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 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).

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 14.2
Example 3:
Use Theorem 2 to evaluate
where 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 non-trivial only for a non-simply connected domain (otherwise all contour integrals 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 ≥ 0 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

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 evaluate
where P is the broken line with the vertices z = −1, i, +1 (in this order).