Complex Roots — Question 10

Question 10 – Solution

Strategy. First obtain a phase-independent bound, then sharpen it for one solution by controlling every later extremum.

Step 1: Resolve the data and phase. The roots are $-2\pm 3i$. Writing $y=e^{-2t}(A\cos 3t+B\sin 3t)$ gives $A=3$ and $-2A+3B=6$, hence $B=4$. Thus $$\boxed{y=e^{-2t}(3\cos 3t+4\sin 3t)=5e^{-2t}\cos(3t-\delta)}, \qquad \delta=\arctan(4/3)\in(0,\pi/2).$$

Step 2: Obtain the best certificate for every phase. The envelope condition is $5e^{-2t}\le 0.1$, so $$\boxed{T_{\mathrm{env}}=\tfrac 12\ln 50\approx 1.956012}.$$ It guarantees the tolerance at every later time. For any proposed $0\le T<T_{\mathrm{env}}$, choose a phase with $3T-\delta$ an integer multiple of $2\pi$. That solution has $|y(T)|=5e^{-2T}>0.1$. Thus no earlier time works for every phase with amplitude 5.

Step 3: Check the fixed phase more sharply. Let $\theta=\arctan(2/3)$. Stationary times are $t_k=(\delta-\theta+k\pi)/3$, and their magnitudes are $(15/\sqrt{13})e^{-2t_k}$. In particular, $$t_1\approx 1.160295<1.9<t_2\approx 2.207493, \qquad |y(t_2)|\approx 0.050317<0.1.$$ Also $y(1.9)=e^{-3.8}(3\cos 5.7+4\sin 5.7)\approx 0.006742$. Between 1.9 and $t_2$ there is no stationary point, so the response is monotone and its absolute value is bounded by the larger endpoint magnitude. Each subsequent stationary magnitude is multiplied by $e^{-2\pi/3}<1$. Monotonicity between those extrema then proves $\boxed{|y(t)|\le 0.1\text{ for all }t\ge 1.9}$.

Step 4: Distinguish a certificate from the exact last crossing. The envelope treats all phases, whereas the specified phase can have small values while the envelope is still above 0.1. The preceding negative minimum has magnitude greater than 0.1, so a final tolerance crossing occurs between $t_1$ and 1.9. Task 3 proves a sufficient time, not that 1.9 is the earliest one; locating the exact last crossing requires solving the corresponding trigonometric-exponential level equation.