Complex integrals

Let $f$ be an analytic function, $C^1$ on an open and connected set $U$.

for all $z\in U$ : $|f(z)-1|<1$.

Prove that $\int_\gamma \frac{f'(z)}{f(z)}dz =0$ for every closed curve (closed curve- $\gamma(a)=\gamma(b)$) in $U$.

Any hints?

Solution 1:

The other answer, which I can't understand why is downvoted, is correct. You can also argue as follows:

The condition $\;|f(z)-1|<1\;,\;\;\forall\,z\in U\;$, makes sure the function $\;f\;$ can not vanish in $\;U\;$ (why?), and thus $\;\cfrac{f'}f\;$ is analytic in $\;U\;$ , so the claim follws directly from CIT.

Solution 2:

Hint: The complex logarithm admits an analytic branch on $U_1(1)$.

Edit: Alternatively, by the definition of the line integral we have $$\int_\gamma \frac{1}{f(x)} f'(x) \, dx = \int_a^b \frac{1}{f(\gamma(x))} f'(\gamma(x)) \gamma'(x) \, dx = \int_{\beta} \frac{1}{x} \, dx,$$ where $\beta = f \circ \gamma$. Now use the Cauchy integral theorem.

Solution 3:

Hint: To be able to use the fact that Log(f) is the antiderivative of f′/f, i.e. to be able to say that (Log(f))' = f'/f, you need to make sure that Log(f) is differentiable.

Yet, the complex logarithm Log is not continuous everywhere (and hence not differentiable everywhere).

The condition |f(z)−1|<1 can be used to guarantee that Log is continuous and differentiable.

Does this make sense ?