Line Integration in the Complex Plane

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Lorraine Ashley Freeman
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1 න f z dz C Engineering Math EECE

2 A Review of Integration of Functions of Real Numbers: a b f x dx is the area under the curve in the figure below: The interval of integration is the path along the x-axis from a to b. In the complex plane, the interval of integration can be much more challenging to determine or describe. Engineering Math EECE

3 The path of integration in the complex plane is The Contour (C). This corresponds to the interval of integration in real valued calculus. This must be expressed as actual limits of integration so an integral can be evaluated. We will do this by combining two distinct functions into a single parametric function (of t) in order to evaluate the integral: 1) The function tobe integrated: f(z) 2) The contour path ofintegration: z t 3) Combine them to create f(t) f z, z(t) and f t are complex functions and, as such, are two dimensional. Engineering Math EECE

4 IMPORTANT: Both given functions must be converted to parametric functions of t, then combined to create f(t) then integrated. Engineering Math EECE

5 Seven Step Process for Solving Line Integrals in the Complex Plane: 1) Get a description of the f(z). (This is often given). 2) Get a description of the contour (path) over which the integral is to be found. 3) Convert the contour into a parametric function of t called z t = x t + iy t valid for t over some interval (a, b). (This is often the hardest part). 4) Find dz t dt use the notation, the derivative of z(t) with respect to t. Note: for simplicity, we will zሶ t = dz t 5) Create f z t from f z and z t. dt. (Note the dot ሶ over z.) 6) Create f t = zሶ t f z t then simplify f t as much as possible. 7) Construct and solve a b f t dt. Engineering Math EECE

6 Steps Involved in Solving Line Integrals in the Complex Plane: The last step of the seven step process is to solve the integral of the parametric function f t : b න f t dt a This is a function of t, a and b which are real numbers. The function itself is (in general) complex. It will have complex coefficients, complex exponentials, etc. Engineering Math EECE

7 Example: Integrate f z = Re z 2 on the line segment that extends from i to 1. Step 1) Get the function to be integrated: DONE Step 2) Get a description of the contour of integration: DONE Step 3) Get z t = x t + iy t for the contour. Engineering Math EECE

8 In order to determine z(t), we must find the relationship between x & y on the contour. We are told that we are going to integrate on the line segment from i to 1. We take these numbers and superimpose them as points on the x-y plane. We then find the path of integration as a contour (path) on the x-y plane. This path is a segment of a straight line connecting the two end points. So, we begin by plotting those points and drawing an extended straight line through them: Engineering Math EECE

9 on the line segment from i to 1 Engineering Math EECE

10 We know that the equation of a straight line is y = mx + b, where m is the slope of the line and b is the y-intercept. By inspection, we can see that the y-intercept is 1 and the slope is -1. The equation of this line is y = x + 1. The actual path of integration lies along this line, starting at (0, 1) and ending at (1, 0). Draw the path. Be sure to indicate the direction of integration. Engineering Math EECE

11 Engineering Math EECE

12 The path must be re-described as a parametric complex function of t over some interval. We are free to choose any interval but it would behoove us to choose an interval that makes developing z(t) as easy as possible. Looking at the path, we can see that x goes from 0 to 1. The easiest interval is t: 0 1. This means t will go from 0 to 1 as we traverse the contour. (0 1) will be our ultimate interval of integration. 1 න f t dt 0 Engineering Math EECE

13 (0 1) will be our ultimate interval of integration. Since x goes from 0 to 1 as we traverse the contour, x t = t In addition (from our straight line): y = 1 x, substitute y t for y substitute x t for x We get: y t = 1 x t and therefore y t = 1 t Engineering Math EECE

14 Step 3 (continued) Because x + iy = z, we substitute z t for z x t for x y t for y So we now write: x t + iy t = z t = t + i 1 t END OF STEP 3 Engineering Math EECE

15 Step 4) Find zሶ t From z t = t + i 1 t we get zሶ t = 1 i (This is used in step 6). Engineering Math EECE

16 Step 5) Find f(z t ) Recall from step 1: f z = Re z 2 = x 2 To get f(z(t)), we replace z with z(t). Because z = x + iy and z(t) = x(t) + iy(t), we replace x with x(t) and y with y(t). Since we have a nice expression for z t already, we will substitute z directly with z(t): f z t = Re t + i 1 t 2 = t 2 Engineering Math EECE

17 Step 5) (Slightly different perspective) Find f(z t ) From step 1: f z = Re z 2 = x 2 We use the second expression by recalling that x(t) = t, therefore: f z t = x t 2 = t 2 Which is the same result as before. Engineering Math EECE

18 Step 6) Construct f t = f z t zሶ t = t 2 1 i No simplification necessary. Since we intend to integrate this function, we leave it unexpanded to make integration easier. Engineering Math EECE

19 Step 7) Solve: a b f t dt Recall that the limits of integration (a, b) are the limits of t that we established at the same time that we determined z(t). 1 න 0 t 2 1 i dt = 1 i න 0 1 t 2 dt = 1 i 1 t = 1 3 t3 ቮ t = 0 = i Engineering Math EECE

20 Example #2 Consider the integral of f z = 3izҧ on a path that lies along the parabola y = x 2 from 2 + 4i to 2 + 4i Engineering Math EECE

21 (a) Show that f(z) is not analytic (b) Find z(t). Sketch the contour. (c) Find C f z dz Engineering Math EECE

22 (a) Show that f(z) is not analytic: f z = 3iz ҧ = 3i x iy = 3y + 3ix = u x, y + iv x, y u = Re f z = 3y, v = Im f z = 3x u x = 0 = v y, so it could be analytic, but u y = 3 v x = 3, and therefore f z is not analytic Engineering Math EECE

23 (b) Find z(t) and sketch the contour: The contour goes from 2 + 4i to i along y = x 2, so x goes from 2 to 2. Let x t = t, t: It is given that y = x 2 (a parabola). Then y t = x t 2 = t 2 We now have x t and y t. Put them together as a complex function to get: x t + iy t = z t = t + it 2, t: 2 +2 Engineering Math EECE

24 (b) Sketch the contour: z t = t + it 2, t: 2 +2 Engineering Math EECE

25 Step 5: Find f z t We have f z = 3iz ҧ = 3ix + 3y (from step 1) and z t = x t + iy t = t + it 2 (from step 3) x t = t and y t = t 2 f z t = 3ix t + 3y t = 3it + 3t 2 Engineering Math EECE

26 From z t = t + it 2, we get zሶ t = 1 + 2it (step 4) ሶ f t = z t f z t (step 6) = 1 + 2it 3it + 3t 2 = 3it + 3t 2 6t 2 + 6it 3 f t = 3it 3t 2 + 6it 3 The last expression is the integrand. Engineering Math EECE

27 (c) Find C f z dz from a b f t dt +2 න 2 3it 3t 2 + 6it 3 dt = 3it2 2 t3 + 6it ቮ 2 = 3i i i i = 3i i = 16 Engineering Math EECE

28 Looking ahead: There are some functions that have properties that greatly simplify integration. In particular, if f(z) is analytic in a simply connected domain D, then නf z dz = F z 1 F z 0, F z = f(z) C and z 0 and z 1 are the end points for the contour. Engineering Math EECE

Functions of Several Variables

Jim Lambers MAT 280 Spring Semester 2009-10 Lecture 2 Notes These notes correspond to Section 11.1 in Stewart and Section 2.1 in Marsden and Tromba. Functions of Several Variables Multi-variable calculus