First some basics from multivariable calculus directly extended to complex functions.

Transcription

1 Complex Integration Tuesday, October 15, :01 PM Our first objective is to develop a concept of integration of complex functions that interacts well with the notion of complex derivative. A certain kind of line integral (called a contour integral) is of primary interest Area integrals don't have many interesting properties (AFAIK) First some basics from multivariable calculus directly extended to complex functions. Start by just defining the integral of a complex-valued function over an interval [a,b] on the real axis: Now we want to extend the notion of line integral to arbitrary piecewise smooth contours in the complex plane. Given a complex function of a complex variable f(z) and a contour C in the complex plane. We define the contour integral in any of the following equivalent ways (equivalent by rules of multivariable calculus for real variables and the above definition of an integral of a complex function): ComplexAnalysis Page 1

2 The parametric form is usually more useful in practice. For the same reasons as in real-valued multivariable calculus, the value of the contour integral is independent of any good parameterization. But note just as with line integrals, contour integrals have a particular direction associated with them. Again it will be useful to obtain a general bound on contour integrals, and as a direct consequence of our previous integral bound, we have: ComplexAnalysis Page 2

3 Also note that contour integrals are equivalent to ordinary integrals on the realaxis when the contour is a subset of the real axis and the function is real-valued on the real axis. How does the contour integral relate to typical line integrals of twodimensional vector fields, as one often sees in physics? This is not really related to contour integrals in any natural way! How does the contour integral relate to integrals over arclength? Let's do some simple but important contour integrals to see how they work. ComplexAnalysis Page 3

4 It's generally best to write contours in some parametric form (If you just specify the set: Then the orientation isn't specified) Let's compute, for integers n: In these steps, we have used the following fact: If t is a real variable, and z(t) is a complex function on the real axis, and f(z) is an analytic complex function of a complex variable, then one can verify the usual chain rule works: ComplexAnalysis Page 4

5 From the above, we have: Note in particular that We will see later that the reason for this result is that z n has a complex-valued antiderivative over the interior of the circle, so its integral over the closed circle will vanish, provided and is an integer. Now again let's take n to be an integer and consider the integral: With the principal branch for the multivalued integrand. (Note how the parameter is chosen to respect the branch cut to avoid confusion in the calculation. The point is this pushes the discontinuity to the end of the integration interval, which is easy to handle because integrating over a closed interval [a,b] is the same as integrating over an open interval (a,b) as long as you don't have exotic objects like delta functions. If one instead has the parameterization cross the branch cut, one has to explicitly account for the discontinuity incurred by the branch cut.) ComplexAnalysis Page 5

6 Let's try to abstract some of these findings into a more elegant framework for handling contour integrals. The idea is to try and apply a sort of "Fundamental Theorem of Calculus" to contour integrals directly, rather than on parameterizations on the real axis. To prepare for the theory, we first note that in general, an application of the multivariable chain rule would say that for a complex-valued function z(t) on the real axis and a complex-valued function f(z) of a complex variable, then we have the following chain rule: But for analytic functions, This observation helps to explain why analytic functions have properties that are reminiscent of one-dimensional functions even though complex variables involve two real ComplexAnalysis Page 6

7 variables. Let's see why. Let's suppose that we could identify an analytic complex-valued function F(z) such that where f(z) is also analytic. We call such a function F(z) an antiderivative of f(z). Then: And in particular, if the contour C is closed (beginning point and endpoint are the same point in the complex plane) and the antiderivative is single-valued then Note that if f(z) were not analytic, then the above statement need not be true. Consider: ComplexAnalysis Page 7

8 Let's look back on how to use the above observation concerning contour integrals of analytic functions to simplify the previous calculations. Let's revisit f(z) = z n. This is an analytic function, and one can readily verify by product rule or whatever that F(z) = is a single-valued antiderivative for So integrating over the closed contour described above (circle of radius R cetnered at origin, traversed counterclockwise): But for integral of is an antiderivative for but is multivalued, so the over a closed contour need not be 0. For our closed contour: We'll explain how to do this and the fractional roots more carefully next time. Note that we can think of a multi-valued antiderivative as single-valued on its Riemann surface. But then the contour of integration is not closed on the Riemann surface. ComplexAnalysis Page 8