Thread: Proving a Function is Analytic

I would like to prove that this function is analytic. It is really hard for me to see.

We suppose that f is analytic and zero free in a domain D, the function I wish to prove is analytic is:



My professor tells me there is a standard argument where I take a and a small disk then take a fixed path from to and then a line segment from to , but I am not sure where to go from there.

Any help is greatly appreciated!

You understand that this integral, over any path, is just , don't you?

Originally Posted by HallsofIvy

You understand that this integral, over any path, is just , don't you?

I was not aware that the integral was path independent. I was told that this integral is path independent if and only if there is a branch of the logarithm of f. If we have two paths and from to with then,



which is zero if and only if there is a branch of the logarithm of f.

Am I missing something?

Originally Posted by Aryth

I was not aware that the integral was path independent. I was told that this integral is path independent if and only if there is a branch of the logarithm of f.

You are given "f is analytic and zero free in a domain D".
I assume that by domain we know that D is path-wise connected.
There is a subdomain $D_0\subset D~\&~z_0\in D_0$. What do you know about the analyticy of $\log(f(z))~?$

Originally Posted by Plato

You are given "f is analytic and zero free in a domain D".
I assume that by domain we know that D is path-wise connected.
There is a subdomain $D_0\subset D~\&~z_0\in D_0$. What do you know about the analyticy of $\log(f(z))~?$

(I apologize for the late reply, I have been away from the internet for a moment).

Yes, that is what I meant by a domain.

I was told that a branch of a logarithm exists if and only if the domain is simply connected (i.e. no holes). So, would that mean that this integral is the logarithm locally? Say, in an open disc?