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

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?

• Your assumption implies that $f$ is non-vanishing on your domain Commented Nov 22, 2019 at 6:11
• Is $\gamma$ a simple closed curve? Commented Nov 22, 2019 at 6:14
• Rouche Theorem?
– xbh
Commented Nov 22, 2019 at 6:45
• Commented Nov 22, 2019 at 7:59

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 $$0$$.