# Proof using Cauchy's integral formula

1. Jun 20, 2007

### malawi_glenn

1. The problem statement, all variables and given/known data
Let f and g ve analytic functions inside and on a simple connected contour $\Gamma$. If f(z) = g(z) for all z on $\Gamma$, prove that
If f(z) = g(z) for all z inside $\Gamma$

2. Relevant equations

$$f(z_{0}) = \int_{\Gamma}\dfrac{f(z)}{z-z_{0}}dz$$

if f is analytic in a simple connected domain containing $\Gamma$ and $z_{0}$ is a point insinde $\Gamma$.

3. The attempt at a solution

I know I must (?) prove that $f(z_{0}) = g(z_{0})$ for all z_0, but i have no idea how to use the fact that f and g are equal on all point on $\Gamma$, I have no Lemma or Theoreme for this in my book (I am not allowed to use the theory of bounds for analytical functions, just Cauchy's integral formula)

Can someone give me a small hint ? =)

2. Jun 20, 2007

### NateTG

Well, what happens if you write out:
$$f(z_0)-g(z_0)$$
in the format you already have?

3. Jun 20, 2007

### malawi_glenn

okay i try that =) thanks!

4. Jun 20, 2007

### malawi_glenn

$$f(z_{0}) - g(z_{0}) = \dfrac{1}{2\pi i}\int_{\Gamma}\dfrac{f(z)}{z-z_{0}}dz - \dfrac{1}{2\pi i}\int_{\Gamma}\dfrac{g(z)}{z-z_{0}}dz$$

I still dont know how to use the fact that they are equal for all z ON $$\Gamma$$ =(