The answer is "no" in general. As Denis suspects, the problem is a global one, and it involves matrices with nontrivial Jordan blocks. These have, in a sense, "fewer" logarithms than the commoners. Concretely, I clain that the holomorphic function $$f(z) = \begin{pmatrix} e^{2\pi i z} & 1 \\ 0 & 1 \end{pmatrix}$$ has no holomorphic logarithm on $\mathbb C$. If it had one, there would also be a holomorphic square root of $f$ on $\mathbb C$, and not even that exists. Indeed, suppose by contradiction that there was a function $g:\mathbb C \to \mathrm{GL}_2(\mathbb C)$ such that $f(z) = g(z)^2$. The matrix $$f(0) = g(0)^2 = \begin{pmatrix} 1 & 1 1\\ 0 & 1 \end{pmatrix}$$ end{pmatrix} $$has only two square roots (a 2-by-2 matrix with distinct eigenvalues has four square roots!) differing by a sign, so we may suppose$$g(0) = \begin{pmatrix} 1 & 1/2 \\ 0 & 1 \end{pmatrix}$$by changing g to -g if necessary. If we move z on the real line from 0 to 1, we find by continuity of g$$g(z) = \begin{pmatrix} e^{\pi i z} & (e^{\pi i z}+1)^{-1} \\ 0 & 1 \end{pmatrix}$$and run into a pole as z approaches 1, end of story. 1 The answer is "no" in general. As Denis suspects, the problem is a global one, and it involves matrices with nontrivial Jordan blocks. These have, in a sense, "fewer" logarithms than the commoners. Concretely, I clain that the holomorphic function$$f(z) = \begin{pmatrix} e^{2\pi i z} & 1 \ 0 & 1 \end{pmatrix}$$has no holomorphic logarithm on \mathbb C. If it had one, there would also be a holomorphic square root of f on \mathbb C, and not even that exists. Indeed, suppose by contradiction that there was a function g:\mathbb C \to \mathrm{GL}_2(\mathbb C) such that f(z) = g(z)^2. The matrix$$f(0) = g(0)^2 = \begin{pmatrix} 1 & 1 \ 0 & 1 \end{pmatrix}$$has only two square roots (a 2-by-2 matrix with distinct eigenvalues has four square roots!) differing by a sign, so we may suppose$$g(0) = \begin{pmatrix} 1 & 1/2 \ 0 & 1 \end{pmatrix}$$by changing g to -g if necessary. If we move z on the real line from 0 to 1, we find by continuity of g$$g(z) = \begin{pmatrix} e^{\pi i z} & (e^{\pi i z}+1)^{-1} \ 0 & 1 \end{pmatrix} and run into a pole as $z$ approaches $1$, end of story.