= Solution
{c}
Write $|\psi\rangle=a|0\rangle+b|1\rangle$ and $\omega=e^{i\pi/4}$. After the <controlled-NOT gate>, the joint state is
$$
\frac1{\sqrt2}\left[a|0\rangle(|0\rangle+\omega|1\rangle)
+b|1\rangle(\omega|0\rangle+|1\rangle)\right].
$$
An ancilla measurement outcome zero leaves the unnormalized data state $(a|0\rangle+\omega b|1\rangle)/\sqrt2=T|\psi\rangle/\sqrt2$. Outcome one leaves $(\omega a|0\rangle+b|1\rangle)/\sqrt2=\omega T^\dagger|\psi\rangle/\sqrt2$. Both outcomes have <probability> $1/2$. The correction for outcome one is $S$, and $ST^\dagger=T$, so after correction the data is always $T|\psi\rangle$, up to an irrelevant <global phase>. This proves <magic-state injection of a T gate>.
The resource state's <Bloch vector> is $(1/\sqrt2,1/\sqrt2,0)$, outside the <stabilizer octahedron>. Clifford gates alone preserve stabilizer structure and, with standard-basis preparation and measurement, admit efficient classical simulation by the <Gottesman--Knill theorem>. Adding the non-Clifford <T gate> supplies a <universal quantum gate set> together with Clifford gates: arbitrary finite-qubit unitaries can be approximated to arbitrary precision. The assertion is approximate universality, not exact representation of every unitary by a finite circuit.
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