 Solution

For the first Hopf boundary of scalar delayed negative feedback, first, if a root has nonnegative real part then $|\lambda +\beta |=\gamma e^{-\mathrm{Re}\lambda }\le \gamma$. For $\beta \ge \gamma$ this is impossible, apart from the potential equality case $\lambda =0$, already excluded. Thus $\beta <\gamma$ is necessary for instability, although not sufficient by itself.

At an imaginary root $\lambda =i\omega$ with $\omega >0$, separate real and imaginary parts:The first crossing uses ${\omega }_c\in (\pi /2,\pi )$ for $\beta >0$, and satisfiesThe period is in scaled time; its physical value is $\tau {\mathcal{T}}_c$. Higher-frequency branches have larger critical gains. For fixed $\beta$, stability at zero delayed gain and absence of any earlier imaginary or zero-root crossing identify this lowest branch as the first stability boundary. Its crossing is transverse: implicit differentiation of $\lambda +\beta +\gamma e^{-\lambda }=0$ gives at the imaginary rootFor very small $\beta$, ${\omega }_c=\pi /2+2\beta /\pi +O({\beta }^2)$, so ${\mathcal{T}}_c\to 4$ and ${\gamma }_c=\pi /2+2\beta /\pi +O({\beta }^2)$. Therefore ${\gamma }_c=\pi /2$ is exact in the limit $\beta =0$ and is the leading approximation for small positive $\beta$. At that limit $\lambda e^{\lambda }=-\gamma$ first has an imaginary pair at $\lambda =\pm i\pi /2$; subsequent pairs occur at $\pi /2+2n\pi$ and cannot be the first loss of stability. A nonlinear Hopf theorem, including its nondegeneracy conditions, is needed to turn the linear imaginary mode into a claim about a particular finite-amplitude periodic branch.