Why Are Eigenvalues of Unitary Operators Pure Phases?

[black_hole]

Homework Statement

We only briefly mentioned this in class and now its on our problem set...

Show that all eigenvalues i of a Unitary operator are pure phases.
Suppose M is a Hermitian operator. Show that e^iM is a Unitary operator.

Homework Equations

The Attempt at a Solution

Uf = λf where is is an eigenfunction, U dagger = U inverse
multiply by either maybe...

 

[qbert]

Uf = λf

denote your inner product (a,b)

a.
(f, U^\dagger U f) = ?
calculate this two ways (in terms of λ and λ^*)
(f, U^\dagger U f ) = (Uf, Uf) = ?
(f, U^\dagger U f ) = (f, f) [since U^\dagger = U^{-1} ]
so what does this say about λ and λ^*?

b.
The second part follows from

1. \left( e^A \right)^\dagger<br /> = \left( 1 + A + \frac{1}{2!}A^2 + \cdots \right)^\dagger<br /> = \left( 1 + A^\dagger + \frac{1}{2!}(A^\dagger)^2 + \cdots \right)<br /> = e^{(A^\dagger)}.

2. For commuting matrices (operators)
e^A e^B = e^{(A+B)} .

now you need to show show U = e^(iM) satisfies UU^(dagger) = 1.

can you fill in the rest?