# Math Help - complex analytic functions

1. ## complex analytic functions

Prove there exists no function F(z) analytic in the annulus D: 1<|z|<2 such that F'(z) = 1/z.
I assumed there was such a function.
let F(z)=Log(z) + c
F'(z)= 1/z
I don't know how to do this.

2. If $f(z)$ is analytic on an (non-trivial) open set $S$ and $e^{f(z)} = z$ for all point $z\in S$ then $f(z) = \ln |z| + i\arg(z)$ for some particular definition of $\arg z$. Thus, our goal is to prove that $F(z)$ as you defined it above has the property that $e^{F(z)} = z$ but then that would mean $F(z) = \ln |z| + i \arg(z)$ for some particular definition of of the argument, but then this contradicts the fact that $F$ is analytic on $1<|z|<2$ because the branch of $\arg z$ will have to hit the annulus as it leaves the origin. There is just one detail I do not know how to show, maybe it is not needed: there is a point $z_0$ on the annulus so that $e^{F(z_0)} = z_0$. With that we can complete the proof. Suppose that $F'(z) = 1/z$ and consider $g(z) = ze^{-F(z)}$ then $g'(z) = e^{-F(z)} - z(1/z)e^{-F(z)} = 0$ and this means $g(z) = k$ for some complex number $k$ because the annulus is a region (open connected set). This means, $ze^{-F(z)} = k$ for all point on the annulus. In particular for $z_0$, which yields. $z_0e^{-F(z_0)} = k \implies k=1$. Thus, $ze^{-F(z)} = 1 \implies e^{F(z)} = z$. We have show $F(z)$ is the inverse function of $e^z$ and this completes the proof using the begining comments.