Let $A,B$ $\in GL_{n}(\mathbb{C})$ and $[A,B] = AB-BA = 0$. My question is about the existence of a $b \in M_{n}(\mathbb{C})$ such that $B = \exp(b) $ and $[A,b] =0 $.
Note that in general $[A,\exp(b)]=0$ does not imply $[A,b]=0$. As an example consider the matrices $A = \begin{bmatrix}1&i\\i&2\end{bmatrix}$ $B= I$ and $b = \begin{bmatrix}2 \pi i&0\\0&0\end{bmatrix} $. We have $[A, \exp(b)] = 0$ but $[A,b] \neq 0$. However, we can replace $b$ with $\widetilde{b}=0$.
I believe in general this could have something to do with the locus in $M_{n}(\mathbb{C})$ where $\exp$ is a local isomorphism. This is the locus of matrices $X$ which have $\lambda_{i} - \lambda_{j} \neq 2 \pi i k $ for $k \in \mathbb{Z} - 0 $ for any two eigenvalues $\lambda_{i} , \lambda_{j}$ of $X$ . Note that $\exp$ is still surjective when restricted to this locus. Namely, perhaps if we restrict to preimages of $B$ under the exponential in this locus then perhaps the matrices commute. In particular, in the example above $b$ is not in this locus.