3 Contour integrals and Cauchy s Theorem

Transcription

1 3 ontour integrals and auchy s Theorem 3. Line integrals of complex functions Our goal here will be to discuss integration of complex functions = u + iv, with particular regard to analytic functions. Of course, one way to think of integration is as antidifferentiation. But we also have the definite integral. For a function f(x) of a real variable x, we have the integral b f(x) dx. For vector fields F = (P, Q) in the plane we have the line a integral P dx + Q dy, where is an oriented curve. Let us begin by recalling the basics of line integrals in the plane:. The vector field F = (P, Q) is a gradient vector field g, which we can write in terms of -forms as P dx + Q dy = dg, if and only if P dx+q dy only depends on the endpoints of, equivalently if and only if P dx+q dy = for every closed curve. If P dx+q dy = dg, and has endpoints z and z, then we have the formula P dx + Q dy = dg = g(z ) g(z ). 2. If D is a plane region with oriented boundary D =, then ( Q P dx + Q dy = x P ) dxdy. y 3. If D is a simply connected plane region, then F = (P, Q) is a gradient vector field g if and only if F satisfies the mixed partials condition Q x = P y. (Recall that a region D is simply connected if every simple closed curve in D is the boundary of a region contained in D. Thus a disk {z : z < } is simply connected, whereas a ring such as {z : < z < 2} is not.) Suppose that P and Q are complex-valued. Then all of the above still makes sense, and in particular Green s theorem is still true. We will be interested in the following integrals. Let dz = dx + idy, a complex -form, and let = u + iv. Then we can define dz for any reasonable closed oriented curve. If is a parametrized curve given D

2 by r(t), a t b, then we can view r (t) as a complex-valued curve, and then b dz = f(r(t)) r (t) dt, a where the indicated multiplication is multiplication of complex numbers (and not the dot product). Another notation which is frequently used is the following. We denote a parametrized curve in the complex plane by z(t), a t b, and its derivative by z (t). Then b dz = f(z(t))z (t) dt. a For example, let be the curve parametrized by r(t) = t + 2t 2 i, t, and let = z 2. Then z 2 dz = (t + 2t 2 i) 2 ( + 4ti) dt = (t 2 4t 4 + 4t 3 i)( + 4ti) dt = [(t 2 4t 4 6t 4 ) + i(4t 3 + 4t 3 6t 5 )] dt = t 3 /3 4t 5 + i(2t 4 8t 6 /3)] = /3 + ( 2/3)i. For another example, let let be the unit circle, which can be efficiently parametrized as r(t) = e it = cos t + i sin t, t 2π, and let = z. Then r (t) = sin t + i cos t = i(cos t + i sin t) = ie it. Note that this is what we would get by the usual calculation of d dt eit. Then z dz = 2π e it ie it dt = 3.2 auchy s theorem 2π e it ie it dt = 2π i dt = 2πi. Suppose now that is a simple closed curve which is the boundary D of a region in. We want to apply Green s theorem to the integral dz. Working this out, since dz = (u + iv)(dx + idy) = (u dx v dy) + i(v dx + u dy), we see that dz = D ( v x u ) ( u da + i y D x v ) da. y 2

3 Thus, the integrand is always zero if and only if the following equations hold: v x = u y ; u x = v y. Of course, these are just the auchy-riemann equations! This gives: Theorem (auchy s integral theorem): Let be a simple closed curve which is the boundary D of a region in. Let be analytic in D. Then dz =. Actually, there is a stronger result: Theorem (auchy s integral theorem): Let D be a simply connected region in and let be a closed curve contained in D. Let be analytic in D. Then dz =. Example: let D = and let be the function z 2 + z +. Let be the unit circle. Then as before we use the parametrization of the unit circle given by r(t) = e it, t 2π, and r (t) = ie it. Thus dz = 2π (e 2it + e it + )ie it dt = i 2π (e 3it + e 2it + e it ) dt. It is easy to check directly that this integral is, for example because terms such as 2π cos 3t dt (or the same integral with cos 3t replaced by sin 3t or cos 2t, etc.) are all zero. On the other hand, again with the unit circle, 2π 2π z dz = e it ie it dt = i dt = 2πi. The difference is that /z is analytic in the region {} = {z : z }, but this region is not simply connected. (Why not?) Actually, the converse to auchy s theorem is also true: if dz = for every closed curve in a region D (simply connected or not), then is analytic in D. However, we will not discuss this. 3

