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1 Week 5 2. Cauchy’s Integral Theorem (continued) Proof: Kreyszig, section 14.2 Theorem 1: Cauchy’s Integral Theorem (light version) ۞ A simple path on.

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Presentation on theme: "1 Week 5 2. Cauchy’s Integral Theorem (continued) Proof: Kreyszig, section 14.2 Theorem 1: Cauchy’s Integral Theorem (light version) ۞ A simple path on."— Presentation transcript:

1 1 Week 5 2. Cauchy’s Integral Theorem (continued) Proof: Kreyszig, section 14.2 Theorem 1: Cauchy’s Integral Theorem (light version) ۞ A simple path on a plane is such that does not intersect or touch itself. 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,

2 2 Proof: Kreyszig, Appendix 4 (non-examinable) 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, 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 3 Example 1: the importance of D being simply connected Evaluate the integral where C is the unit circle, centred at the origin, traversed in the clockwise direction. Observe that no simply connected domain exists, where the integrand of this integral would be analytic (because of the singularity at z = 0 ). Comment:

4 4 Example 2: Formulate an equivalent of Cauchy’s Theorem for multiply connected domains. This theorem is formulated and proven in Kreyszig, section 14.2, but don’t read it. Think instead in terms of topologically equivalent contours. Hint:

5 5 Proof: Kreyszig, Section 14.2 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). Example 3: Use Theorem 2 to evaluate where P is the broken line with the vertices z = −1, i, +1 (in this order).

6 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 7 Proof: Kreyszig, Section 14.3 Theorem 3: Cauchy’s Integral Formula Let f(z) be analytic in a simply connected domain D, z 0 be a point in D, and C be a contour in D, such that C is traversed in the counter-clockwise direction and encloses z 0. Then (1)

8 8 Proof: Kreyszig, Section 14.4 (non-examinable) 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 and, in general, (2) (3) (4)

9 9 Proof: Kreyszig, Section 14.4 Comment: Formulae (2)–(4) can be obtained by formally differentiating Cauchy’s formula (1) for f(z 0 ) with respect to z 0. Theorem 5: Liouville’s Theorem If f(z) is an entire function, and | f | is bounded, then f = const. ۞ An entire functions is a function that is analytic for all z. Example 4: The functions z n ( n is an integer), exp z, sin z, and cos z are entire.

10 10 Proof: Kreyszig, Section 14.1 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 z 1 and finishing at z 2, Obviously, F(z) is analytic in D.

11 11 Example 5 (Example 3 revisited): Use Theorem 6 to evaluate where P is the broken line with the vertices z = −1, i, +1 (in this order).


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