Let $A_1,A_2\in GL(n,\mathbb C)$ commutes, show that $\log(A_1)$ commutes with $\log(A_2)$ (for some pair) 
Let $A_1,A_2\in GL(n,\mathbb C)$ be two commuting invertible matrix, we want to show $\log(A_1)$ commutes with $\log(A_2)$ (for some pair).

To be more precise, let $A\in GL(n,\mathbb C)$, we define $\log(A)$ to be the preimage of $A$ under exponential map, i.e., the set of matrix $B\in M_{n\times n}(\mathbb C)$ such that $\exp(B):=\sum_{k\ge 0}B^k/k!=A$. We want to show there exists $B_i\in \log(A_i)$, $i=1,2$, such that $B_1B_2=B_2B_1$.
Note that when $A_1, A_2$ are unipotent, the proof is easy: The canonical logarithm is given by the Taylor series $$B_i=\sum_{k\ge 1}(-1)^{k+1}\frac{(Id-A_i)^k}{k}$$ which is actually a finite sum, and $B_1$ commutes with $B_2$ follows from $(Id-A_1)^k$ commutes with $(Id-A_2)^k$.
In the general situation, the difficulty is that the representation of matrix logarithm is complicated.
To find a logarithm for $A\in GL(n,\mathbb C)$, first by the identity $S^{-1}\exp(B)S=\exp(S^{-1}BS)$, it reduce to assume that $A=\mathrm {diag}(J_1,...,J_p)$ is in the standard Jordan form, where $$J_i=\begin{bmatrix}
    \lambda_i       & 1 &    &  \\
           & \lambda_i &   \ddots  &  \\
           &  &  \ddots  & 1\\
           &  &      & \lambda_i
\end{bmatrix}$$ 
is a standard Jordan block with eigenvalue $\lambda_i$ $(1\le i\le p)$.
Note that $\lambda_i\neq 0$, $J_i/\lambda_i$ is unipotent, so its logarithmic is (multivalued) $ \log(\lambda_i)\log(J_i/\lambda_i)$.
So the possible logarithm of $A\in GL(n,\mathbb C)$ is 
$$B=S^{-1}\begin{bmatrix}
    \log(\lambda_1)\log(J_1/\lambda_1)       &  &    &  \\
           & \log(\lambda_2)\log(J_2/\lambda_2) &   \ddots  &  \\
           &  &  \ddots  & \\
           &  &      & \log(\lambda_p)\log(J_i/\lambda_p)
\end{bmatrix}S$$ 
I stopped here... Hope someone could help.
 A: Let $\theta\in\mathbb{R}$ s.t. for every $\lambda_1\in spectrum(A_1),\lambda_2\in spectrum(A_2)$, $arg(\lambda_1)-\theta\notin 2\pi \mathbb{Z},arg(\lambda_2)-\theta\notin 2\pi\mathbb{Z}$. 
Note that $\log_{\theta}: re^{i\alpha}\in\{re^{i\alpha}\in\mathbb{C};r>0,\alpha\in(\theta,\theta+2\pi)\}\rightarrow \log(r)+i\alpha$ is a holomorphic function; then $\log_{\theta}(A_i)$ can be defined using the Hermite's interpolation (cf. Higham, functions of matrices); therefore $e^{\log_{\theta}(A_i)}=A_i$. Then $\log_{\theta}(A_i)$ is a polynomial in $A_i$ and, finally, $\log_{\theta}(A_1)$ and $\log_{\theta}(A_2)$ commute. 
Remark. That does not imply that $\log_{\theta}(A_1)+\log_{\theta}(A_2)$ can be written in the form  $\log_{\beta}(A_1A_2)$. For example, let $A_1=A_2=\begin{pmatrix}0&1\\-1&0\end{pmatrix}$. Then $\log_{-\pi}(A_i)=\begin{pmatrix}0&\pi/2\\-\pi/2&0\end{pmatrix}$ and $\log_{-\pi}(A_1)+\log_{-\pi}(A_2)=\begin{pmatrix}0&\pi\\-\pi&0\end{pmatrix}$; on the  other hand, $A_1A_2=-I_2$ and any $\log_{\beta}(-I)$ ($\beta\notin -\pi+2\pi\mathbb{z}$) is a scalar matrix.
