Solution to linear $2 \times 2$ system with complex eigenvalues When solving $y' = Ay$ where $A$ is a $2 \times 2$ (real) matrix with complex eigenvalues eigenvalues $\sigma, \overline{\sigma}$ (and corresponding eigenvectors $u$ and $\overline{u}$), we write $y(t)$ as $$y(t) = c_1 e^{\sigma t}u + c_2 e^{\overline{\sigma}t}\overline{u}  $$ $$= c_1 e^{\lambda t + i \omega t}(v+iw) + c_2e^{\lambda t - i\omega t}(v-iw)  $$ $$= e^{\lambda t}\left(((c_1 + c_2)\cos(\omega t) + i(c_1-c_2)\sin(\omega t))v + (i(c_1-c_2)\cos(\omega t) - (c_1+c_2)\sin(\omega t) )w \right) $$ $$= e^{\lambda t}((A\cos(\omega t) + B\sin(\omega t))v + (B\cos(\omega t)-A\sin(\omega t))w) $$
Supposedly, this can be written as $$y(t) = Ce^{\lambda t}(\cos(\omega t + \theta)v-\sin(\omega t + \theta)w), $$ with $C = \sqrt{A^2 + B^2}$.
I don't understand that last way of writing $y(t)$. Neither the part about $C$ nor the part about $\cos(\omega t + \theta)$.  
 A: $$e^{\lambda t}((A\cos(\omega t) + B\sin(\omega t))v + (B\cos(\omega t)-A\sin(\omega t))w)$$
Now for 
$$A\cos(\omega t) + B\sin(\omega t)$$
We can turn this into its equivalent 'harmonic' form by letting it be equal to $R\cos{(\omega t+\alpha)}$ for some constants $R, \alpha$. i.e. Let
$$A\cos(\omega t) + B\sin(\omega t)=R\cos{(\omega t+\alpha)}$$
$$=R(\cos{(\omega t)}\cos{(\alpha)}-\sin{(\omega t)}\sin{(\alpha)})$$
$$=R\cos{(\omega t)}\cos{(\alpha)}-R\sin{(\omega t)}\sin{(\alpha)}$$
So by comparing coefficients of $\cos{(\omega t)}$ and $\sin{(\omega t)}$ we then have
$$(1)\,\,R\cos{(\alpha)}=A$$
$$(2)\,\,R\sin{(\alpha)}=-B$$
$$(1)^2+(2)^2:R^2=A^2+B^2 \Rightarrow R=\sqrt{A^2+B^2}$$
$$(2)\,/\,(1): \tan{(\alpha)}=-\frac{B}{A}\Rightarrow \alpha=\arctan{(-\frac{B}{A})}$$
So we have that
$$A\cos(\omega t) + B\sin(\omega t)=R\cos{(\omega t+\alpha)}=\sqrt{A^2+B^2}\cos{(\omega t+\arctan{(-\frac{B}{A})})}$$
A similar method will give the other identity
$$B\cos(\omega t) - A\sin(\omega t)=-\sqrt{A^2+B^2}\sin{(\omega t+\arctan{(-\frac{B}{A})})}$$
A: That comes from the classical problem of writing a linear combination of $\cos$ and $\sin$ of the same angle $\omega t$ as a sine or a cosine of $\omega t$ plus a phase difference, thanks to the addition formulæ. It proceeds as follows:
Set $\;\alpha=\dfrac A{\sqrt{A^2+B^2}}$,  $\;\beta=\dfrac B{\sqrt{A^2+B^2}}\;$ and note that $\alpha^2+\beta^2=1$, so there exist a unique number $\theta$ $\;(0\le\theta<2\pi)$ such that \begin{cases}
\sin \theta=\dfrac A{\sqrt{A^2+B^2}},\\
\cos\theta= \dfrac A{\sqrt{A^2+B^2}}.
\end{cases}
Now we can rewrite the trigonometric part of the solution as
\begin{align}
A\cos(\omega t) + B\sin(\omega t)&=\sqrt{A^2+B^2}(\sin\theta\cos\omega t+ \cos\theta\sin\omega t)=C\sin(\omega t+\theta),\\
B\cos(\omega t) - A\sin(\omega t)&=\sqrt{A^2+B^2}(\cos\theta\cos\omega t - \sin\theta\sin(\omega t)=C\cos(\omega t+\theta)
\end{align}
