How find the general solution to a linear differential system with complex coefficients Find the general solution to the following system:
$$\dot{\textbf{x}}=\begin{pmatrix}
0 & \mathrm{i}\\
\mathrm{-i} & \mathrm{2i}
\end{pmatrix}\textbf{x}$$
Let $\textbf{A}=\begin{pmatrix}
0 & \mathrm{i}\\
\mathrm{-i} & \mathrm{2i}
\end{pmatrix}$. I want to solve this system by finding the eigenvalues $\lambda_1,\lambda_2$ and the corresponding eigenvectors $\textbf{v}_1, \textbf{v}_2$ of $\textbf{A}$, but when I take a step further to write the solution, I have no idea whether I can write the solution into real and imaginary parts and omit the imaginary unit $\mathrm{i}$ of the imaginary part as we often do for linear differetial system with real coefficients.
If I can't, is it correct to reserve $\mathrm{i}$? Or what should I do?
Any help appreciated!
 A: There are no nonzero real-valued solutions because the differential equation would imply that the derivative of such a solution is nonreal. You should leave your answer as a complex-valued function.
Bear in mind that this matrix is not diagonalizable, so finding eigenvectors and eigenvalues alone is not enough.
A: You cannot take the real part. 
Your matrix is not diagonalizable, so you have to look into the generalized eigenspaces. But in this case  $(A-iI)^2=0$, which simplifies things a lot. 
The solution of the system is $\mathbf x(t)=e^{At}\mathbf x_0 $. In this particular case, since $(A-i)^2=0$, we have
$$
e^{(A-i)t}=I+(A-iI)t,
$$
so
$$
e^{At}=e^{(A-i)t}e^{it}=e^{it}\,I+t\,e^{it}(A-iI)=(1-it)e^{it}\,I+te^{it}A
=\begin{bmatrix}(1-it)e^{it}&ite^{it}\\-ite^{it}&(1+it)e^{it}\end{bmatrix}.
$$
So now you write $\mathbf x_0=\begin{bmatrix}c_1\\ c_2\end{bmatrix}$, and the solution becomes 
$$
\mathbf x(t)=e^{At}\mathbf x_0=\begin{bmatrix}(1-it)e^{it}&ite^{it}\\-ite^{it}&(1+it)e^{it}\end{bmatrix}\begin{bmatrix}c_1\\ c_2\end{bmatrix}=\begin{bmatrix}
c_1(1-it)e^{it}+c_2ite^{it}\\ -c_1ite^{it}+c_2(1+it)e^{it}.
\end{bmatrix}
$$
