem On Tuesday, we learned how to integrate a function along a contour in the complex plane. If that function is analytic, as we will see, such integrals are much easier to evaluate. This is because of a key result called the auchy-goursat theorem and its consequences. The two main results will be: For a closed curve, if a function is analytic on and everywhere in the region bounded by, the integral dz of that function around that closed curve is zero. 5

6 If two curves 1 and connect the same endpoints, and you can deform 1 into without leaving the region in which is analytic, the contour integrals 1 dz and dz have the same value. Let s see why these results are true, and how they can help us evaluate contour integrals..1 Integral Around a losed Loop First consider the integral around a closed curve. emember that we noted at the end of Tuesday s class that if = fx + iy) = ux, y) + ivx, y), the contour integral of around a closed curve is related to the flux and circulation of the Pólya vector field H = uˆx vŷ around that curve: irculation of ) H Flux of ) H dz = H ˆt ds + i H ˆn ds = + i.1) around through There is a result from vector calculus known as Green s theorem, which has two equivalent statements, one relating the circulation of a vector field to its curl and the other relating the flux to the divergence. In both cases, let the closed curve be the boundary of a region. Then Green s theorem in its two forms says F x dx + F y dy) = F ˆt ds = F x dy F y dx) = F ˆn ds = curl F ) d A = div F ) d A = Fy x F x y Fx x + F y y ) dx dy ) dx dy Applying this to the Pólya vector field in.1) gives us dz = [u dx v dy] + i [u dy + v dx] = H ˆt ds + i H ˆn ds = curl H ) d A + i div H ) d A = v x u ) dx dy + i y u x v ) dx dy y.a).b).3) But recall that the divergence and curl of the Pólya vector field, which appear in the integrands of the integrals over, vanish when the auchy-iemann equations are satisfied. So if is analytic everywhere in, then the divergence and curl of H = uˆx vŷ are zero everywhere in, which means the circulation of H around the boundary and the flux of H through that boundary are zero, which means the integral of around is zero. For an illustration of the relationship between flux and circulation and the integral of a function around a closed curve, see 6

7 As specific example, consider = z; this is analytic everywhere. Let be the circle of radius a centered at the origin; one parametrization is which has z = ae it t : π.4) We expect the integral around to be zero. Let s check: π π dz = z dz = a e it iae it )dt = ia e it dt = a dz = iae it dt.5) eit π = a e i4π e i) =.6) As expected, the integral is zero. Now, consider the function = 1. It is analytic everywhere except the origin z =. z Since the region bounded by, the unit disk, includes the origin, the auchy-goursat theorem doesn t apply, and the integral need not be zero. Let s check: dz π dz = z = iae it π dt = ia dt = πi.7) a eit Notice two things: Even though is analytic not just differentiable) all along the curve, the auchy- Goursat theorem does not apply because is not analytic at all points in the region bounded by. And in fact dz The integral dz doesn t depend on the radius of the curve.. Independence of Path for Analytic Functions Now to the second big consequence of the auchy-goursat theorem, concerning the deformation of contours. Suppose 1 and are two different curves that connect the same endpoints z I and z F. Then if we make the closed curve by following 1 from z I and z F, then following backwards from z F back to z I, we have dz = dz 1 dz.8) where the contribution from integrating backwards from z F to z I is minus the integral dz which results from integrating forwards along from z I to z F. This means that if is analytic everywhere in the region between 1 and, which is the region of which is the boundary, then the integral around the closed loop is zero by the auchy-goursat theorem, and = dz 1 dz.9) 7

