= Solution
For the <first Hopf boundary of scalar delayed negative feedback>, first, if a root has nonnegative <real part> then $|\lambda+\beta|=\gamma e^{-\operatorname{Re}\lambda}\le\gamma$. 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:
$$
\beta+\gamma\cos\omega=0,\qquad\omega=\gamma\sin\omega.
$$
The first crossing uses $\omega_c\in(\pi/2,\pi)$ for $\beta>0$, and satisfies
$$
\boxed{\beta=-\omega_c\cot\omega_c,\qquad\gamma_c=\frac{\omega_c}{\sin\omega_c}=\sqrt{\beta^2+\omega_c^2},\qquad\mathcal T_c=\frac{2\pi}{\omega_c}.}
$$
The <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 root
$$
\operatorname{Re}\frac{d\lambda}{d\gamma}=\frac{\beta(1+\beta)+\omega_c^2}{\gamma[(1+\beta)^2+\omega_c^2]}>0.
$$
For very small $\beta$, $\omega_c=\pi/2+2\beta/\pi+O(\beta^2)$, so $\mathcal T_c\to4$ and $\gamma_c=\pi/2+2\beta/\pi+O(\beta^2)$. Therefore \b[$\gamma_c=\pi/2$ 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.
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