18.04 S18 Topic 3: Line Integrals and Cauchy's Theorem

Topic 3 Notes Jeremy Orloff 3 Line integrals and Cauchy’s theorem 3.1 Introduction The basic theme here is that complex line integrals will mirror much of what we’ve seen for multivariable calculus line integrals. But, just like working with ��� is easier than working with sine and cosine, complex line integrals are easier to work with than their multivariable analogs. At the same time they will give deep insight into the workings of these integrals. To define complex line integrals, we will need the following ingredients: • The complex plane: � = � + �� • The complex differential �� = �� + ��� • A curve in the complex plane: �(�) = �(�) + ��(�), defined for � ≤ � ≤ �. • A complex function: � (�) = �(�, �) + ��(�, �) 3.2 Complex line integrals Line integrals are also called path or contour integrals. Given the ingredients we define the complex line integral ∫� � (�) �� by � (�) �� ∶= ∫� � ∫� � (�(�))� ′ (�) ��. (1a) 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 �2 . Also, make sure you understand that the product � (�(�))� ′ (�) is just a product of complex numbers. An alternative notation uses �� = �� + ��� to write ∫� � (�) �� = ∫� (� + ��)(�� + ���) (1b) Let’s check that Equations 1a and 1b are the same. Equation 1b is really a multivariable calculus expression, so thinking of �(�) as (�(�), �(�)) it becomes ∫� � (�) �� = � ∫� ( ) [�(�(�), �(�)) + ��(�(�), �(�)] �′ (�) + ��′ (�) �� But, �(�(�), �(�)) + ��(�(�), �(�)) = � (�(�)) and �′ (�) + ��′ (�) = � ′ (�) 1 3 LINE INTEGRALS AND CAUCHY’S THEOREM 2 so the right hand side of this equation is � ∫� � (�(�))� ′ (�) ��. That is, it is exactly the same as the expression in Equation 1a. Example 3.1. Compute ∫� �2 �� along the straight line from 0 to 1 + �. Solution: We parametrize the curve as �(�) = �(1 + �) with 0 ≤ � ≤ 1. So � ′ (�) = 1 + �. The line integral is 1 2�(1 + �) �2 (1 + �)2 (1 + �) �� = . �2 �� = ∫ ∫0 3 Example 3.2. Compute ∫� � �� along the straight line from 0 to 1 + �. Solution: We can use the same parametrization as in the previous example. So, 1 ∫� Example 3.3. Compute ∫� � �� = ∫0 �(1 − �)(1 + �) �� = 1. �2 �� along the unit circle. Solution: We parametrize the unit circle by �(�) = e�� , where 0 ≤ � ≤ 2�. We have � ′ (�) = �e�� . So, the integral becomes 2� ∫� �2 �� = Example 3.4. Compute ∫ ∫0 e2�� �e�� �� = ∫0 e�3� || = 0. 3 ||0 2� 2� �e3�� �� = � �� along the unit circle. Solution: Parametrize �: �(�) = e�� , with 0 ≤ � ≤ 2�. So, � ′ (�) = �e�� . Putting this into the integral gives 2� ∫� 3.3 � �� = ∫0 e�� � e�� �� = 2� ∫0 � �� = 2��. Fundamental theorem for complex line integrals This is exactly analogous to the fundamental theorem of calculus. Theorem 3.5. (Fundamental theorem of complex line integrals) If � (�) is a complex analytic function on an open region � and � is a curve in � from �0 to �1 then ∫� � ′ (�) �� = � (�1 ) − � (�0 ). 3 LINE INTEGRALS AND CAUCHY’S THEOREM 3 Proof. This is an application of the chain rule. We have �� (�(�)) = � ′ (�(�)) � ′ (�). �� So ∫� � ′ (�) �� = � ∫� � ′ (�(�)) � ′ (�) �� = � ∫� |� �� (�(�)) �� = � (�(�))|| = � (�1 ) − � (�0 ). �� |� Another equivalent way to state the fundamental theorem is: if � has an antiderivative � , i.e. � ′ = � then � (�) �� = � (�1 ) − � (�0 ). ∫� Example 3.6. Redo ∫� �2 ��, with � the straight line from 0 to 1 + �. Solution: We can check by inspection that �2 has an antiderivative � (�) = �3 ∕3. Therefore the fundamental theorem implies �3 || 3 ||0 1+� ∫� Example 3.7. Redo ∫� �2 �� = = (1 + �)3 2�(1 + �) = . 