3 Contour integrals and Cauchy s Theorem

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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 there is also the definite integral. For a function f(x) of a real variable x, we have the integral b a f(x) dx. In case f(x) = u(x) + iv(x) is a complex-valued function of a real variable x, the definite integral is the complex number obtained by integrating the real and imaginary parts of f(x) separately, i.e. b b b a f(x) dx = u(x) dx + i v(x) dx. For vector fields F = (P, Q) in the plane we have a a the line integral P dx+q dy, where is an oriented curve. In case P and Q are complex-valued, in which case we call P dx + Q dy a complex -form, we again define the line integral by integrating the real and imaginary parts separately. Next we recall 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 < } D

2 is simply connected, whereas a ring such as {z : < z < 2} is not.) In case P dx + Q dy is a complex -form, 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 (with P = and Q = i), and let = u + iv. The expression dz = (u + iv)(dx + idy) = (u + iv) dx + (iu v) dy = (udx vdy) + i(vdx + udy) is also a complex -form, of a very special type. Then we can define dz for any reasonable closed oriented curve. If is a parametrized curve given 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 = 2π 2 e it ie it dt = 2π i dt = 2πi.

3 One final point in this section: let = u + iv be any complex valued function. Then we can compute f, or equivalently df. This computation is important, among other reasons, because of the chain rule: if r(t) = (x(t), y(t)) is a parametrized curve in the plane, then d dt f(r(t)) = f r (t) = f dx x dt + f dy y dt. (Here means the dot product.) We can think of obtaining d f(r(t)) roughly dt by taking the formal definition df = f f dx + dy and dividing both sides x y by dt. Of course we expect that df should have a particularly nice form if is analytic. In fact, for a general function = u + iv, we have df = ( u x + i v x ) dx + ( u y + i v y ) dy and thus, if is analytic, ( u df = x + i v ) ( dx + v ) x x + i u dy x ( u = x + i v ) ( u dx + x x + i v ) ( u idy = x x + i v ) (dx + idy) = f (z) dz. x Hence: if is analytic, then df = f (z) dz and thus, if z(t) = (x(t), y(t)) is a parametrized curve, then d dt f(z(t)) = f (z(t))z (t) This is sometimes called the chain rule for analytic functions. For example, if α = a + bi is a complex number, then applying the chain rule to the analytic function = e z and z(t) = αt = at + (bt)i, we see that d dt eαt = αe αt. 3

4 3.2 auchy s theorem 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 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, which we shall prove in the next section: Theorem (auchy s integral theorem 2): Let D be a simply connected region in and let be a closed curve (not necessarily simple) 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 4 2π (e 3it + e 2it + e it ) dt.

5 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. We will see this later. 3.3 Antiderivatives If D is a simply connected region, is a curve contained in D, P, Q are defined in D and Q x = P, then the line integral P dx+q dy only depends y on the endpoints of. However, if P dx + Q dy = df, then P dx + Q dy only depends on the endpoints of whether or not D is simply connected. We see what this condition means in terms of complex function theory: 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 dx v dy) + i(v dx + u dy) = Equating terms, this says that ( ) U U dx + x y dy u = U x = V y v = U y = V x. ( ) V V + i dx + x y dy. In particular, we see that F satisfies the auchy-riemann equations, and its complex derivative is F (z) = U x + i V x = u + iv =. 5

6 We say that F (z) is a complex antiderivative for, i.e. F (z) =. In this case u x = 2 U x 2 = 2 V x y = v y ; u y = 2 U y 2 = 2 V x y = v x. It follows that, if has a complex antiderivative, then satisfies the auchy-riemann equations: is necessarily analytic. Thus we see: Theorem: If the -form dz is of the form df, or equivalently the vector field (u + iv, v + iu) is a gradient vector field (U + iv ), then both F (z) and are analytic, and F (z) is a complex antiderivative for : F (z) =. onversely, if F (z) is a complex antiderivative for, then F (z) and are analytic and dz = df. The theorem tells us a little more: Suppose that F (z) is a complex antiderivative for, i.e. F (z) =. 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. When does an analytic function have a complex antiderivative? From vector calculus, we know that dz = df if and only if dz only depends on the endpoints of, if and only if dz = for every closed curve. In particular, if dz = for every closed curve then is analytic (converse to auchy s theorem). 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 = df, in other words that has an antiderivative. Hence: 6