8 i.e. dz = dz.1) 1 This is a very handy result. It means that for analytic functions we can deform a potentially inconvenient contour into one along which it s easier to calculate the integral. Example: Evaluate cos z dz.11) where is the curve y = sin x from z I = to z F = π + i. Solution: Since cos z is analytic everywhere, we can evaluate the integral along any curve from z I = to z F = π + i. In particular, consider the curve which goes along the real axis from to π, and then straight up in the imaginary direction from π to π + i. The first piece is parametrized by x from to π with zx) = x and dz = dx; the second piece is parametrized by y from to 1 with z = π + iy and dz = i dy, so, recalling that cosx + iy) = cos x cosh y i sin x sinh y cos z dz = cos z dz = π/ cos x cosh i sin x sinh ) dx + = π/ cos x dx cos π cosh y i sin π sinh y ) i dy) sinh y dy = sin x x=π/ x= + cosh y y=1 y= = 1 ) + [cosh 1] 1) = cosh 1 = e 4 + e 1.3 Deformation of losed ontours ) We saw that the integral dz was not necessarily zero, even if was analytic all along, if there were one or more points in the region bounded by where was not analytic. We ll now show that in a case like that, you can still deform without changing the value of the integral, i.e., dz = dz.13) 1 as long as you can smoothly deform 1 into while staying inside the region where is analytic. The idea is to construct a closed contour which goes like this: 1. ounter-clockwise around 1 from some point P 1 back to that same point. From P 1 to a point P on, along some curve A 3. lockwise i.e., backwards) around from P back to P 4. From P back to P 1, following A backwards we can call this the curve A) 8

9 The integral around this curve is the sum of the contributions from each piece: dz = dz + dz + dz + dz 1 A A = dz + dz dz dz 1 A A.14) Since the two integrals along A in the opposite direction cancel out, we re left with dz = dz dz.15) 1 If is analytic in the region between 1 and, which is just the region inside the closed contour, then dz is zero, and dz = dz.16) 1 So this idea of path independence works for closed curves as well. The overall theme is that you can deform the contour around which a contour integral is evaluated, without changing the result of the integral, as long as you don t change the endpoints of the curve, or the overall direction for a closed curve, and you only move the curve through regions where the function is analytic..4 The Antiderivative In real single-variable calculus we can think of the derivative in two ways. The indefinite integral is the operation that undoes the derivative: F x) = f x) F x) = fx) dx.17) while the definite integral is the area under the function between two points x 1 and x : x x 1 fx) dx = F x ) F x 1 ).18) To evaluate the definite integral of fx), we only need to know the antiderivative F x) evaluated at the two endpoints. For general complex functions, we only really have the equivalent of the definite integral. The contour integral dz.19) doesn t just depend on the endpoints of the curve, but it is a property of the curve itself. However, if a function is analytic in some domain simply connected open set) D, then we 9

10 know that the contour integral of that function along any curve in D depends only on the endpoints of the curve. It is thus reasonable to write zf z I dz = dz any curve in D from z I to z F ; analytic in D.) This then allows us to talk about an antiderivative F z) such that = F z) everywhere in D, and write zf z I dz = F z I ) F z F ).1) For example, let = cos z; then F z) = sin z, and π i π 4 cos z dz = sinπ i) sinπ/).) Since sinx + iy) = sin x cosh y + i cos x sinh y.3) we know sinπ i) = icos π) sinh ) = i 1) sinh ) = i sinh.4) and of course sinπ/) = 1 so π i π 4 cos z dz = i sinh 1 = 1 + ie e i.5) Practice Problems 18..9, , , 18..1, , , , , Tuesday 4 October 11 eview for exam Thursday 6 October 11 First Prelim Exam Tuesday 11 October 11 3 auchy s Integral Formulas We can use the auchy-goursat theorem, and the ability to deform closed contours, to deduce some remarkable properties of analytic functions, which will form the basis for all 1

11 of the considerations of complex power series to come in the following weeks. In particular, an analytic function will be seen to be infinitely differentiable, and the function and each of its derivatives at a point will determine the value of various integrals along contours around that point. To start with, consider the contour integral dz 3.1) z z where is any curve which goes counter-clockwise around z, and is a function which is analytic in some domain D which completely contains the curve. Now, the function z z is not necessarily analytic at z, but by the quotient rule, it s differentiable everywhere else that is. That means that it s analytic on a region which consists of D with the point z removed. So we don t change the value of the integral if we deform into a circle ρ z ) of radius ρ centered on z, as long as ρ is small enough that ρ z ) lies completely within D: dz = dz 3.) z z z z Now, we can make ρ as small as we like as long as it remains positive, so to as good an approximation as we like, we can replace the in the integrand with the constant fz ): fz ) fz ) = dz = dz 3.3) z z z z z z z z ρz ) ρz ) and then, since fz ) is a constant, we can take it outside the integral, and get fz ) dz = fz ) 3.4) z z z z ρz ) We can evaluate the contour integral using the parametrization z = z + ρe it, dz = iρe it dt and find dz π iρe it dt = = πi 3.5) ρz ) z z ρe it which means that dz = z z ρz ) ρz ) ρz ) fz ) z z dz = πifz ) 3.6) Strictly speaking, to show this we need to show that fz ) dz = 3.7) z z ρz ) which is done formally in Zill and Wright by showing that fz ) dz z z ρz ) )

