# Some Complex Analysis help

1. Mar 16, 2006

### Warr

Here's my question:

Let f and g be analytic inside and on the smple loop $$\Gamma$$. Prove that if f(z)=g(z) for all z on $$\Gamma$$, then f(z)=g(z) for all z inside $$\Gamma$$.

Don't really know where to start on this one. This comes from the section 'Cauchy's Integral Formula'.

2. Mar 16, 2006

### cepheid

Staff Emeritus
What does Cauchy's Integral Formula tell you about the value of a function at point z0 **inside** Gamma in terms of the integral of that function around Gamma?

3. Mar 16, 2006

### Warr

well it states that for a point z0 inside gamma,

$$f(z_0)=\frac{1}{2\pi\i}{\int_\Gamma}\frac{f(z)}{z-z_0}dz$$

But where do I go from there. I think i'm missing something big and obvious :x

k here is my guess:

since $$f(z_0)=g(z_0)$$ for all $$z_0$$ on $$\Gamma$$, then we know

$$\frac{1}{2\pi\i}{\int_\Gamma}\frac{f(z)}{z-z_0}dz=\frac{1}{2\pi\i}{\int_\Gamma}\frac{g(z)}{z-z_0}dz$$, and therefore f(z)=g(z) for all z in $$\Gamma$$ since this formula is ambiguous whether z_0 is on $$\Gamma$$ or inside of it.

Problem is this seems like i'm missing something. For this case, even if I write that shouldn't it still only apply to the loop itself and not what is inside. Is there some subtle thing I am missing that would allow me to actually say what I just said?

Last edited: Mar 16, 2006
4. Mar 16, 2006

### HallsofIvy

Staff Emeritus
That's a perfectly valid proof. However, it's not true that "this formula is ambiguous whether z_0 is on or inside of it". It requires that z0 be inside the loop so that $\frac{f(z)}{z-z_0}$ be defined for all x on the loop.

5. Mar 17, 2006

### cepheid

Staff Emeritus
How can you start the proof with this statement? That is what you are trying to prove!

Ummmm......NO. As Halls pointed out, z0 is a point INSIDE gamma, not on it.

I think this is the statement you start the proof with. It follows immediately from the fact that f(z) = g(z) on Gamma. From there the proof is as straightforward as applying the formula.

6. Mar 17, 2006

### Warr

Sorry, I don't think I was thinking too clearly when I wrote that. Thanks for the advice.