7 Theorem: Let D be a simply connected region and let be an analytic function in D. Then there exists a complex antiderivative F (z) for. Fixing a base point p D, a complex antiderivative F (z) for is given by dz, where is any curve in D joining p to z. As a consequence, we see that, if D is simply connected, is analytic in D and is a closed curve in D, then dz = (auchy s integral theorem 2), since dz = df, where F is a complex antiderivative for, and hence dz = df =, by the Fundamental Theorem for line integrals. 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, z 2 dz = 2π 2π e 2it ie it dt = i e it dt =. This calculation can be done somewhat differently as follows. Let r(t) = e αt, where α is a nonzero complex number. Then, by the chain rule for analytic functions, an antiderivative for the complex curve r(t) is checked to be s(t) = e αt dt = α eαt. 7

8 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 i dt = [e (n+)it] 2π i(n + ) = i i(n + ) ( e 2(n+)πi e ) = ( ) =. n + 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 ) = d ln r. But the imaginary part corresponds to the vector field ( ) y F = x 2 + y 2, x x 2 + y 2, which is a standard example of a vector field F for which Green s theorem fails, because F is undefined at the origin. In fact, in terms of -forms, y dx x 2 + y 2 + x dy x 2 = d arg z = dθ. + y2 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 (in other words, is simply connected). 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 > ). We always orient by viewing it as R and using the orientation coming from the statement of Green s theorem. 8

9 Theorem (auchy s integral formula): Let D be a simply connected region in and let be a simple 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. For example, if is a circle of radius 5 about, then e z2 z 2 dz = 2πie4. e z2 But if is instead the unit circle, then dz =, as follows from z 2 auchy s integral theorem. Before we discuss the proof of auchy s integral formula, 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 independent of r. The fact that 2π re it ireit dt = i dt = 2πi, 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 9

10 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 : dz =. z z R r 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 = f(z ) 2π 2π re it ireit dt = i dt = 2πi, r z z dz = 2πif(z ), 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 z z dz = 2πif(z ). Dividing through by 2πi gives auchy s formula. The main theoretical 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

11 that the z in the formula is a dummy variable. Thus we could equally well write: = f(w) 2πi 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 real variable. In fact, every analytic function can be expressed as a power series. This fact can be seen by rewriting auchy s formula above as = 2πi f(w) w( z/w) dw, and then expanding as a geometric series. The fact that every z/w analytic function is given by a convergent power series is yet another way of characterizing analytic functions. 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 D be a region in and let u(x, y) be a real-valued function on D. We seek another real-valued function v(x, y) such that = u + iv is analytic, i.e. satisfies the auchy-riemann equations. Equivalently, we want to find a function v such that v x = u y and v y = u x, ( u ). Show that which says that v is the vector field F = y, u x F satisfies the mixed partials condition exactly when u is harmonic. onclude that, if D is simply connected, then F is a gradient vector field v and hence that u is the real part of an analytic function.

12 4. Let be a circle centered at 4+i of radius. Without any calculation, explain why dz =. z 5. 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! 6. Let D be a simply connected region in and let be a simple 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 7. Use auchy s formula to evaluate dz, where is a circle of z + radius 4 centered at the origin (and oriented counterclockwise). 8. Let be the unit circle centered at in the complex plane and oriented counterclockwise. Evaluate each of the following integrals, and be sure that you can justify your answer by a calculation or a clear and concise explanation. (d) (a) z 4 dz; z 2 /3 z + 5 (b) dz; (e) e z2 z i/2 dz; (c) dz; (f) (2z 5) 2 z 5 dz; e 2z 3z + 2 dz. 9. Let D be a simply connected region in and let be a simple 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. 2

13 (c) Now apply auchy s formula to conclude that f (z ) = 2πi (z z ) 2 dz. 3

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