Topic 3 Notes Jeremy Orloff

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1 Topic 3 Notes Jerem Orloff 3 Line integrals and auch s theorem 3.1 Introduction The basic theme here is that comple line integrals will mirror much of what we ve seen for multivariable calculus line integrals. But, just like working with e iθ is easier than working with sine and cosine, comple line integrals are easier to work with than their multivariable analogs. At the same time the will give deep insight into the workings of these integrals. 3.2 Ingredients The comple plane: z = + i The comple differential dz = d + id A curve in the comple plane: (t) = (t) + i(t), defined for a t b. A comple function: f(z) = u(, ) + iv(, ) 3.3 omple line integrals Note. Line integrals are also called path or contour integrals. Given the ingredients we define the comple line integral f(z) dz b f(z) dz = b a f((t)) (t) dt. (1) You should note that this notation looks just like integrals of a real variable. We don t need the vectors and dot products of line integrals in R 2. Also, make sure ou understand that the product f((t)) (t) is just a product of comple numbers. An alternative notation uses dz = d + id to write f(z) dz = (u + iv)(d + id) (2) Let s check that Equations 1 and 2 are the same. Equation 2 is reall a multivariable calculus epression, so thinking of (t) as ((t), (t)) it becomes b f(z) dz = (u((t), (t)) + iv((t), (t))( (t) + i (t))dt a But, u((t), (t)) + iv((t), (t)) = f((t)) and (t) + i (t) = (t) so the right hand side of this equation is b a f((t)) (t) dt. 1

2 3 LINE INTEGRALS AND AUHY S THEOREM 2 That is, it is eactl the same as the epression in Equation 1. Eample 3.1. ompute z 2 dz along the straight line from to 1 + i. answer: We parametrize the curve as (t) = t(1 + i) with t 1. So (t) = 1 + i. The line integral is 1 z 2 dz = t 2 (1 + i) 2 2i(1 + i) (1 + i) dt =. 3 Eample 3.2. ompute z dz along the straight line from to 1 + i. answer: We can use the same parametrization as in the previous eample. So, 1 zdz = t(1 i)(1 + i) dt = 1. Eample 3.3. ompute z 2 dz along the unit circle. answer: We parametrize the unit circle b (θ) = e iθ, where θ 2π. (θ) = ie iθ. So, the integral becomes 2π 2π z 2 dz = e 2iθ ie iθ dθ = ie 3iθ dθ = ei3θ 2π 3 =. Eample 3.4. ompute zdz along the unit circle. We have answer: Parametrize : (t) = e it, with t 2π. So, (t) = ie it. Putting this into the integral gives 2π 2π z dz = e it i e it dt = i dt = 2πi. 3.4 Fundamental theorem for comple line integrals This is eactl analogous to the fundamental theorem of calculus. Theorem 3.5. (Fundamental theorem of comple line integrals) If f(z) is a comple analtic function and is a curve from z to z 1 then f (z) dz = f(z 1 ) f(z ). Proof. This is an application of the chain rule. We have So f (z) dz = b a df((t)) dt f ((t)) (t) dt = = f ((t)) (t). b a df((t)) dt dt = f((t)) b a = f(z 1) f(z ).

3 3 LINE INTEGRALS AND AUHY S THEOREM 3 Another wa to state the fundamental theorem is: if f has an antiderivative F, i.e. F = f then f(z) dz = F (z 1 ) F (z ). Eample 3.6. Redo z 2 dz, with the straight line from to 1 + i. answer: We can check b inspection that z 2 has an antiderivative F (z) = z 3 /3. Therefore the fundamental theorem implies Eample 3.7. Redo z 2 dz = z3 3 1+i = (1 + i)3 3 z 2 dz, with the unit circle. = 2i(1 + i). 3 answer: Again, since z 2 had antiderivative z 3 /3 we can evaluate the integral b plugging the endpoints of into the z 3 /3. Since the endpoints are the same the resulting difference will be! 3.5 auch-riemann all the wa down We ve defined an anltic function as one having a comple derivative. The following theorem shows that if f is analtic then so is f. Thus, there are derivatives all the wa down! Theorem 3.8. If f(z) is analtic then so is f (z). Proof. To show this we have to prove that f (z) satisfies the auch-riemann equations. If f = u + iv we know u = v ; u = v ; f = u + iv. Let s write f = U +iv, so U = u = v and V = v = u. We want to show that U = V and U = V. We do them one at a time. U = V : U = v and V = v. Since v = v, we have U = V. U = V : U = u and V = u. So, U = V. QED. Technical point. We ve assumed as man partials as we need. So far we can t guarantee that all the partials eist. Soon we will have a theorem which sas that an analtic function has derivatives of all order. We ll just assume that for now. In an case, in most eamples this will be obvious. 3.6 Path independence We sa the integral f(z) dz is path independent if it has the same value for an two paths with the same endpoints.

