Show that $\exists g$ holomorphic st $e^{g(z)} = f(z)$ with given $f$ holomorphic and $\int_\gamma \frac{f'(z)}{f(z)}dz = 0$

Question

Suppose $f : C^* \to C^* $ is a holomorphic function. Define the loop $\gamma :[0,2\pi] \to C^*$ by $\gamma(t) = e^{it}$. If we have $\int_\gamma \frac{f'(z)}{f(z)} dz = 0 $ then show that there exists a holomorphic function $g : C^* \to C^*$ such that $e^{g(z)} = f(z) , \forall z \in C^*$.

I really don't have a clue how to solve this. Thanks in advance.

Answer 1

If $f=\exp(g)$, then $f'=g'f$. So we want to define $\displaystyle g(z) = w_0+\int_{z_0}^z \frac{f'(w)}{f(w)} dw$, where:

• $z_0$ is a fixed point in $\mathbb C^*$,
• the integral is taken along any path that connects $z_0$ and $z$ and does not go through the origin,
• $w_0$ is such that $f(z_0)=\exp(w_0)$, which exists since $f(z_0)\ne0$ and the image of $\exp$ is $\mathbb C \setminus \{ {0}\}$.

The key point is to prove that the integral does not depend on the chosen path.

Let $\displaystyle I_r = \int_{\gamma_r} \frac{f'(z)}{f(z)} dz$, where $\gamma_r$ is the circle of radius $r$ centered at the origin. Then $I_1=0$ implies $I_r=0$ for all $r>0$, because the integrand is holomorphic in the annulus between $\gamma_1$ and $\gamma_r$.

This is enough to prove that $g$ is well defined because a path from $z_0$ to $z$ that does not go through the origin does not cross $\gamma_r$ for sufficiently small $r>0$.