Given a unitary matrix $U$, how do I find $A$ such that $U=e^{iA}$? A unitary matrix $U \in \mathbb C^{n \times n}$ can always be written in exponential form
$$U = e^{iA} \tag{1}$$
where $A$ is Hermitian. My goal is to find the Hermitian matrix $A$, given the unitary matrix $U$. I figured out a way by diagonalizing $U$, in the following form:
$$U = V^{\dagger} [e^{ia_{kk}}] V$$
Therefore, we get
$$A = V^{\dagger} [a_{kk}] V$$
Is this the standard way for finding the Hermitian matrix $A$ in equation (1)? 
If I'd like to learn more about the exponentiation of unitary operators, and their general properties, what topics should I read?

Given $U=e^{iH}$, assume V diagonalizes H: 
$e^{V^{-1} iH V} = V^{-1} e^{iH} V = V^{-1} U V$
implying that V also diagonalizes U; 
hence, $V$ is easily found.
Let $\alpha_i$ denote the $i^{th}$ diagonal element of $V^{-1} U V$, then
$\alpha_i = e^{i \theta_i}$ where $e^{i \theta_i}$ is the $i^{th}$ diagonal element of $e^{V^{-1} iH V}$. Then, it easily follows that
$\theta_i = \tan^{-1}(Im(\alpha_i)/Re(\alpha_i))$ since the eigenvalues of U lie on the unit circle.
