Assume that $f(z)$ is holomorphic and satisfies the inequality $|f(z)-1|<1$ in a region $|z|<R$. Show that $$\int_\gamma\frac{f'(z)}{f(z)}dz=0$$ for all closed curves $\gamma$ in $|z|<R$. (The continuity of $f'(z)$ is taken for granted.)
The anti-derivative of $\dfrac{f'(z)}{f(z)}$ is just $\ln(f(z))$, which is holomorphic as long as $\ln(f(z))$ avoids the log branch cut. The log branch cut is the negative real axis, $a\leq 0$.
I want to use the fact that: If $f$ is the derivative of a holomorphic function in the region $\omega$, then the integral $\int_\gamma fdz$ depends only on the endpoints of $\gamma$.
To use this, we only need the condition that $f(z)$ is not a non-positive real number. The condition $|f(z)-1|<1$ seems like an overkill. (Of course it does imply that $f(z)$ is not a non-positive real number.) But am I misunderstanding something?
