Cayley Transform and Eigenvalues I have a particular operator, namely $A=-i\frac{d}{dx}$ that I would like to Cayley transform. $A$ is defined on the Hilbert space $L^{2}[0,1]$ and has domain $\mathcal{D}_{\alpha}=\{g:g \in C^{\infty}, \, g(1)=e^{i \alpha}g(0)\}$. 
I can find the eigenfunctions and eigenvalues for $A$:
Eigenfunctions: $\, f^{\alpha}_{n}(x)=e^{i(2 \pi n + \alpha)x}, \, n \in \mathbb{Z},$
Eigenvalues: $\, 2 \pi n + \alpha, \, n \in \mathbb{Z}.$
As previously stated, my interest is in the Cayley transform of $A$, namely
$$U(A)=(A-i)(A+i)^{-1}.$$
The problem that I am having is computation of the inverse $(A+i)^{-1}$. I have only been able to write down the trivial steps 
$$(A+i)^{-1}=(-i\frac{d}{dx}+i)^{-1}=(-i(\frac{d}{dx}-1))^{-1},$$
and am stuck at this point. To elucidate my motivation for the calculation I am searching for the eigenvalues of the Cayley transform $U(A).$

If $$(-i\frac{\partial}{\partial x}+i)=g$$
then 
$$(1-\frac{\partial}{\partial x})f=-ig$$
$$f=\sum_{k=0}^{\infty}-i\frac{\partial^n}{\partial x^n}g$$
Now $h(t)=e^{-it\alpha}g(t)$ satisfies $h(0)=h(1)$. Writing $h$ as a Fourier series gives you $g$ as a sum of exponentials for which the expression for $(A+i)^{-1}$ above is easy to compute.
