Let $H_0=-d^2/dx^2+x^2$. The Hermite expansion gives
$$
\langle H_0f,f\rangle
=\sum_{m=n+1}^\infty(2m+1)|c_m|^2
=\|f'\|^2+\|xf\|^2.
$$
Since $T=-d^2/dx^2+ix^2$,
$$
\langle Tf,f\rangle=\|f'\|^2+i\|xf\|^2,
$$
and therefore
$$
\boxed{\langle Tf,f\rangle
=(i-1)\|xf\|^2
+\sum_{m=n+1}^\infty(2m+1)|c_m|^2}.
$$

Write $X=\|xf\|^2$ and $S=\sum_{m=n+1}^\infty(2m+1)|c_m|^2$. Then $\operatorname{Re}\langle Tf,f\rangle=S-X$, $\operatorname{Im}\langle Tf,f\rangle=X$, and
$$
\operatorname{Re}\langle Tf,f\rangle
+\operatorname{Im}\langle Tf,f\rangle=S\geq2n+3.
$$
If $|z|\leq n$, then $|\operatorname{Re}z+\operatorname{Im}z|\leq\sqrt2n<2n+3$, so
$$
\boxed{|z|\leq n\Longrightarrow z\notin W(Q_nTQ_n^*)}.
$$
Every fixed compact set is eventually excluded from the tail numerical ranges. Part (b) therefore implies
$$
\boxed{W_e(T)=\varnothing}.
$$

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 358 / 2 / c / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-358/2/c/solution