= Solution
For each mode in <Cartesian coordinates>, $[a_i,a_j^\dagger]=\delta_{ij}$ and $a_i^\dagger a_i=H_i/(\hbar\omega)-1/2$. Consequently the <two-dimensional isotropic harmonic oscillator> has
$$
\boxed{H=\hbar\omega(N_x+N_y+1)},\qquad N_i=a_i^\dagger a_i,\qquad E_{kl}=\hbar\omega(k+l+1).
$$
The normalized actions of the <creation and annihilation operators> are
$$
a_x|k,l\rangle=\sqrt{k}\,|k-1,l\rangle,\qquad
a_y^\dagger|k,l\rangle=\sqrt{l+1}\,|k,l+1\rangle,\qquad
a_x|0,l\rangle=0.
$$
Invert their definitions as $x=\sqrt{\hbar/(2m\omega)}(a_x+a_x^\dagger)$ and $p_x=i\sqrt{m\hbar\omega/2}(a_x^\dagger-a_x)$, and similarly for $y$. <linear operators> belonging to different coordinates commute, so substitution into $L_z=xp_y-yp_x$ gives
$$
L_z=\frac{i\hbar}{2}\big[(a_x+a_x^\dagger)(a_y^\dagger-a_y)-(a_y+a_y^\dagger)(a_x^\dagger-a_x)\big]
=\boxed{-i\hbar(a_x^\dagger a_y-a_y^\dagger a_x)}.
$$
Each term transfers one excitation between the two modes: $[N_x+N_y,a_x^\dagger a_y]=(1-1)a_x^\dagger a_y=0$, and likewise for its adjoint. Thus $[H,L_z]=0$ and, with no explicit time dependence, $dL_z/dt=(i/\hbar)[H,L_z]=0$. This is <conservation of angular momentum> under <planar rotations>.
Finally $L_z|1,0\rangle=i\hbar|0,1\rangle$ and $L_z|0,1\rangle=-i\hbar|1,0\rangle$. Hence
$$
\boxed{L_z\big(|1,0\rangle+i|0,1\rangle\big)
=\hbar\big(|1,0\rangle+i|0,1\rangle\big)}.
$$
The <angular momentum> is $+\hbar$; division by $\sqrt2$ performs <wavefunction normalization> on this state without changing its <eigenvalue>. It is one excitation of the positive circular mode.
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