= Solution
The anticommuting unequal <Pauli matrices> cancel the cross terms, leaving
$$
T^2=(\alpha_1^2+\alpha_2^2+\alpha_3^2)I=|\alpha|^2I.
$$
Also $\operatorname{tr}T=0$, so its <eigenvalues> are $\pm|\alpha|$. Consequently
$$
\boxed{\lambda_\pm(U)=e^{\pm i|\alpha|/2}.}
$$
For $|\alpha|>0$, 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 $U=(-1)^kI$.
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