2. Remember that when you diagonalize a matrix B, the diagonal values of the resulting diagonalized matrix will be the eigenvalues of B, and that a matrix P such that $P^{-1}BP$ is diagonal will have as it's columns the eigenvectors corresponding to those eigenvalues. Thus, we require that B have real eigenvectors. As our matrix is real, real eigenvalues are a necessary condition for real eigenvectors.
Note that $\det(I\lambda-B)=0$ gives $(\lambda-a)^2+b^2=0$.