4 3 LINE INTEGRALS AND AUHY S THEOREM 4 More precisel, if f(z) is defined on a region A then it has the same value for an two paths in A with the same endpoints. f(z) dz is path independent in A, if Theorem 3.9. If f(z) has an antiderivative in an open region A, then the path integral f(z) dz is path independent for all paths in A. Proof. This is the same argument as in Eample 3.7. Since f(z) has an antiderivative of f(z), the fundamental theorem tells us that the integral onl depends on the endpoint of, i.e. f(z) dz = F (z 1 ) F (z ) where z and z 1 are the beginning and end point of. An alternative wa to epress path independence uses closed paths. Theorem 3.1. The following two things are equivalent. (1) The integral f(z) dz is path independent. (2) The integral f(z) dz around an closed path is. Proof. This is essentiall identical to the equivalent multivariable proof. We have to show two things: (i) Path independence implies the line integral around an closed path is. (ii) If the line integral around all closed paths is then we have path independence. To see (i), assume path independence and consider the closed path shown in figure (i) below. Since the starting point z is the same as the endpoint z 1 the line integral f(z) dz must have the same value as the line integral over the curve consisting of the single point z. Since that is clearl we must have the integral over is. To see (ii), assume f(z) dz = for an closed curve. onsider the two curves 1 and 2 shown in figure (ii). Both start at z and end at z 1. B the assumption that integrals over closed paths are we have f(z) dz =. So, 1 2 f(z) dz = f(z) dz. 1 2 That is, an two paths from z to z 1 have the same line integral. This shows that the line integrals are path independent.

5 3 LINE INTEGRALS AND AUHY S THEOREM 5 P = Q c Q 2 P 1 Figure (i) Figure (ii) 3.7 Eamples Eample Wh can t we compute z dz using the fundamental theorem. answer: Because z doesn t have an antiderivative. We can also see this b noting that if z had an antiderivative, then its integral around the unit circle would have to be. But, we saw in Eample 3.4 that this is not the case. Eample ompute 1/z dz over several contours (i) The line from 1 to 1 + i. (ii) The circle of radius 1 around z = 3. (iii) The unit circle. answer: For parts (i) and (ii) there is no problem using the antiderivative log(z) because these curves are contained in a simpl connected region that doesn t contain the origin. (i) answer: log(1 + i) log(1) = log( 2) + iπ/4. (ii) answer: (since the beginning and end points are the same). (iii) answer: We parametrize the unit circle b (θ) = e iθ with θ 2π. We compute (θ) = ie iθ. So the integral becomes 1 2π z dz = 1 2π e iθ ieiθ dt = i dt = 2πi. Notice that we could use log(z) if we were careful to let the argument increase b 2π as it went around the origin once. 1 Eample ompute dz, where is the unit circle in two was. z2 (i) Using the fundamental theorem, Theorem?? (ii) Directl from the definition. answer: (i) Let f(z) = 1/z. Since f (z) = 1/z 2, the fundamental theorem sas 1 z 2 dz = f (z) dz = f(endpoint) f(start point) =. (It equals because the start and endpoints are the same.)

6 3 LINE INTEGRALS AND AUHY S THEOREM 6 (ii) As usual, we parametrize the unit circle as (θ) = e iθ with θ 2π. So, (θ) = ie iθ and the integral becomes 1 2π z 2 dz = 1 2π e 2iθ ieiθ dθ = ie iθ dθ = e iθ 2π =. 3.8 auch s theorem auch s theorem is analogous to Green s theorem for curl free vector fields. Theorem (auch s theorem) Suppose A is a simpl connected region, f(z) is analtic on A and is a simple closed curve in A. Then the following three things hold: (i) f(z) dz = (i ) We can drop the requirement that is simple in part (i). (ii) Integrals of f on paths within A are path independent. That is, two paths with same endpoints integrate to the same value. (iii) f has an antiderivative in A. Proof. We will prove (i) using Green s theorem, We could give a proof that didn t rel on Green s, but it would be quite similar in flavor to the proof of Green s theorem. Let R be the region inside the curve. And write f = u + iv. Now we write out the integral as follows f(z) dz = (u + iv) (d + id) = (u d v d) + i(v d + u d). Let s appl Green s theorem to the real and imaginar pieces separatel. piece. u d v d = v u d d =. We get because the auch-riemann equations sa u = v, so v u =. Likewise for the imaginar piece: v d + u d = R R u v d d =. We get because the auch-riemann equations sa u = v, so u v =. First the real To see part (i ) ou should draw a few curves that intersect themselves and convince ourself that the can be broken into a sum of simple closed curves. Thus, (i ) follows from (i). Note: In order to trul prove part (i ) we would need a more technicall precise definition of simpl connected so we could sa that all closed curves within A can be continuousl deformed to each other. Part (ii) follows from (i) and Theorem 3.1 To see (iii), pick a base point z A and let F (z) = z z f(w) dw.

