I'm working on a research paper on Delay Differential Equations and have arrived at the same characteristic equation for the linearised system:
\begin{align} \lambda^2+p\lambda+r+(s\lambda+q)e^{-\lambda\tau}=0 \end{align}
Defining $\tau\neq0$, we can use $\lambda=i\omega$, where $\omega\in\Re$, as a root of this characteristic equation.
Substituting this into the characteristic equation, I have:
\begin{align} 0&=i^2\omega^2+pi\omega+r+(si\omega+q)e^{-i\omega\tau} \\ &=-\omega^2+qe^{-i\omega\tau}+pi\omega+r+is\omega e^{-\omega\tau} \end{align}
Using the Euler equations, I rewrote $e^{\pm i\omega\tau}$ as
\begin{align} e^{i\omega\tau}&=\cos(\omega\tau)+i\sin(\omega\tau) \\ e^{-i\omega\tau}&=\cos(\omega\tau)-i\sin(\omega\tau). \end{align}
This gives me \begin{align} 0&=-\omega^2+q\left(\cos(\omega\tau)-i\sin(\omega\tau)\right)+pi\omega+r+is\omega(\cos(\omega\tau)-i\sin(\omega\tau)) \\ &=-\omega^2+q\cos(\omega\tau)-qi\sin(\omega\tau)+i\omega p+r +is\omega\cos(\omega\tau)-i^2s\omega\sin(\omega\tau) \\ &=-\omega^2+q\cos(\omega\tau)-qi\sin(\omega\tau)+i\omega p+r +is\omega\cos(\omega\tau)+s\omega\sin(\omega\tau) \end{align}
I'm then confused about how they have evaluated \begin{align} F(i\omega)&= r-\omega^2-s\omega\sin(\omega\tau)+q\cos(\omega\tau)=0 \hspace{5mm} (5.8)\\ G(i\omega)&= p\omega + s\omega\cos(\omega\tau)+q\sin(\omega\tau)=0 \hspace{12mm} (5.9) \end{align}
I've tried taking out the real and imaginary components of this equation: \begin{align} F(i\omega)&= r-\omega^2+s\omega\sin(\omega\tau)+q\cos(\omega\tau)=0 \\ G(i\omega)&= p\omega + s\omega\cos(\omega\tau)-q\sin(\omega\tau)=0 \end{align}
If anyone can explain what the step I'm missing is? What have I done wrong? Any help would be much appreciated!