Let $n,k\geq 1$ be integers, let $U \subseteq \mathbb C^n$ be a contractible open subset, and let $f:U\to \mathrm{GL}_k(\mathbb C)$ be a holomorphic function. Does there exist a holomorhpic function $F:U\to \mathrm{M}_k(\mathbb C)$ such that $\exp(F(u))= f(u)$ holds for all $u\in U$?
Here, $\mathrm{M}_k(\mathbb C)$ means complex $k$ by $k$ matrices. The answer is of course "yes" if $k=1$.
As soon as $k\geq 2$, the problem is that for some invertible matrices $A \in \mathrm{GL}_k(\mathbb C)$ the set of matrices $B\in \mathrm{M}_k(\mathbb C)$ with $\exp(B)=A$ is not discrete. This happens for example if $A$ is diagonalisable and has a double eigenvalue. If in the question we require that for all $u\in U$ the eigenvalues of $f(u)$ are pairwise distinct, the answer would again be yes.