So I have been reading the paper "Thermalization for Perturbations of Dynamical Systems" by G. Barrera and M. Jara (you can find the paper here) and there is a step that I don't quite understand in the proof of collorary 2.12 which states:
Write $\gamma=\lambda +i \theta$, where $\lambda>0$ with $\theta \neq 0$. To the eigenvalue $\gamma$ we associated an [complex] eigenvector $u_1,u_2\in \mathbb{R}^2$. An straightforward computation shows: $$e^{\lambda t }e^{-Qt}x_0=(c_1 cos(\theta t) -c_2 sin (\theta t))u_1 + (c_1 sin(\theta t) +c_2 cos(\theta t))u_2$$ for any $t\geq 0$, where $c_1:=c_1(x_0)$ and $c_2:=c_2(x_0)$ are not both zero. Notice that $c:=\sqrt{c_1^2+c_2^2}>0$ and let $cos(\alpha)=c_1/c$ and $sin(\alpha)=c_2/c$. Then $$e^{\lambda t }e^{-Qt}x_0=c cos(\theta t + \alpha)u_1+c sin(\theta t + \alpha)u_2$$ for any $t\geq 0$.
Now, following what I believe the paper refers to as a "a straightforward computation" is that for dimention 2, if $Q$ has two complex conjugate eigen vectors $v_1=u_1-i u_2$ and $v_2=u_1+i u_2$, then $$e^{\lambda t }e^{-Qt}x_0=\sum_{k=1}^2 e^{i\theta_k t}v_k=e^{i\theta t}(u_1-i u_2)+e^{-i\theta t}(u_1+i u_2)$$ But with this I conclude that $c_1=2$ and $c_2=0$ always. My question is, would this work if I had different coefficients in the sum? Say: \begin{align} \sum_{k=1}^2 d_k e^{i\theta_k t}v_k=&d_1 e^{i\theta t}(u_1-i u_2)+d_2 e^{-i\theta t}(u_1+i u_2)\\ =&d_1(cos(\theta t)+i sin(\theta t))(u_1-i u_2)+d_2 (cos(\theta t)+i sin(\theta t))(u_1+i u_2)\\ =&[(d_1+d_2)cos(\theta t)-i(d_2-d_1) sin (\theta t)]u_1\\ &+[(d_1+d_2)sin(\theta t)+i(d_2-d_1) cos (\theta t)]u_2 \end{align}
then, would it be posible to choose $c_1=d_1+d_2$, $c_2=i(d_2-d_1)$ and $c$ and $\alpha$ as they did and continue the proof normally? In particular, having to deal with the $i$ in $c_2$ is the part confuses me.