a) Consider an example:

$A = \begin{bmatrix} cos\theta & -sin\theta \\ sin\theta & cos\theta \end{bmatrix}$

The eigen value of A are found by solving $D = |A - \lambda I_{2}| = (cos\theta - \lambda)^2 + sin^2\theta = 0 $

Giving eigen values $\lambda = cos\theta + isin\theta = e^{i\theta} \\ \lambda ^* = cos\theta -iIsin\theta = e^{-i\theta} $

The modules of $\lambda$ and $\lambda^*$ is 1.

b) If A is unitary the $A^* A = I$

Thus if $Ax = \lambda x$ then $x^* A^*= \lambda^* x^*$

Hence $x^* x = x^* A * A * x = X*X x*x$

Since $x^*x \neq 0$ we obtain $x^*\lambda = 1$

Hence $|\lambda| = 1$