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Published byBrianna Preston Modified over 4 years ago

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**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

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**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.

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**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. Comment: Observe that no simply connected domain exists, where the integrand of this integral would be analytic (because of the singularity at z = 0).

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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.

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

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**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).

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**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

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**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)

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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 Theorem If f(z) is an entire function, and | f | is bounded, then f = const. Proof: Kreyszig, Section 14.4

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**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

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**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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