 Solution

Write the normalized qubit as $|\varphi ⟩=c_0|0⟩+c_1|1⟩$, with $|c_0{|}^2+|c_1{|}^2=1$. Only the $|1⟩$ amplitude changes between the two phase gates, soSince $|e^{i\delta }-1|=2|\sin (\delta /2)|\le |\delta |$,The phase-gate discretization error is uniform over all input states; its exact operator norm is $2|\sin (\delta /2)|$. To choose an approximating gate, round $N\theta /(2\pi )$ to the nearest integer and reduce it modulo $N$. Taking the shortest circular angle difference gives $|\delta |\le \pi /N$, and hence a uniform state-vector error at most $\pi /2^n$. The circular choice handles the identified endpoints $0$ and $2\pi$ correctly.