My teacher has given us that
If $f$ is normal then it is diagonalizable
The thing is that I've studied all 4 operators, orthogonal, unitary, symmetric and self-adjoint operators and I've been told that all of them are normal operators, since all satisfy that $ff^{*}=f^{*}f$. But I have also been told that the orthogonal operators, which are the ones that satisfy $f^{*}=f^{-1}$ (in the real plane) and are not always diagonalizable. So what's going on here? Is it that the theorem is "if $f$ is normal and complex then it is diagonalizable"? Or what is going on so that both statement don't correlate?
Consider $$O=\begin{pmatrix} \cos\theta&-\sin\theta\\ \sin\theta&\cos\theta \end{pmatrix} $$
Then $O^t O = O^* O = O O^* = O O^t = \mathbf{1}$. So $O$ is normal. $$ O \begin{pmatrix} 1\\ i \end{pmatrix} = \begin{pmatrix} e^{-i\theta}\\ i e^{-i\theta} \end{pmatrix} = e^{-i\theta} \begin{pmatrix} 1\\ i \end{pmatrix} $$
$$ O \begin{pmatrix} 1\\ -i \end{pmatrix} = \begin{pmatrix} e^{i\theta}\\ -i e^{i\theta} \end{pmatrix} = e^{i\theta} \begin{pmatrix} 1\\ -i \end{pmatrix} $$ The eigenvalues $e^{i\theta}, e^{-i\theta}$ are non-real (except for $\theta\in\{0,\pi\}$) and associated eigenvectors have non-real entries. $O$ is not diagonalizable over $\Bbb{R}$. For the cases $\theta=0$ or $\theta=\pi$, $O=\pm \mathbf{1}$ so all vectors (including real-valued vectors) are eigenvectors for it.