Let real $A$ have eigenvalue $a+ib$ with $b\ne0$ and eigenvector $u+iv$. Then $Au=au-bv$ and $Av=bu+av$. With $P=[u\ v]$, $P^{-1}AP=\begin{bmatrix}a&b\\-b&a\end{bmatrix}$. This is a real block form, not a real diagonalization. Over $\mathbb C$, the eigenvectors $u+iv,u-iv$ diagonalize $A$ with diagonal entries $a+ib,a-ib$.
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