$$ A=\begin{bmatrix} 4 & -15 & 6 \\ 1 & -4 &2 \\ 1 & -5 &3 \end{bmatrix} $$ Find natural logarithm of $A$.
I notice $Sp_A=\{1,1,1\}$. So $(A-E)^3$ must be zero.
$\ln(E+(A-E))=(A-E)-\frac{(A-E)^2}{2}$
Calculating, I get just a regular matrix with integers, but the key in the book says
$$\begin{bmatrix} 3 & -15 & 6 \\ 1 & -5 &2 \\ 1 & -5 &2 \end{bmatrix} + 2\pi in E, n\in \mathbb{Z}.$$
Where do $\pi$ and $i$ come from?
On
The principal $\log$ of $A$ is $A-I_3$. Yet, there are many other logarithms of $A$.
$A$ is similar to $B=diag(I_2+J_2,1)$ where $J_k$ is the nilpotent Jordan block of dimension $k$. Then the $\log$'s of $B$ may have two types
Type 1. Primary functions (cf. Higham, Functions of matrices); they are polynomials in $B$.
$B-I_3+2i\pi k I_3$, $k\in\mathbb{Z}$.
Type 2. Non-primary functions; they are not polynomials in $B$.
EDIT. $Udiag(J_2+2i\pi lI_2,2i\pi k)U^{-1}$ where $UB=BU$ and $k,l\in \mathbb{Z}$.
Let's first look at $1\times 1$ matrices instead. Those are some times more intuitive. For instance, the natural logarithm of the matrix $[e]$ is $[1]+2\pi i n[1]$ because $e^{1+2\pi in}=1$ for any integer $n$.
In your case it's exactly the same thing happening, only your matrices are $3\times 3$.