3 3 �2 ��, with � the unit circle. Solution: Again, since �2 had antiderivative �3 ∕3 we can evaluate the integral by plugging the endpoints of � into the �3 ∕3. Since the endpoints are the same the resulting difference will be 0! 3.4 Path independence We say the integral ∫� � (�) �� is path independent if it has the same value for any two paths with the � (�) �� is path independent ∫� in �, if it has the same value for any two paths in � with the same endpoints. same endpoints. More precisely, if � (�) is defined on a region � then The following theorem follows directly from the fundamental theorem. The proof uses the same argument as Example 3.7. Theorem 3.8. If � (�) has an antiderivative in an open region �, then the path integral path independent for all paths in �. ∫� � (�) �� is Proof. Since � (�) has an antiderivative of � (�), the fundamental theorem tells us that the integral only depends on the endpoints of �, i.e. ∫� � (�) �� = � (�1 ) − � (�0 ) where �0 and �1 are the beginning and end point of �. An alternative way to express path independence uses closed paths. Theorem 3.9. The following two things are equivalent. 3 LINE INTEGRALS AND CAUCHY’S THEOREM 1. The integral ∫� 2. The integral ∫� 4 � (�) �� is path independent. � (�) �� around any closed path is 0. Proof. This is essentially identical to the equivalent multivariable proof. We have to show two things: (i) Path independence implies the line integral around any closed path is 0. (ii) If the line integral around all closed paths is 0 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 �0 is the same as the endpoint �1 the line integral � (�) �� must have the same ∫� value as the line integral over the curve consisting of the single point �0 . Since that is clearly 0 we must have the integral over � is 0. � (�) �� = 0 for any closed curve. Consider the two curves �1 and �2 shown ∫� in figure (ii). Both start at �0 and end at �1 . By the assumption that integrals over closed paths are To see (ii), assume 0 we have ∫�1 −�2 � (�) �� = 0. So, ∫�1 � (�) �� = ∫�2 � (�) ��. That is, any two paths from �0 to �1 have the same line integral. This shows that the line integrals are path independent. y y P =Q Q Cc C2 C1 P x x Figure (ii) Figure (i) 3.5 Examples Example 3.10. Why can’t we compute ∫� � �� using the fundamental theorem. Solution: Because � doesn’t have an antiderivative. We can also see this by noting that if � had an antiderivative, then its integral around the unit circle would have to be 0. But, we saw in Example 3.4 that this is not the case. 1 �� over each of the following contours Example 3.11. Compute ∫� � (i) The line from 1 to 1 + �. 3 LINE INTEGRALS AND CAUCHY’S THEOREM 5 (ii) The circle of radius 1 around � = 3. (iii) The unit circle. Solution: For parts (i) and (ii) there is no problem using the antiderivative log(�) because these curves are contained in a simply connected region that doesn’t contain the origin. (i) √ 1 � �� = log(1 + �) − log(1) = log( 2) + � . ∫� � 4 (ii) Since the beginning and end points are the same, we get 1 �� = 0 ∫� � (iii) We parametrize the unit circle by �(�) = e�� with 0 ≤ � ≤ 2�. We compute � ′ (�) = �e�� . So the integral becomes 2� 2� 1 1 �� �� = � �� = 2��. �� = �e ∫� � ∫0 ∫0 e�� Notice that we could use log(�) if we were careful to let the argument increase by 2� as it went around the origin once. 1 ��, where � is the unit circle in two ways. ∫� �2 (i) Using the fundamental theorem. Example 3.12. Compute (ii) Directly from the definition. Solution: (i) Let � (�) = −1∕�. Since � ′ (�) = 1∕�2 , the fundamental theorem says 1 �� = � ′ (�) �� = � (endpoint) − � (start