I'm looking for a simpler proof for the following exercise :
I have $5$ matrices, denoted by $M_i$, of size $2\times 2$ (with real coefficients) and I need to find, for each matrix $M_i$, an invertible matrix such that $$S_i^{-1}M_i S_i = \begin{pmatrix} a_i & -b_i \\ b_i & a_i \end{pmatrix}$$ where $a_i$ and $b_i$ are real numbers. My idea is the following one : I get $a_i$ since $\text{tr } M_i=2a_i$ and then I can find $b_i$ using the determinant. Finally, I set $S_i = \begin{pmatrix} e_i & f_i \\ g_i & h_i\end{pmatrix}$ where $e_i$, $f_i$, $g_i$ and $h_i$ are unknowns and then writing the previous equation as $$ M_iS_i = S_i\begin{pmatrix} a_i & -b_i \\ b_i & a_i \end{pmatrix} $$ I can get a linear system of equations to finally get the matrix $S_i$. Do you see a simpler way to solve this exercise ?
If $M$ has an eigenvector $a+bi$ (where $i=\sqrt{-1}$) with corresponding eigenvector $v+iw$ (where vectors $v$ and $w$ have real entries) then let $S$ be the matrix whose columns are $v$ and $w$.
EDIT: For example: suppose $$M=\pmatrix{1&2\cr-4&5\cr}$$ The characteristic equation is $\lambda^2-6\lambda+13=0$, so $\lambda=3+2i$ is an eigenvalue. Then $$M-\lambda I=\pmatrix{-2-2i&2\cr-4&2-2i\cr}$$ and an eigenvector is given by $$(1,1+i)=(1,1)+i(0,1)$$ So we can take $$S=\pmatrix{1&0\cr1&1\cr}$$ See what you get when you calculate $S^{-1}MS$.