= Solution
The <angular momentum commutation relations> are $[L_i,L_j]=i\hbar\epsilon_{ijk}L_k$. The <commutator> product rule gives
$$
[L_i,L^2]=i\hbar\sum_{j,k}\epsilon_{ijk}(L_kL_j+L_jL_k)=0,
$$
since the bracketed expression is symmetric in $j,k$ and $\epsilon_{ijk}$ is antisymmetric. Also
$$
L_-L_+=L_1^2+L_2^2+i[L_1,L_2]=L_1^2+L_2^2-\hbar L_3,
$$
proving \b[$\boxed{L^2=L_-L_++L_3^2+\hbar L_3}$].
For <orbital angular momentum>, $L_i=-i\hbar\epsilon_{ijk}x_j\partial_k$. A differentiable radial function has $\partial_kf(r)=f'(r)x_k/r$, so its contraction with the antisymmetric symbol vanishes: $Lf(r)=0$. Put $s=x_1+ix_2$. The coordinate differential operators give $L_3s=\hbar s$, while
$$
L_+=\hbar\left[x_3(\partial_1+i\partial_2)-s\partial_3\right]
$$
annihilates every power of $s$, because $(\partial_1+i\partial_2)s=0$. Since all $L_i$ annihilate the radial factor, the product rule yields
$$
\boxed{L_3[s^nf(r)]=n\hbar s^nf(r),\qquad L_+[s^nf(r)]=0.}
$$
Substitution into the expression for $L^2$ gives the <highest-weight complex-coordinate orbital wavefunction> result
$$
\boxed{L^2[s^nf(r)]=\hbar^2n(n+1)s^nf(r).}
$$
As $[L^2,L_-]=0$, lowering preserves this <eigenvalue> whenever the lowered function is nonzero. Explicitly,
$$
\boxed{L_-[s^nf(r)]=-2n\hbar x_3s^{n-1}f(r),\qquad
\text{unchanged eigenvalue }\hbar^2n(n+1).}
$$
For $n=0$ the lowered function is zero, so it is not an <eigenfunction>. For $n\geq0$ the original nonzero function has the usual regular angular dependence. The PDF's “any integer” differential identities also hold locally for negative $n$ where $s\ne0$, but those negative powers are singular on the axis and are not square-integrable regular angular eigenstates. The converted TeX incorrectly writes $L_4$ in place of the PDF's $L_+$.
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