Step 1: Check whether all spherical harmonics are complex.
Spherical harmonics \(Y_l^m(\theta,\phi)\) carry the angular factor \(e^{im\phi}\), which is complex whenever \(m \neq 0\). But when \(m=0\), \(e^{i\cdot0\cdot\phi}=1\), and \(Y_l^0\) reduces to a real function built from the Legendre polynomial \(P_l(\cos\theta)\), for example \(Y_0^0 = \frac{1}{\sqrt{4\pi}}\) and \(Y_1^0 \propto \cos\theta\), both real. So not all spherical harmonics are complex; statement (A) is wrong.
Step 2: Check the \(\hat{L}^2\) eigenvalue property.
Spherical harmonics are, by construction, the angular eigenfunctions of the total angular momentum squared operator:
\[ \hat{L}^2 Y_l^m = l(l+1)\hbar^2 Y_l^m \]
This holds for every \(l\) and \(m\), so statement (B) is correct.
Step 3: Check the \(\hat{L}_z\) eigenvalue property.
They are simultaneously eigenfunctions of the z-component of angular momentum:
\[ \hat{L}_z Y_l^m = m\hbar Y_l^m \]
This is exactly why the two labels \(l\) and \(m\) are used to index them. Statement (C) is correct.
Step 4: Check whether \(Y_1^1\) and \(Y_1^{-1}\) are degenerate.
For a system with a central, spherically symmetric potential (such as a rigid rotor or the angular part of the hydrogen atom), the energy depends only on \(l\), not on \(m\). \(Y_1^1\) and \(Y_1^{-1}\) share the same \(l=1\), so in the absence of an external field (like a magnetic field, which would lift this degeneracy) they have the same energy. So they are degenerate, and statement (D) is correct.
Final Answer:
Statements (B), (C) and (D) are correct; (A) is wrong because \(m=0\) harmonics are real.
\[ \boxed{\text{B, C, D}} \]