 Solution

False; in fact no matrix with this spectrum is diagonalizable over $\mathbb{R}$. If a real invertible $P$ gave $P^{-1}AP=D$ with $D$ a real diagonal matrix, all eigenvalues of $A$ would be the real diagonal entries of $D$. But two of the required eigenvalues have nonzero imaginary parts. This contradiction provesThe real block form in the initial example is consistent with this: a two-dimensional rotation block is not a real diagonalization.