The Taylor expansion of the exponential function is given by
$$ e^{-ix} = \sum_n^N \dfrac{(-ix)^n}{n!} = 1-ix -\dfrac{x^2}{2!} + \ ... $$
suppose the $x=25$, so if I want to use the above equation to estimate
$$ e^{-25i} = \sum_n^N \dfrac{(-25i)^n}{n!} $$
for converging I need around $100$ terms i.e. $N=100$. But if I use the following trick $$ e^{-25i} = e^{-6 \pi i} e^{-6.15i} = e^{-6.15i} $$ I would need a much fewer number of orders to converge $N=10$.
Now my question is whether is there any similar way to do when we have a matrix instead of a number? $$ e^{-iA} = ? $$ where $A$ is a matrix, in my case it is a Hermitian matrix.
I think it is difficult, but maybe someone has a good way.
For any integer $k$, we have $$ e^{iA} = e^{i(2 \pi k I)} \cdot e^{i (A - 2 \pi k I)} = e^{i(A - 2 \pi k I)}. $$ With that, it would suffice to select $k$ so that the eigenvalues of $A - 2 \pi k I$ are smaller in magnitude. Note that for any eigenvalue $\lambda$ of $A$, $\lambda - s$ is an eigenvalue of $A - s I$.