7 3 LINE INTEGRALS AND AUHY S THEOREM 7 Here the integral is over an path in A connecting z to z. B part (ii), F (z) is well defined. If we can show that F (z) = f(z) then we ll be done. Doing this amounts to managing the notation to appl the fundamental theorem of calculus and the auch-riemann equations. Let s write f(z) = u(, ) + iv(, ) and F (z) = U(, ) + iv (, ). Then we can write f = u + iv, etc. We can formulate the auch-riemann equations for F (z) as F (z) = F = 1 F i, i.e. F (z) = U + iv = 1 i (U + iv ) = V iu. (3) For reference we note that using the path (t) = (t) + i(t), with () = z and (b) = z we have z z F (z) = f(w) dw = (u(, )+iv(, ))(d+id) = (u((t), (t))+iv((t), (t))( (t)+i (t)) dt. z z (4) Our goal now is to prove that the auch-riemann equations given in Equation 4 hold for F (z). The figure below shows an arbitrar path from z to z, which can be used to compute F (z). To compute the partials of F we ll need the straight lines that continue to z + h or z + ih. b z A z + ih z z + h Paths for proof of auch s theorem To prepare the rest of the argument we remind ou that the fundamental theorem of calculus implies h g(t) dt lim = g(). (5) h h (That is, the derivative of the integral is the original function.) First we ll look at F. So, fi z = + i. Looking at the paths in the figure above we have F (z + h) F (z) = f(w) dw f(w) dw = f(w) dw. +

8 3 LINE INTEGRALS AND AUHY S THEOREM 8 The curve is parametrized b (t) = + t + i, with t h. So, F = lim F (z + h) F (z) = lim f(w) dw h h h h = lim h h u( + t, ) + iv( + t, ) dt = u(, ) + iv(, ) = f(z). (6) h (The second to last equalit follows from Equation 5.) Similarl, we get (remember: w = z + it, so dw = i dt) 1 F i = lim F (z + ih) F (z) = lim f(w) dw h ih h ih = lim h Together Equations 6 and 7 show h u(, + t) + iv(, + t) i dt = u(, ) + iv(, ) = f(z). (7) ih f(z) = F = 1 F i B Equation 3 we have shown that F is analtic and F = f. QED 3.9 Etensions of auch s theorem auch s theorem requires that the function f(z) be analtic on a simpl connected region. In cases where it is not, we can etend it in a useful wa. Suppose R is the region between the two simple closed curves 1 and 2. Note, both 1 and 2 are oriented in a counterclockwise direction. Theorem (Etended auch s theorem) If f(z) is analtic on R then 1 2 f(z) dz =. Proof. The proof is based on the following figure. We cut both 1 and 2 and connect them b two copies of 3, one in each direction. (In the figure we have drawn the two copies of 3 as separate curves, in realit the are the same curve traversed in opposite directions.)

9 3 LINE INTEGRALS AND AUHY S THEOREM 9 With 3 acting as a cut, the region enclosed b is simpl connected, so auch s theorem (Theorem 3.14 applies. We get f(z) dz = The contributions of 3 and 3 cancel, which leaves f(z) dz =. QED 1 2 Note. This clearl implies f(z) dz = f(z) dz. 1 2 Eample Let f(z) = 1/z. f(z) is defined and analtic on the punctured plane. Question: oriented) in the plane? Punctured plane: {} What values can f(z) dz take for a simple closed curve (positivel answer: We have two cases (i) 1 not around (ii) 2 around 2 R 1 In case (i) auch s theorem applies directl because the interior does not contain the problem point at the origin. Thus, 1 f(z) dz =. For case (ii) we will show that f(z) dz = 2πi. 2

10 3 LINE INTEGRALS AND AUHY S THEOREM 1 R 3 2 Figure for part (ii) Let 3 be a small circle of radius a centered at and entirel inside 2. B the etended auch theorem we have f(z) dz = f(z) dz. 2 3 Thus, fdr = fdr. 2 3 Using the usual parametrization of a circle we can easil compute that the line integral is 3 f(z) dz = 2π i dt = 2πi. QED. Answer to the question: The onl possible values are and 2πi. We can etend this answer in the following wa: If is not simple, then the possible values of f(z) dz are 2πni, of times goes (counterclockwise) around the origin. where n is the number Definition. n is called the winding number of around. (n also equals the number of times crosses the positive -ais, counting +1 for crossing from below and 1 for crossing from above.) A curve with winding number 2 around the origin. Eample A further etension: using the same trick of cutting the region b curves to make it simpl connected we can show that if f is analtic in the region R shown below then f(z) dz =

11 3 LINE INTEGRALS AND AUHY S THEOREM 11 R That is, is the boundar of the region R. Orientation. It is important to get the orientation of the curves correct. One wa to do this is to make sure that the region R is alwas to the left as ou traverse the curve. In the above eample. The region is to the right as ou traverse 2, 3 or 4 in the direction indicated. This is wh we put a minus sign on each when describing the boundar.