= Solution
In the position representation, the <momentum operators> are $\hat p_x=-i\hbar\partial_x$ and $\hat p_y=-i\hbar\partial_y$. Therefore
$$
\hat L_z=\hat x\hat p_y-\hat y\hat p_x=-i\hbar(x\partial_y-y\partial_x)=\boxed{-i\hbar\partial_\phi},
$$
since $\partial_\phi=-y\partial_x+x\partial_y$ at fixed radius. Solving the eigenvalue equation on the circle and imposing $2\pi$ periodicity gives
$$
\boxed{u_m(\phi)=\frac{e^{im\phi}}{\sqrt{2\pi}},\qquad L_z=\hbar m,\quad m\in\mathbb Z.}
$$
Their normalization uses the angular measure $d\phi$ on $[0,2\pi)$.
For the <half-ring sine-squared angular state>, write $C=\sqrt{8/(3\pi)}$. The stated <wavefunction> is normalized because $C^2\int_0^\pi\sin^4\phi\,d\phi=1$. It and its first derivative match continuously to zero at the boundaries, so the integrations by parts have no boundary contributions. The <angular momentum expectation> and second moment are
$$
\langle L_z\rangle=-i\hbar\int\psi\psi'\,d\phi=0,\qquad\langle L_z^2\rangle=\hbar^2\int|\psi'|^2d\phi=C^2\hbar^2\int_0^\pi\sin^2(2\phi)\,d\phi=\frac43\hbar^2.
$$
Hence \b[$\operatorname{Var}(L_z)=4\hbar^2/3$]. The jumps in the second derivative do not invalidate this result: the periodic <wavefunction> has a square-integrable second derivative and a continuous first derivative.
The probability of $L_z=\hbar$ is the squared projection onto $u_1$:
$$
c_1=\frac C{\sqrt{2\pi}}\int_0^\pi\sin^2\phi\,e^{-i\phi}d\phi=\frac2{\sqrt3\pi}\left(-\frac{4i}3\right)=-\frac{8i}{3\sqrt3\pi},\qquad\boxed{P(L_z=\hbar)=\frac{64}{27\pi^2}\simeq0.2402.}
$$
The odd sine term gives the imaginary overlap; a real <wavefunction> need not have zero probability of nonzero angular <momentum>. Its positive and negative angular-<momentum> probabilities are equal.<Quantum Mechanics by David Tong>[Chapters 3–4: operator expectation values, angular momentum and eigenstate projections]{ref}
\Image[/past-exam-of-the-physics-course-of-the-year-4-of-the-natural-sciences-course-of-the-university-of-cambridge/2016/paper-1-ring-momentum.png]
{title=A real state confined to half a ring has zero mean angular momentum and symmetric nonzero measurement probabilities}
{source=paper-1-ring-momentum.py}
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