From a delay system, I obtain the following as part of a characteristic equation:
$$f(\lambda) = \lambda - a + be^{-c\lambda},$$
where $a, b,$ and $c$ are positive number and $a<b, ac<1$. **My goal is to find the sign of the real part of the root to $f(\lambda) = 0$**. Taking $\lambda = x + iy$, I obtain:
\begin{align}
x &= a - be^{-cx}cos(cy)\\
y &= be^{-cx}sin(cy).
\end{align}
I am looking for a condition for $x$ to be negative, but I fail to obtain anything meaningful based on the given constraints (for instance, if $a<0$, then $x<0$). I also narrow down $y \in [-\frac{\pi}{2c},\frac{\pi}{2c}],$ but fail to get something from that as well. Any help would be much appreciated!


All of my simulations show that under the given conditions, $x$ is always negative.



Conditions to determine sign of real roots
I think your equation is $f(\lambda) = \lambda - a + be^{-c \lambda}$. Let us consider the parametrized family $f\_{\varepsilon}(\lambda) = \lambda - a + b e^{-\varepsilon c \lambda}$, where $\varepsilon \in [0,1]$. First of all note that the number of roots $f\_{\varepsilon}(\lambda)=0$ in every half-plane $\operatorname{Re}\lambda > \beta$, $\beta \in \mathbb{R}$, is finite and uniformly (in $\varepsilon$) bounded. Suppose we know that for all $\varepsilon$ there is no roots $f\_{\varepsilon}(\lambda)=0$ lying on the imaginary axis. Then we can consider a closed contour in $\operatorname{Re}\lambda \geq 0$, which encloses all the roots with $\operatorname{Re}\lambda > 0$ for all $\varepsilon$. From the Rouché theorem it follows that all equations $f\_{\varepsilon}(\lambda)=0$ has the same number of roots in this contonur for all $\varepsilon$. It is clear that for $\varepsilon=0$ there is only one root $\lambda = -b + a$, which is negative due to $a < b$. Therefore, $f\_{1}(\lambda)=f(\lambda) = \lambda - a + be^{-c \lambda}$ has no roots with $\operatorname{Re}\lambda > 0$.


So now we should provide conditions for the equations $f\_{\varepsilon}(\lambda)=0$ to not have purely imaginary roots. I leave this for you.


