Show that $\int_\gamma \frac{f'(z)}{f(z)}=0$ for every closed curve $\gamma$ in $\Omega$

Question

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.

Answer 1

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

$$\int_{\gamma} \frac{f^{\prime}(z)}{f(z)}dz=\ln f(z_0)-\ln f(z_0)=0$$

Answer 2

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 $\ldots$

Answer 3

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, $$N_{zeros}=N_{poles}=0$$ $$\int\limits_{\gamma}\frac{f'(z)}{f(z)}dz=0.$$