 Solution

The anticommuting unequal Pauli matrices cancel the cross terms, leavingAlso $\mathrm{tr}T=0$, so its eigenvalues are $\pm |\alpha |$. ConsequentlyFor $|\alpha |>0$, the matrix exponential for a scalar square also gives $U=I\cos (|\alpha |/2)+iT\sin (|\alpha |/2)/|\alpha |$. At $\alpha =0$, its continuous limit is $U=I$. The two exponential eigenvalues coincide when $|\alpha |=2\pi k$, making $U=(-1{)}^{k}I$.