4 3.3 Antiderivatives We now look at the other method for showing that a line integral depends only on the endpoints. Let = u + iv and suppose that dz = df, where we write F in terms of its real and imaginary parts as F = U + iv. This says that ( ) ( ) U U V (u dx v dy) + i(v dx + u dy) = dx + x y dy V + i dx + x y dy. Equating terms, this says that u = U x = V y v = U y = V x. In particular, we see that F satisfies the auchy-riemann equations, and its complex derivative is Thus we see: F (z) = U x + i V x = u + iv =. Theorem: The -form dz is of the form df, or equivalently the vector field (u + iv, v + iu) is a gradient vector field (U + iv ), if and only if F is also analytic, and in this case F (z) is an antiderivative for : F (z) =. onversely, if F (z) is an antiderivative for, then dz = df. The theorem tells us a little more: if has endpoints z and z, and is oriented so that z is the starting point and z the endpoint, then we have the formula dz = df = F (z ) F (z ). For example, we have seen that, if is the curve parametrized by r(t) = t+2t 2 i, t and = z 2, then z 2 dz = /3+( 2/3)i. But z 3 /3 is clearly an antiderivative for z 2, and has starting point and endpoint + 2i. Hence z 2 dz = ( + 2i) 3 /3 = ( + 6i 2 8i)/3 = ( 2i)/3, which agrees with the previous calculation. 4

5 If is analytic in a simply connected region D, then the fact that dz = P dx + Q dy satisfies Q/ x = P/ y (here P and Q are complex valued) says that (P, Q) is a gradient vector field, or equivalently that dz = dg, in other words that has an antiderivative. From this point of view, we can see why dz = 2πi, where z is the unit circle. The antiderivative of /z is log z, and so the expected answer (viewing the unit circle as starting at = e and ending at e 2πi = is log log. But log is not a single-valued function, and in fact as z = e it turns along the unit circle, the value of log changes by 2πi. So the correct answer is really log log, viewed as log e 2πi log e = 2πi = 2πi. Of course, /z is analytic except at the origin, but {z : z } is not simply connected, and so /z need not have an antiderivative. The real point, however, in the above example is something special about log z, or /z, but not the fact that /z fails to be defined at the origin. We could have looked at other negative powers of z, say z n where n is a negative integer less than, or in fact any integer. In this case, z n has an antiderivative z n+ /(n + ), and so by the fundamental theorem for line integrals z n dz = for every closed curve. To see this directly for the case n = 2 and the unit circle, 2π z 2 dz = e 2it ie it dt = i 2π e it dt =. This calculation can be done somewhat more efficiently as follows. Let r(t) = e αt, where α is a nonzero complex number. Then an antiderivative for the complex curve r(t) is checked to be s(t) = e αt dt = α eαt. Hence, b e αt dt = ( e αb e αa). a α In general, we have seen that z n dz = for every integer n, where is a closed curve. To verify this for the case of the unit circle, we have 2π 2π z n dz = e nit ie it dt = i e (n+)it dt = i n + ( e 2(n+)πi e ) = 5 i ( ) =. n +