point) = 0. ∫� ∫� �2 It equals 0 because the start and endpoints are the same. (ii) As usual, we parametrize the unit circle as �(�) = e�� with 0 ≤ � ≤ 2�. So, � ′ (�) = �e�� and the integral becomes 1 �� = ∫� �2 ∫0 3.6 2� 1 e2�� �� �e �� = 2� ∫0 |2� �e−�� �� = −e−�� | = 0. |0 Cauchy’s theorem Cauchy’s theorem is analogous to Green’s theorem for curl free vector fields. Theorem 3.13. (Cauchy’s theorem) Suppose � is a simply connected region, � (�) is analytic on � and � is a simple closed curve in �. Then the following three things hold: (i) ∫� � (�) �� = 0 (i′ ) We can drop the requirement that � is simple in part (i). (ii) Integrals of � on paths within � are path independent. That is, two paths with the same endpoints integrate to the same value. 3 LINE INTEGRALS AND CAUCHY’S THEOREM 6 (iii) � has an antiderivative in �. Proof. We will prove (i) using Green’s theorem – we could give a proof that didn’t rely on Green’s, but it would be quite similar in flavor to the proof of Green’s theorem. Let � be the region inside the curve. And write � = � + ��. Now we write out the integral as follows ∫� � (�) �� = ∫� (� + ��) (�� + ���) = ∫� (� �� − � ��) + �(� �� + � ��). Let’s apply Green’s theorem to the real and imaginary pieces separately. First the real piece: ∫� � �� − � �� = ∫� (−�� − �� ) �� �� = 0. We get 0 because the Cauchy-Riemann equations say �� = −�� , so −�� − �� = 0. Likewise for the imaginary piece: ∫� � �� + � �� = ∫� (�� − �� ) �� �� = 0. We get 0 because the Cauchy-Riemann equations say �� = �� , so �� − �� = 0. To see part (i′ ) you should draw a few curves that intersect themselves and convince yourself that they can be broken into a sum of simple closed curves. Thus, (i′ ) follows from (i).1 Part (ii) follows from (i) and Theorem 3.9. To see (iii), pick a base point �0 ∈ � and let � (�) = � ∫�0 � (�) ��. Here the integral is over any path in � connecting �0 to �. By part (ii), � (�) is well defined. If we can show that � ′ (�) = � (�) then we’ll be done. Doing this amounts to managing the notation to apply the fundamental theorem of calculus and the Cauchy-Riemann equations. So, let’s write � (�) = �(�, �) + ��(�, �), � (�) = � (�, �) + �� (�, �). Then we can write �� = �� + ��� , etc. �� We can formulate the Cauchy-Riemann equations for � (�) as � ′ (�) = i.e. �� 1 �� = �� � �� 1 � ′ (�) = �� + ��� = (�� + ��� ) = �� − ��� . � (2a) (2b) 1 In order to truly prove part (i′ ) we would need a more technically precise definition of simply connected so we could say that all closed curves within � can be continuously deformed to each other. 3 LINE INTEGRALS AND CAUCHY’S THEOREM 7 For reference, we note that using the path �(�) = �(�) + ��(�), with �(0) = �0 and �(�) = � we have � (�) = � ∫�0 � (�) �� = � ∫�0 � = ∫0 (�(�, �) + ��(�, �))(�� + ���) (�(�(�), �(�)) + ��(�(�), �(�))(�′ (�) + ��′ (�)) ��. (3) Our goal now is to prove that the Cauchy-Riemann equations given in Equation 3 hold for � (�). The figure below shows an arbitrary path from �0 to �, which can be used to compute � (�). To compute the partials of � we’ll need the straight lines that continue � to � + ℎ or � + �ℎ. z + ih A Cy z Cx z + h C z0 Paths for proof of Cauchy’s theorem To prepare the rest of the argument we remind you that the fundamental theorem of calculus implies ℎ ∫0 �(�) �� = �(0). ℎ (That is, the derivative of the integral is the original function.) �� First we’ll look at . So, fix � = � + ��. Looking at the paths in the figure above we have �� lim ℎ→0 � (� + ℎ) − � (�) = ∫�+�� � (�) �� − ∫� � (�) �� = ∫�� (4) � (�) ��. The curve �� is parametrized by �(�) = � + � + ��, with 0 ≤ � ≤ ℎ. So, ∫� � (�) �� � (� + ℎ) − � (�) �� = lim = lim � ℎ→0 ℎ ℎ �� ℎ→0 ℎ ∫ �(� + �, �) + ��(� + �, �) �� = lim 0 ℎ→0 ℎ = �(�, �) + ��(�, �) = � (�). The second to last equality follows from Equation 4. (5) 3 LINE INTEGRALS AND CAUCHY’S THEOREM 8 Similarly, we get (remember: � = � + ��, so �� = � ��) � (� + �ℎ) − � (�) 1 �� = lim = lim ℎ→0 �ℎ � �� ℎ→0 = lim ∫� � (�) �� � �ℎ ℎ ∫0 �(�, � + �) + ��(�, � + �) � �� ℎ→0 = �(�, �) + ��(�, �) �ℎ = � (�). Together Equations 5 and 6 show � (�) = (6) �� 1 �� = �� � �� By Equation 2a we have shown that � is analytic and � ′ = � . □ 3.7 Extensions of Cauchy’s theorem Cauchy’s theorem requires that the function � (�) be analytic on a simply connected region. In cases where it is not, we can extend it in a useful way. Suppose � is the region between the two simple closed curves �1 and �2 . Note, both �1 and �2 are oriented in a counterclockwise direction. Theorem 3.14. (Extended Cauchy’s theorem) If � (�) is analytic on � then ∫�1 −�2 � (�) �� = 0. Proof. The proof is based on the following figure. We ‘cut’ both �1 and �2 and connect them by two copies of �3 , one in each direction. (In the figure we have drawn the two copies of �3 as separate curves, in reality they are the same curve traversed in opposite directions.) With �3 acting as a cut, the region enclosed by �1 + �3 − �2 − �3 is simply connected, so Cauchy’s Theorem 3.13 applies. We get ∫�1 +�3 −�2 −�3 � (�) �� = 0 3 LINE INTEGRALS AND CAUCHY’S THEOREM The contributions of �3 and −�3 cancel, which leaves Note. This clearly implies ∫�1 � (�) �� = ∫�2 9 ∫�1 −�2 � (�) �� = 0. QED � (�) ��. Example 3.15. Let � (�) = 1∕�. � (�) is defined and analytic on the punctured plane. y x Punctured plane: � − {0} Question: What values can the plane? ∫� � (�) �� take for � a simple closed curve (positively oriented) in Solution: We have two cases (i) �1 not around 0, and (ii) �2 around 0 y R C1 C2 x Case (i): Cauchy’s theorem applies directly because the interior does not contain the problem point at the origin. Thus, ∫�1 � (�) �� = 0. Case (ii): we will show that ∫�2 � (�) �� = 2��. Let �3 be a small circle of radius � centered at 0 and entirely inside �2 . y R C3 C2 x Figure for part (ii) 3 LINE INTEGRALS AND CAUCHY’S THEOREM 10 By the extended Cauchy theorem we have 2� ∫�2 � (�) �� = ∫�3 � (�) �� = ∫0 � �� = 2��. Here, the line integral for �3 was computed directly using the usual parametrization of a circle. Answer to the question: The only possible values are 0 and 2��. We can extend this answer in the following way: If � is not simple, then the possible values of ∫� � (�) �� are 2���, where � is the number of times � goes (counterclockwise) around the origin 0. Definition. � is called the winding number of � around 0. � also equals the number of times � crosses the positive �-axis, counting +1 for crossing from below and −1 for crossing from above. y x C A curve with winding number 2 around the origin. Example 3.16. A further extension: using the same trick of cutting the region by curves to make it simply connected we can show that if � is analytic in the region � shown below then ∫�1 −�2 −�3 −�4 � (�) �� = 0. R C2 C3 C1 C4 That is, �1 − �2 − �3 − �4 is the boundary of the region �. Orientation. It is important to get the orientation of the curves correct. One way to do this is to make sure that the region � is always to the left as you traverse the curve. In the above example. The region is to the right as you traverse �2 , �3 or �4 in the direction indicated. This is why we put a minus sign on each when describing the boundary.