Holomorphic function with analytic square root but no holomorphic logarithm

Question

Give an example of a domain $U$ and a holomorphic function $f:U \to \Bbb C$ which has a holomorphic square-root but for which there does not exist a holomorphic function $F$ such that $f = \exp F$.

I am struggling to come up with an appropriate example. Could I have a hint?

Answer 1

Just take $U=\mathbb C$ and $f(z)=z^2$.

Answer 2

$\exp$ misses the value $0$, so if $f$ has a zero in $U$, you can't write $f=\exp{} \circ F$ for a holomorphic $F$.

On the other hand, the square root has a branch point at zero, so you need a function that has a way to avoid this. For example, $(\sqrt{z})^2 = z$ is holomorphic, so $z^2$ is suitable for this.

Answer 3

Take any holomorphic function $g$ that does not have a holomorphic logarithm and take $f=g^2$.