Trace of Hermitian Matrix

Question

please tell me where I am wrong in the following chain of reasoning
1. trace of a matrix isn't affected by unitary change of basis
2. therefore trace of hermitian matrix is the sum of eigenvalues
3. if $U=e^{iH}$ where $H$ is hermitian, then $U$ is unitary
4. $\det U=e^{i\times Tr{H}}$
5. But $\det U=1$.
6. therefore $Tr(H) = 2n\pi$ which I think is false

Answer 1

An unitary does not necessarialy have $\det U = 1$, we can only say that $|\det U| = 1$, which implies $\mathrm{tr}\, H \in \mathbf R$, which is correct, due to the fact that all diagonal entries of $H$ are real.