In the context of DDEs of the type, $$\dot{x}(t)=f(x(t),x(t-\tau_1), ..., x(t-\tau_N)),$$ I am interested in the following question:
For a fixed N, but arbitrary $\tau_n>0$ and $a_n\in\mathbb{R}$, how many distinct imaginary roots can the following quasipolynomial have? $$h(s) = s - a_0 - \sum_{n=1}^{N} a_n e^{-s\tau_n} = 0 \qquad (\star)$$
I found a statement (without proof) that for any $\{s_m := i\omega_m\}_{m=1}^{2N-1}$ (counting only $\omega_m>0$) we can find appropriate delays $\tau_n$ and coefficients $a_n$ such that the $\{s_m\}$ are zeros of $(\star)$.
This would mean that we can always find $\tau_n>0$ and $a_n\in\mathbb{R}$ such that, $$i\omega_m = a_0 + \sum_{n=1}^{N} a_n e^{-i\omega_m \tau_n}\qquad ,\forall m=1,...,M=2N-1. \quad (\star\star) $$
While this is obviously true for $N=1$, I could not yet reproduce it for $N>1$. Am I missing something here?
A much weaker result I could derive is that for $N=2M$ and $\omega_m = m \omega, m=1,...,M$ and $\tau_n = \frac{2\pi n}{\omega (M+1)}, n=1,...,N$ we can find coefficients $a_n$ such that $(\star)$ holds. The idea here is that for our choice of $\omega_m, \tau_n\quad(\star\star)$ can be seen as the discrete fourier transform (DFT) of the signal $a_n$. Then using the inverse DFT we get, $$(a_0, ..., a_N) = DFT^{-1} (0, i\omega_1, ... i\omega_M, -i\omega_M, ..., -i\omega_1). $$ Where the complex conjugate symmetric fourier transform signal is needed to ensure that all $a_n$ are real numbers.
I would be really glad about any hints whether the former statement is true for $M = 2N-1$ or even for $M=N$, which intuitively seems a bit more reasonable to me.
Thanks in advance!
By the way: In order to be more specific, I have completely rewritten this question. I hope that is ok.