Classical Electrodynamics Third Edition solves the Laplace equation on spherical coordinate and cylindrical coordinate. I wonder why the image part of the general solution of the azimuthal angle is missing in the cylindrical coordinate.
Spherical coordinates
Here is the process to derive the solution on spherical coordinates in the book.
Assume the target function.
$$\Phi = \frac{U(r)}{r}P(\theta)Q(\phi)$$
And get these equations.
$$\begin{align*} \frac{1}{Q}\frac{\partial^2 Q}{\partial\phi^2} &= -m^2 \\ \frac{r^2}{U}\frac{\partial^2U}{\partial r^2} &= l(l+1) \\ \frac{1}{P\sin\theta}\frac{\partial}{\partial\theta} \left( \sin\theta \frac{\partial P}{\partial\theta} \right) -\frac{m^2}{\sin^2\theta} &= - l(l+1) \end{align*}$$
$$\begin{align*} Q &= e^{\pm im \phi} = \cos(m\phi) \pm \sin(m\phi) \\ U &= Ar^{l + 1} + Br^{-l} \\ P &= P^m_l \end{align*}$$
Cylindrical coordinate
$$\Phi = R(r)Q(\phi)Z(z)$$
$$\begin{align*} \frac{1}{Z}\frac{d^2 Z}{dz^2} &= k^2 \\ \frac{1}{Q}\frac{d^2Q}{d \phi^2} &= -\nu^2 \\ \frac{1}{R}\left[ \frac{d^2R}{d\rho^2} + \frac{1}{\rho}\frac{dR}{d\rho} \right] &= \frac{\nu^2}{\rho^2} - k^2 \end{align*}$$
$$\begin{align*} Z(z) &= e^{\pm kz} = \sinh kz \\ Q(\phi) &= e^{\pm i\nu \phi} = A \sin m\phi + B \cos m\phi \\ R(\rho) &= CJ_\nu(k\rho) + DN_\nu(k\rho) \\ \end{align*}$$
Why there is no image part of $Q$ in the solution of the cylindrical one compared to the spherical one?