How to write the General solution of the homogeneous system with complex entries in the Eigan vectors in terms of real functions?

Question

I have a function:

So I know that

$ e^{it} = cos(t) + isin(t) $

I'm solving a homogeneous system and it has complex entries in 2 of the eigan vectors. So I need to re-write these in terms of real functions. So far I have this:

$ y = c_1e^{t} \begin{bmatrix} 0 \\ 2 \\1 \end{bmatrix} + c_2e^{(1+i)t}\begin{bmatrix} i \\ 1 \\1 \end{bmatrix} + c_3e^{(1-i)t}\begin{bmatrix} -i \\ 1 \\1 \end{bmatrix} $

For the $c_2$ part,

$ e^{t}e^{it}\begin{bmatrix} i \\ 1 \\1 \end{bmatrix} = e^{t}(cos(t) + isin(t))\begin{bmatrix} i \\ 1 \\1 \end{bmatrix} $

$ =e^t(cos(t)\begin{bmatrix} 0 \\ 1 \\1 \end{bmatrix} + isin\begin{bmatrix} 1 \\ 0 \\0 \end{bmatrix}) $

For the $c_3$ part, I get $e^{-it}$. How do I convert that to a real function? It looks like it would be:

$(e^{it})^{-1} = (cos(t) + isin(t))^{-1}$

Am I on the right track? Is there any easier way to do this?

Answer 1

Hint:

$$e^{-it} = \cos(-t) + i\sin(-t)=\cos(t) - i\sin(t)$$