Fundamental Set of Complex Solutions I am confused on how to do this problem, it states:
Find a fundamental set of solutions and put it in general form for the given system.
$x' = \begin{pmatrix} -1/2 & 1\\ -1 & -1/2  \end{pmatrix}x$
I got the eigen values to be $\lambda_1$ = [ (-1/2) + i ] , $\lambda_2$ = [ (-1/2) - i ] so the corresponding eigen vectors are 
For $\lambda_1 = v_1 = \begin{pmatrix} 1 \\ i \end{pmatrix}$
For $\lambda_2 = v_2 = \begin{pmatrix} 1\\ -i  \end{pmatrix}$
But here is where I get confused on how to write the general solution using Euler's formula for $e^{it} = cost + isint$, thus we have that 
$x_1(t) = e^{-t/2}(cost + isint)\begin{pmatrix} 1\\ i  \end{pmatrix}$
The answer is below, but how did they get that?
$x_1(t) = \begin{pmatrix} e^{-t/2} cost\\ -e^{-t/2}sint \end{pmatrix}  +    i\begin{pmatrix} e^{-t/2}sint\\
e^{-t/2}cost \end{pmatrix} = u(t) + iw(t)$
$u(t) = \begin{pmatrix} e^{-t/2} cost\\ -e^{-t/2}sint \end{pmatrix}  ,  w(t) =   \begin{pmatrix} e^{-t/2}sint\\
e^{-t/2}cost \end{pmatrix}$
 A: Based on your derivations, the general solution has the form
$$ x(t) = c_1 \begin{pmatrix} 1 \\ i \end{pmatrix} {\rm e}^{(-\frac{1}{2}+i)t} + c_2 \begin{pmatrix} 1\\ -i  \end{pmatrix} {\rm e}^{(-\frac{1}{2}-i)t}\,, $$
where $c_1,c_2$ are arbitrary constants. Simplifying further
$$ x(t) = c_1 {\rm e}^{-\frac{1}{2}t} \begin{pmatrix} 1 \\ i \end{pmatrix} {\rm e}^{it} + c_2 {\rm e}^{-\frac{1}{2}t} \begin{pmatrix} 1\\ -i  \end{pmatrix} {\rm e}^{-it} $$
$$= c_1{\rm e}^{-\frac{1}{2}t} \begin{pmatrix} 1 \\ i \end{pmatrix}(\cos(t)+i\sin(t)) + c_2 {\rm e}^{-\frac{1}{2}t} \begin{pmatrix} 1\\ -i  \end{pmatrix} (\cos(t)-i\sin(t))\,. $$
Now, make the above as a linear combination of $\cos(t){\rm e}^{-\frac{1}{2}t}$ and $\sin(t){\rm e}^{-\frac{1}{2}t}$. 
$$ A\cos(t){\rm e}^{-\frac{1}{2}t} + B \sin(t){\rm e}^{-\frac{1}{2}t}\,, $$
where $A$ and $B$ are two constant vectors given by
$$ A = c_1\begin{pmatrix} 1 \\ i \end{pmatrix}+c_2 \begin{pmatrix} 1\\ -i  \end{pmatrix} \,,\quad B = ic_1\begin{pmatrix} 1 \\ i \end{pmatrix}+ic_2 \begin{pmatrix} 1\\ -i  \end{pmatrix} \,. $$
Or, you can write them in terms of new constants,
$$ A= \begin{pmatrix} a_1 \\ a_2 \end{pmatrix}\,, \quad B = \begin{pmatrix} b_1 \\ b_2 \end{pmatrix} \,.$$
