 Solution

If $Av=\lambda v$, then $A^{n}v={\lambda }^{n}v$, and the matrix exponential series gives $Uv=e^{\lambda }v$. Also $(I+A{)}^{-1}v=(1+\lambda {)}^{-1}v$, since $1+\lambda \ne 0$. ThusThe assumed diagonalization ensures that these account for all eigenvalues, including multiplicities. In particular, for $\lambda =i\theta$,both of modulus one, in agreement with the unitary matrix property.