Let $M \in \mathbb{C}^{n \times n}$ be a complex-symmetric $n \times n$ matrix. That is, $M$ is equal to its own transpose (without conjugation). If the real part of $M$ is positive-definite, then is $M$ necessarily diagonalizable?
Of course, if we drop the positive-definiteness requirement, then there are examples of non-diagonalizable complex-symmetric matrices. For example,
$$\begin{bmatrix} 1 & i \\ i & -1 \end{bmatrix}$$
is one such example. However, its real part $\begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}$ is clearly not positive-definite.
Best Answer
The answer is no. Based on your example, $$ \pmatrix{3&i\\i&1} $$ will not be diagonalizable, but has a positive definite real part.