Assuming $U = A + iB$, where $A$, $B$ are hermitian and $U$ is unitary, show that $A^2 + B^2 = I$ Let $A$ and $B$ be n x n hermitian matrices and $U$ an n x n unitary matrix.
Assuming that $U = A + iB$, I'm being asked to show the following:
(i) $A^2 + B^2 = I$
(ii) $[A,B] = AB - BA = 0$
My problem is that I can't seem to show (i) to be true without first assuming (ii). My working is as follows:
$$UU^\dagger = (A+iB)(A+iB)^\dagger = (A+iB)(A^\dagger - iB^\dagger) = (A+iB)(A-iB) = A^2 + B^2 + i(BA-AB) = I$$
So clearly (i) holds provided that the commutator $[A,B]=0$. But how do I show (i) without needing to assume that $A$ and $B$ commute? Alternatively, how can I show that they commute without having to assume (i)? 

$$\begin{aligned}
(A+iB)(A^\star-iB^\star)
  &=AA^\star+iBA^\star-iAB^\star + BB^\star\\
  &=AA^\star +i(BA^\star-AB^\star) + BB^\star\\
  &=A^2+B^2+i(BA^\star-AB^\star)
\end{aligned}$$
Also
$$\begin{aligned}
(A^\star-iB^\star)(A+iB)
  &=AA^\star-iB^\star A+iA^\star B + BB^\star\\
  &=AA^\star-i(B^\star A-A^\star B) + BB^\star\\
  &=A^2+B^2-i(B A^\star-A B^\star)
\end{aligned}$$
Since $U = A+iB$ is unitary, we know that $U^\star$ is its inverse and so we know that $I=UU^\star=U^\star U$ and so
$$(A+iB)(A^\star-iB^\star)=(A^\star-iB^\star)(A+iB).$$
Therefore we can combine these pieces of information with the above derivations (subtract both) to conclude that
$
2i (BA^\star-AB^\star) = 0.
$
Then we find
$$
(A+iB)(A^\star-iB^\star) = A^2 + B^2.
$$
Combine with $I=UU^\star$ and you have the result.