12 is smaller than any positive number you choose, using the ML limit. Example: integrate e z dz 3.9) z + i where is a counter-clockwise circle of radius centered on the origin. 3.1 auchy s Integral Formula for Derivatives The formula fz ) = 1 πi dz 3.1) z z is the n = case of a general formula for the nth derivative of f n) z) evaluted at z = z : f n) z ) = n! dz valid for any non-negative integer n 3.11) πi z z ) n+1 Note that if we knew could be described by a Taylor series f k) z ) = z z ) k 3.1) k! then it would be trivial to show n! n! dz = πi z z ) n+1 πi k= k= f k) z ) k! because it s easy to show by deformation of contours that { z z ) k n 1 k n dz = πi k = n z z ) k n 1 dz = f k) z ) 3.13) 3.14) However, this would be completely cheating, because the formula 3.1) will be used to show that an analytic function is infinitely differentiable, and also that is equal to its Taylor series. So instead, to avoid circular logic, there s a proof by induction, where the integral formula for f z ) can be derived using the formula for fz ), then the formula for f z ) can be derived using the formula for f z ), etc. The fundamentals of the method go like this: if we ve already shown that f n) z ) = n! dz 3.15) πi z z ) n+1 for some n, we can write f n+1) f n) z + z) f n) z ) z ) = lim z z 1 n! = lim z = lim z z n! πi n! dz πi z z z) n+1 πi z z ) z z ) z) n+1) z z ) n+1) dz z 1 ) dz n )

13 where is some curve that lies within the domain where is analytic, and goes counterclockwise around both z and z + z. We can show, for example using l Hôpital s rule, that z z z) n+1) z z ) n+1) lim = n + 1)z z ) n+) 3.17) z z and so f n+1) z ) = n!n + 1) πi n + 1)! dz = z z ) n+ πi dz 3.18) z z ) n+ which is indeed the auchy integral formula for f n+1) z ). Example: Zill & Wright problem As noted a moment ago, auchy s integral formula for derivatives shows that if a function is analytic at a point, we can construct any derivative we like by defining a contour integral on a contour which winds counterclockwise around that point. This means that if a function is analytic which required that it be differentiable in some neighborhood containing that point, and that the relevant first derivatives be continous in that neighborhood), it s actually infinitely differentiable. This is definitely not the case for real functions. onsider for example the function { x x fx) = 3.19) 1 cos x x For this function fx) and f x) are continuous at the origin, as are f x) and f 3) x), but f 4) x) is discontinuous at the origin, and f 4) ) is not defined. 3. onsequences of auchy s Integral Formulas 3..1 auchy s Inequality If an analytic function is bounded along a circle ρ z ) of radius ρ of radius ρ centered on a point z M when z z = ρ 3.) we can use auchy s Integral Formulas together with the ML bound to set a bound on the derivatives of at any point z inside. Specifically, f n z ) = = n! dz πi z z ) n+1 n! M n!m πρ) = 3.1) π ρn+1 ρ n 3.. Liouville s Theorem ρz ) auchy s inequality seems sort of academic, but it can be used to demonstrate something more surprising: any function which is analytic everywhere which we call an entire function), 13

14 and also bounded everywhere, is a constant. We can do this by using auchy s inequality with n = 1 to show the derivative at a point z obeys the inequality f z ) max ρz ) ρ max ρ 3.) but if the function is analytic everywhere, we can make ρ arbitrarily large, and the right-hand side arbitrarily small. That means f z ) must be zero, which means f z ) must be zero. But we can do this at any point, so f z) = everywhere and therefore is a constant. Note that no such restriction exists for real functions: fx) = 1 is infinitely differentiable for every real x, and obeys fx) 1 for every real 1+x x. Practice Problems , , , , , , , ,