Presentation on theme: "Week 5 2. Cauchy’s Integral Theorem (continued)"— Presentation transcript:
1 Week 5 2. Cauchy’s Integral Theorem (continued) ۞ A simple path on a plane is such that does not intersect or touch itself.Theorem 1: Cauchy’s Integral Theorem (light version)Let f(z) be analytic and its derivative f '(z) be continuous in a simply connected domain D, then, for every simple contour C in D,Proof: Kreyszig, section 14.2
2 Theorem 1a: Cauchy’s Integral Theorem (heavy version) Let f(z) be analytic in a simply connected domain D. Then, for any simple contours C in D,Proof: Kreyszig, Appendix 4 (non-examinable)Comment:The difference between the light and heavy versions of Cauchy’s Theorem is that the former assumes f '(z) to be continuous. This assumption is, of course, correct (the derivative of an analytic function is continuous), but it still needs to be proved.
3 Example 1: the importance of D being simply connected Evaluate the integralwhere C is the unit circle, centred at the origin, traversed in the clockwise direction.Comment:Observe that no simply connected domain exists, where the integrand of this integral would be analytic (because of the singularity at z = 0).
4 Example 2:Formulate an equivalent of Cauchy’s Theorem for multiply connected domains.Hint:This theorem is formulated and proven in Kreyszig, section 14.2, but don’t read it. Think instead in terms of topologically equivalent contours.
5 Theorem 2: Independence of path If f(z) is analytic in a simply connected domain D, then the integral of f along a path in D doesn’t depend on the path’s shape (only on its endpoints).Proof: Kreyszig, Section 14.2Example 3:Use Theorem 2 to evaluatewhere P is the broken line with the vertices z = −1, i, +1 (in this order).
6 Comment: Principle of deformation of path The path of an integral of a function f(z) can be deformed as long as: (a) all of it, at all times, remains in the region where f is analytic, and (b) the endpoints (if any, i.e. if it’s not a contour) are not moved.In application to contours, the principle of deformation of path is trivial (all contour integrals of analytic functions are zero anyway).
7 Theorem 3: Cauchy’s Integral Formula Let f(z) be analytic in a simply connected domain D, z0 be a point in D, and C be a contour in D, such that C is traversed in the counter-clockwise direction and encloses z0.Then(1)Proof: Kreyszig, Section 14.3
8 Theorem 4: derivatives of an analytic function If f(z) is analytic in a domain D, it has derivatives of all orders in D. These derivatives are given by(2)(3)and, in general,(4)Proof: Kreyszig, Section 14.4 (non-examinable)
9 Comment:Formulae (2)–(4) can be obtained by formally differentiating Cauchy’s formula (1) for f(z0) with respect to z0.۞ An entire functions is a function that is analytic for all z.Example 4:The functions zn (n is an integer), exp z, sin z, and cos z are entire.Theorem 5: Liouville’s TheoremIf f(z) is an entire function, and | f | is bounded, then f = const.Proof: Kreyszig, Section 14.4
10 Theorem 6: Existence of indefinite integral If f(z) is analytic in a simply connected domain D, it has an indefinite integral, i.e. a function F(z) exists, such that F ' = f and, for any path P beginning from z1 and finishing at z2,Obviously, F(z) is analytic in D.Proof: Kreyszig, Section 14.1
11 Example 5 (Example 3 revisited): Use Theorem 6 to evaluatewhere P is the broken line with the vertices z = −1, i, +1 (in this order).