= Solution
The azimuthal derivative is $\partial_\phi\psi=-i\psi$, so $\hat J_z\psi=(-i\hbar)(-i)\psi=-\hbar\psi$. For the polar part of the angular Laplacian,
$$
\frac1{\sin\theta}\partial_\theta(\sin\theta\,\partial_\theta\psi)
=\frac{\cos^2\theta-\sin^2\theta}{\sin\theta}e^{-i\phi},
\qquad
\frac1{\sin^2\theta}\partial_\phi^2\psi=-\frac1{\sin\theta}e^{-i\phi}.
$$
Their sum is $-2\sin\theta e^{-i\phi}=-2\psi$, hence $\hat J^2\psi=2\hbar^2\psi$. Comparing these two <eigenvalues> with $M\hbar$ and $J(J+1)\hbar^2$ gives \b[$J=1$, $M=-1$]. The sign of $M$ follows from both minus signs in the $\hat J_z$ calculation.
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