Show that $\int_\gamma \frac{f'(z)}{f(z)}=0$ for every closed curve $\gamma$ in $\Omega$ I have just started taking complex analysis course,The following problem is given in my class.Please help me solving it. Thnx in advance.

Suppose $f(z)$ is analytic and satisfies the relation $|f(z)-1| < 1 $ in a region $\Omega$
  Show that $\displaystyle \int_\gamma \frac{f'(z)}{f(z)}=0$ for every closed curve $\gamma$ in $\Omega$ 

I am only taught upto Cauchy's Theorem which I think is applicable for disc. Now here nothing is mentioned about $\Omega$, only thing I know that it is open connected.So how can I apply Cauchy's Theorem here?
Being very new to complex analysis I appologise if I am missing something very simple or doing something very much wrong.
Please help me to solve this question. Thnx again.
 A: This is a straightforward logarithmic integration, in your region $|f(z)-1|<1$, which means that the values of $f$ lie in a ball of center $1$ and radius $1$. There is a well defined branch of the log defined on this region, (in fact on any simply connected region not containing the origin). 
Therefore 
$$\int \frac{f^{\prime}(z)}{f(z)}dz=\ln f(z)$$ a well defined and single valued function.
Now over a closed curve the initial and final points, $z_0$ are the same so by the fundamental theorem
$$\int_{\gamma} \frac{f^{\prime}(z)}{f(z)}dz=\ln f(z_0)-\ln f(z_0)=0$$
A: A hint:
The quotient ${f'\over f}$ rings a bell from real calculus. Maybe one can make the analogy work. By the way: There has to be a reason for the "technical assumption" in the problem $\ldots$
A: Use principle of argument:
$$\int\limits_{\gamma}\frac{f'(z)}{f(z)}dz=2\pi i (N_{zeros}-N_{poles}) .$$
As was said,
$$|f(z)-1|<1 \Leftrightarrow -1<f(z)-1<1\Leftrightarrow 0 <f(z)<2 $$
Since $f(z)>0$, it has no zeros in any $\operatorname{Int}\gamma$. As $f(z)<\infty$, it has no poles. Therefore,
$$N_{zeros}=N_{poles}=0$$
$$\int\limits_{\gamma}\frac{f'(z)}{f(z)}dz=0.$$
