My guess is no, because the equations for $L^2$ and $L_z$ have no $r$-dependence:

$$L^2 = \hbar^2 \ell(\ell+1) f_m^\ell ( \theta , \phi )$$

$$L_z = \hbar m f_m^\ell ( \theta , \phi )$$

Is this the correct way of thinking about it? Or would such a function still be an eigenstate with eigenvalues both = 0?

My guess is no, because the equations for $L^2$ and $L_z$ have no $r$-dependence:

$$L^2 f_m^\ell ( \theta , \phi ) = \hbar^2 \ell(\ell+1) f_m^\ell ( \theta , \phi )$$

$$L_z f_m^\ell ( \theta , \phi )= \hbar m f_m^\ell ( \theta , \phi )$$

Is this the correct way of thinking about it? Or would such a function still be an eigenstate with eigenvalues both = 0?

My guess is no, because the equations for $L^2$ and $L_z$ have no r-dependence:

$L^2$ = $\hbar^2$ $l(l+1)$ $f_m^l$($\theta$, $\phi$)

$L_z$ = $\hbar$$m$ $f_m^l$($\theta$, $\phi$)

Is this the correct way of thinking about it? Or would such a function still be an eigenstate with eigenvalues both = 0?

My guess is no, because the equations for $L^2$ and $L_z$ have no $r$-dependence:

$$L^2 = \hbar^2 \ell(\ell+1) f_m^\ell ( \theta , \phi )$$

$$L_z = \hbar m f_m^\ell ( \theta , \phi )$$

Is this the correct way of thinking about it? Or would such a function still be an eigenstate with eigenvalues both = 0?