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:

$\displaystyle \int_{z_0}^z \frac{f'(\zeta )}{f(\zeta )} ~d\zeta$

My professor tells me there is a standard argument where I take a $\displaystyle z_1 \in D$ and a small disk $\displaystyle D_r(z_1) \subset D$ then take a fixed path from $\displaystyle z_0$ to $\displaystyle z_1$ and then a line segment from $\displaystyle z_1$ to $\displaystyle z$, 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 $\displaystyle ln\left(\frac{f(z)}{f(z_0)}\right)$, don't you?

Originally Posted by HallsofIvy

You understand that this integral, over any path, is just $\displaystyle ln\left(\frac{f(z)}{f(z_0)}\right)$, 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 $\displaystyle \gamma_1$ and $\displaystyle \gamma_2$ from $\displaystyle z_0$ to $\displaystyle z$ with $\displaystyle \gamma = \gamma_1 \circ \gamma_2^{-1}$ then,

$\displaystyle \int_{\gamma_1} \frac{f'(\zeta )}{f(\zeta )} ~d\zeta - \int_{\gamma_2} \frac{f'(\zeta )}{f(\zeta )} ~d\zeta = \int_{\gamma} \frac{f'(\zeta )}{f(\zeta)} ~d\zeta = 2\pi i \ n(f\circ \gamma , 0) \in 2\pi i \mathbb{Z}$

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?