Eigenvalue for complex variable I was trying to reproduce the results of an exercise where they calculate the normal modes of oscillation.
$$\begin{pmatrix} 
   \dfrac{d}{dt}C   \\ \dfrac{d}{dt}C^{*}  \end{pmatrix}= - \dfrac{1}{i} \begin{pmatrix}1/2+m& 1/2   \\
  -1/2 & -1/2-m   \end{pmatrix} \begin{pmatrix} 
   C   \\
   C^{*}   \\
   \end{pmatrix}.$$
If I suppose the ansatz  $C = A e^{iwt}$ and $C^{*}= A^{*} e^{-iwt}$, i found the following expression:
$$ i \begin{pmatrix}1/2+m-w& 1/2   \\
  -1/2 & -1/2-m+w   \end{pmatrix} \begin{pmatrix} 
   A   \\
   A^{*}   \\
   \end{pmatrix} = 0.$$
The problem is diagonalizing a 2x2 matrix. On the other hand, in the book, it is assumed an ansatz $C = A e^{iwt}$ and $C^{*}= B e^{iwt}$ that have different eigenvalue, what is the correct way form found the eigenvalue of the problem?
 A: Another way of finding the solution here is to write your system of equations as $\frac{d}{dt} \mathbf{x}(t) = \mathbb{M} \mathbf{x}(t)$, where
$$
\mathbf{x}(t) := \left[ \begin{matrix} C(t) \\ C^{\ast}(t) \end{matrix} \right] \quad \mathrm{and} \quad \mathbb{M} := - \frac{i}{2} \left[ \begin{matrix} 1 + 2m & 1 \\ -1 & - 1 - 2 m \end{matrix} \right] \ .
$$
Now diagonalize the matrix $\mathbb{M}$ ie. find the eigenvalues $\lambda_{\pm}$ and eigenvectors $\mathbf{v}_{\pm}$ such that $\mathbb{M} \mathbf{v}_{\pm} = \lambda_{\pm}\mathbf{v}_{\pm}$ . After some work, the eigenvalues end up being
$$
\lambda_{\pm} = \pm i \sqrt{ m^2 + m }
$$
and the corresponding eigenvectors are 
$$
\mathbf{v}_{\pm} = \left[ \begin{matrix} - 1 - 2 m \pm 2\sqrt{m^2 +m} \\ 1 \end{matrix} \right] .
$$
You can now write your solution in an ansatz of the form $\mathbf{x}(t) = c_{-} \mathbf{v}_{-} e^{\lambda_{-} t} + c_{+} \mathbf{v}_{+} e^{\lambda_{+} t}$ for some yet-to-be determined integration constants $c_{\pm}$ (you should check this explicitly: it's easy to prove that this expression satisfies $\frac{d}{dt} \mathbf{x}(t) = \mathbb{M} \mathbf{x}(t)$).
Impose the initial condition $\mathbf{x}(0) = \left[ \begin{matrix} C(0) \\ C^{\ast}(0) \end{matrix} \right]$ and using the ansatz you get an equation $\mathbf{x}(0) = c_{-} \mathbf{v}_{-} + c_{+} \mathbf{v}_{+}$, which is explicitly:
$$
\left[ \begin{matrix} C(0) \\ C^{\ast}(0) \end{matrix} \right] \ = \ c_{-} \left[ \begin{matrix} - 1 - 2 m - 2\sqrt{m^2 +m} \\ 1 \end{matrix} \right] + c_{+} \left[ \begin{matrix} - 1 - 2 m + 2\sqrt{m^2 +m} \\ 1 \end{matrix} \right]
$$
which determines the coefficients $c_{\pm}$. When you have these, you can plug these back into the ansatz $\mathbf{x}(t) = c_{-} \mathbf{v}_{-} e^{\lambda_{-} t} + c_{+} \mathbf{v}_{+} e^{\lambda_{+} t}$ to get your functions of time.
At the end of the day you get the rather ugly solution
$$
C(t) = C(0) \left[ \cos( \sqrt{m^2 +m} \; t ) -  \frac{i(1+2m)}{2\sqrt{m^2+m}} \sin(\sqrt{m^2 +m} \; t) \right] - C^{\ast}(0) \frac{i}{2\sqrt{m^2+m}} \sin(\sqrt{m^2 +m} \; t) 
$$
and you just need to take the complex conjugate of this to get your solution $C^{\ast}(t)$.
