# Holomorphic function and primitives

I need to prove that $\int_\gamma f'(z)/f(z)dz=0$ for any closed curve. It is given that f is holomorphic and satisfies $|f(z)-1|\lt1$ in the region. And we can assume $f'(z)$ is continuous.

I think that first I need to prove that $f'(z)/f(z)$ has a primitive, so that I can use the proposition to prove the integral is equal to 0.

But I don't know how to prove $f'/f$ has a primitive and how to use that inequality in my proof? Thanks.

• The integrand is holomorphic; can you use the Cauchy integral theorem? – arkeet Sep 30 '16 at 20:33
• So using the Cauchy integral theorem, we know that it's integral over the closed curve would be zero and that means $f'/f$ has a primitive? – J.doe Sep 30 '16 at 20:36

## 1 Answer

The quotient $\;\frac{f'}f\;$ is analytic wherever $\;f\neq0\;$, and since this happens in the integration region (why?) Cauchy's Theorem gives the answer at once (I'm assuming you meant closed, simple and rectifiable curve)

• So using the Cauchy integral theorem, we know that it's integral over the closed curve would be zero and that means $f'/f$ has a primitive? – J.doe Sep 30 '16 at 20:41
• @J.doe Well, yes: the primitive is easy: $\;\log f(z)\;$ , after choosing a suitable branch. Yet you don't even need to mention that when using Cauchy theorem... – DonAntonio Sep 30 '16 at 20:42
• Wait, why do we need $logf(z)$? – J.doe Sep 30 '16 at 20:54
• @J.doe Once again: if you use Cauchy's THeorem you don't even need to mention "primitive function", and of course we do use the inequality! How otherwise could we deduce that $\;f\neq0\;$ ? – DonAntonio Sep 30 '16 at 21:36
• As long as $\gamma$ is contained in a simply connected open subset of the domain (on which $f \ne 0$), it doesn't matter whether it is simple or not. – arkeet Sep 30 '16 at 23:04