6 Finally, returning to /z, a calculation shows that ( x dx z dz = x 2 + y 2 + y dy ) ( y dx x 2 + y 2 + i x 2 + y 2 + x dy ) x 2 + y 2. The real part is the gradient of the function 2 ln(x2 +y 2 ). But the imaginary part corresponds to the vector field ( ) y F = x 2 + y 2, x x 2 + y 2, which as we have seen many times this semester is a vector field F for which Green s theorem fails, because F is undefined at the origin. In the next section, we will see how to systematically use the fact that the integral of /z dz around a closed curve enclosing the origin to get a formula for the value of an analytic function in terms of an integral. 3.4 auchy s integral formula Let be a simple closed curve in. Then = R for some region R. If z is a point which does not lie on, we say that encloses z if z R, and that does not enclose z if z / R. For example, if is the unit circle, then is the boundary of the unit disk B = {z : z < }. Thus encloses a point z if z lies inside the unit disk ( z < ), and does not enclose z if z lies outside the unit disk ( z > ). Theorem (auchy s integral formula): Let D be a simply connected region in and let be a closed curve contained in D. Let be analytic in D. Suppose that z is a point enclosed by. Then f(z ) = 2πi z z dz. Before we discuss the proof, let us look at the special case where is the constant function, is the unit circle, and z =. The theorem says in this case that = f() = 2πi z dz, as we have seen. In fact, the theorem is true for a circle of any radius: if r is a circle of radius r centered at, then r can be parametrized by re it, t 2π. Then r z dz = 2π 2π re it ireit dt = i dt = 2πi, 6

7 independent of r. The fact that r z dz is independent of r also follows from Green s theorem. The general case is obtained from this special case as follows. Let = R, with R D since D is simply connected. We know that encloses z, which says that z R. Let r be a circle of radius r with center z. If r is small enough, r will be contained in R, as will the ball B r of radius r with center z. Let R r be the region obtained by deleting B r from R. Then R r = r, where this is to be understood as saying that the boundary of R r has two pieces: one is with the usual orientation coming from the fact that is the boundary of R, and the other is r with the clockwise orientation, which we record by putting a minus sign in front of r. Now z does not lie in R r, so we can apply Green s theorem to the function /(z z ) which is analytic in D except at z and hence in R r : R r z z dz =. But we have seen that R r = r, so this says that dz dz =, z z z z or in other words that r dz = z z r z z dz. Now suppose that r is small, so that is approximately equal to f(z ) on r. Then the second integral dz is approximately equal to z z r f(z ) dz = f(z ) r z z r z z dz, where r is a circle of radius r centered at z. Thus we can parametrize r by z + re it, t 2π, and as before. Thus r z z dz = 2π 2π re it ireit dt = i dt = 2πi, f(z ) dz = 2πif(z ), r z z 7

8 and so dz = dz is approximately equal to 2πif(z ). In z z r z z fact, this becomes an equality in the limit as r. But dz is z z independent of r, and so in fact dz = 2πif(z ). z z Dividing through by 2πi gives auchy s formula. The main application of auchy s theorem is to think of the point z as a variable point inside of the region R such that = R; note that the z in the formula is a dummy variable. Thus we could equally well write: = 2πi f(w) w z dw, for all z enclosed by. This description of the analytic function by an integral depending only on its values on the boundary curve of R turns out to have many very surprising consequences. For example, it turns out that an analytic function actually has derivatives of all orders, not just first derivatives, which is very unlike the situation for functions of a reaal variable. In fact, every analytic function can be expressed as a power series. 3.5 Homework. Let = x 2 + iy 2. Evaluate dz, where (a) is the straight line joining to 2 + i; (b) is the curve ( + t) + t 2 i, t. Are the results the same? Why or why not might you expect this? 2. Let α = c + di be a complex number. Verify directly that d dt eαt = αe αt. 3. Let be a circle centered at 4+i of radius. Without any calculation, explain why dz =. z 4. Let be the curve defined parametrically as follows: z(t) = t( t)e t + [cos(2πt 3 )]i, t. Evaluate the integral e z2 dz. Be sure to explain your reasoning! 8

9 5. Let D be a simply connected region in and let be a closed curve contained in D. Let be analytic in D. Suppose that z is a point which is not enclosed by. What is dz? 2πi z z e z 6. Use auchy s formula to evaluate dz, where is a circle of z + radius 4 centered at the origin (and oriented counterclockwise). 7. Let D be a simply connected region in and let be a closed curve contained in D. Let be analytic in D. Suppose that z is a point enclosed by. (a) By the usual formulas, show that ( ) d dz z z = f (z) z z (z z ) 2. (b) By using the fact that the line integral of a complex function with an antiderivative is zero and the above, conclude that f (z) dz = z z (z z ) 2 dz. (c) Now apply auchy s formula to conclude that f (z ) = 2πi (z z ) 2 dz. 9