= Solution
In the position representation, the <orbital angular momentum> operator is
$$
\boxed{\widehat{\mathbf L}=\mathbf r\times\widehat{\mathbf p}=-i\hbar\,\mathbf r\times\nabla,\qquad L_i=\epsilon_{iab}x_ap_b.}
$$
On a common smooth domain use $[x_a,p_b]=i\hbar\delta_{ab}$, $[x_a,x_b]=[p_a,p_b]=0$. Expanding the product commutator gives
$$
[x_ap_b,x_cp_d]=-i\hbar\delta_{bc}x_ap_d+i\hbar\delta_{ad}x_cp_b.
$$
Contracting with the two alternating tensors gives the <orbital angular momentum commutation relations>:
$$
\boxed{[L_i,L_j]=i\hbar(x_ip_j-x_jp_i)=i\hbar\epsilon_{ijk}L_k.}
$$
For $L_\pm=L_1\pm iL_2$, these imply $[L_3,L_\pm]=\pm\hbar L_\pm$. Therefore $L_3\psi=0$ yields
$$
L_3(L_\pm\psi)=L_\pm L_3\psi+[L_3,L_\pm]\psi=\pm\hbar L_\pm\psi.
$$
Thus
$$
\boxed{L_\pm\psi\text{ has eigenvalue }\pm\hbar\text{ whenever }L_\pm\psi\ne0.}
$$
The nonzero qualification is necessary: a rotationally invariant wavefunction is annihilated by all three components, and the zero vector is not an <eigenvector>.
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