$|f(z)-1| < 1$ implies $ \int_{\gamma} \frac{f'(z)}{f(z)}dz = 0$ [duplicate]

Question

friends around the world!

I need to show the following:

Let $f(z):\mathbb{C}\rightarrow \mathbb{C}$ an analytic function such that

$$|f(z)-1| < 1$$

in $\Omega \subset \mathbb{C}$

Then

$$ \int_{\gamma} \frac{f'(z)}{f(z)}dz = 0 $$

IDEAS:

There is some result that might be helpful:

Theorem: The integral

$$ \int_{\gamma} pdx + qdy = 0$$

iff the integral defined in $\Omega$ depends only on the end points of $\gamma$

iff there is a function $U(x,y)$ in $\Omega$ with the partial derivatives $\frac{\partial U }{\partial x} = p$, $\frac{\partial U}{\partial y} = q$.

However I am not sure how to use the hypothesis $|f(z)-1| < 1$.

Can anyone help me?

Answer 1

The principal branch of logarithm is analytic in $\{z:|z-1|<1\}$ and its derivative is $\frac 1 z$. Hence the given integral is $\int_{\gamma} h'(z)dz$ where $h(z)=Log (f(z))$. The integral of a derivative over any closed path is $0$.