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

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.

• Hint: Can you think of an anti-derivative for $f'(z)/f(z)$ in $\Omega$? Aug 24, 2014 at 11:49
• Note that the inequality implies that $f(z)\not = 0$ on $\Omega$. Aug 24, 2014 at 11:57
• You do remember what the indefinite integral $$\int\frac{f'(x)}{f(x)}\,dx$$ is, don't you? You have certainly seen it, when $f(x)=x$ and when $f(x)=x^2+1$. Most likely also, when $f(x)=\cos x$. Aug 24, 2014 at 12:05
• Is it lnf(z)?So existence of premitive implies the integral is zero?But can I say that lnf(z) exists in the given region?Sorry again.perhaps I am missing something very much simple. Aug 24, 2014 at 12:18
• Correct. That's one possibility. So is it analytic in the open disk $$B(1,1)=\{z\in\Bbb{C}\mid |z-1|<1\}?$$ If it is, then can you conclude that $\ln (f(z))$ is analytic in $\Omega$? Aug 24, 2014 at 12:34

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 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 $\ldots$

$$\int\limits_{\gamma}\frac{f'(z)}{f(z)}dz=2\pi i (N_{zeros}-N_{poles}) .$$
$$|f(z)-1|<1 \Leftrightarrow -1<f(z)-1<1\Leftrightarrow 0 <f(z)<2$$ Since $f(z)>0$, it has no zeros in any $\operatorname{Int}\gamma$. As $f(z)<\infty$, it has no poles. Therefore, $$N_{zeros}=N_{poles}=0$$ $$\int\limits_{\gamma}\frac{f'(z)}{f(z)}dz=0.$$
• To write $|f(z)-1| < 1$ as $-1 < f(z)-1 < 1$, aren't you implicitly assuming that $f$ is real-valued? Dec 7, 2021 at 3:01