= Solution
If $Av=\lambda v$, then $A^nv=\lambda^nv$, and the <matrix exponential> series gives $Uv=e^\lambda v$. Also $(I+A)^{-1}v=(1+\lambda)^{-1}v$, since $1+\lambda\ne0$. Thus
$$
\boxed{\lambda_U=e^\lambda,\qquad\lambda_V=\frac{1-\lambda}{1+\lambda}.}
$$
The assumed diagonalization ensures that these account for all <eigenvalues>, including multiplicities. In particular, for $\lambda=i\theta$,
$$
\boxed{\lambda_U=e^{i\theta},\qquad\lambda_V=\frac{1-i\theta}{1+i\theta}=e^{-2i\arctan\theta},}
$$
both of <modulus> one, in agreement with the <unitary matrix> property.
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