# For any unitary matrix $U$, there exists a Hermitian matrix $H$ such that $U=e^{iH}$

For any unitary matrix $$U$$, there exists a Hermitian matrix $$H$$ such that $$U=e^{iH}$$.

Is the above statement true? (I guess not)

If not, how "many" unitary matrixes can the expression $$U=e^{iH}$$ cover?

The statement is true. Any unitary matrix $U$ can be written as $U=e^{iH}$ for some $H$. However, $H$ is not unique, since taking $H'=H+2\pi I$, we have $e^{iH'}=e^{iH}=U$.
To see why the statement is true, simply write $U$ in its diagonal form (we can do this because of Spectral Theorem): $$U=VDV^\dagger$$ where $D=\text{diag}(e^{i\theta_1},e^{i\theta_2},\ldots)$, where $\theta_j\in\mathbb{R}$, and $e^{i\theta_j}$ are eigenvalues of $U$ which by definition of unitarity must be complex numbers with modulus 1. Then one choice of $H$ is $$H=\frac{1}{i}\ln U = \frac{1}{i}V \ln(D)V^\dagger = V\text{diag}(\theta_1,\theta_2,\ldots)V^\dagger$$ It isn't hard to check $e^{iH}=U$, and $H^\dagger = H$ is Hermitian since $\theta_j$ are real.