It is easy to see that within the disk algebra $A(D)$
 $$\Delta(z):= \begin{pmatrix} 1&0\\z&1 \end{pmatrix}\; \begin{pmatrix} 1&1\\0&1 \end{pmatrix}=
 \begin{pmatrix} 1&1\\z&1+z
 \end{pmatrix} $$
 is a product of two exponential matrices. Is $\Delta(z)$ itself an exponential matrix of a holomorphic matrix $M(z)$ ?
 I don't think so, but could not come up with a proof. Is there a general method to deal with these questions?



Logarithms of matrices in the disk-algebra
$\Delta(z)$ does not have a nonpositive real eigenvalue for $|z| < 4$, so the principal branch of the logarithm is defined and analytic on a neighbourhood of its spectrum, and thus the holomorphic functional calculus produces the desired analytic logarithm $M(z)$ for $|z| < 4$.


