# Proof Using Cauchy's Formula

1. Apr 25, 2014

### bbdynamite

1. The problem statement, all variables and given/known data
Let D be a simply connected region in C (complex domain) and let C be a simple closed curve contained in D. Let f(z) be analytic in D. Suppose that z0 is a point which is not enclosed by C. What is 1/(2πi)∫f(z)/(z-z0)dz?

2. Relevant equations
Cauchy's formula: f(z0) = 1/(2πi)∫f(z)/(z-z0)dz

3. The attempt at a solution
I have a gut feeling that since z0 is not in enclosed by C, it is also not part of D. Since f(z) is analytic in D, this somehow means that f(z) = 0 outside of D so f(z0) = 0. I know I'm missing a lot of connections and mathematical reasoning, but this is a guess I have because I don't actually know how to show it mathematically. Help appreciated!

2. Apr 26, 2014

### haruspex

What f does outside of D is unknown, so irrelevant.
C is contained in D; z0 is in D but outside C. Are you perhaps confused by the use of C at the start of the question as a reference to the complex plane? Everywhere else it is referring to a closed curve inside D.
For Cauchy's integral formula to apply, what would be the relationship between z0 and the path of the integral?

3. Apr 26, 2014

### HallsofIvy

Staff Emeritus
The whole point of this question is that the integral of an function, analytic inside a given closed curve, around that closed curve, is 0. Did you not know that?

The proof is fairly simple: any analytic function, pretty much by definition, can be written as a Taylor's series. If you integrate term by term, the integral of $(z- a)^n$ about a closed curve containing a, is equivalent to integration around a circle with center at a and that is 0:
Let $$z= a+ Re^{i\theta}$$ and $$\int_C (z- a)^n dz= \int_0^{2\pi} (Re^{i\theta})(iRe^{i\theta}d\theta)= iR^2\int_0^{2\pi} e^{2i\theta}d\theta= \left[(R^2/2)e^{2i\theta}\right]_0^{2\pi}= 0$$