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?