There is no need for the solution $\psi(x)$ to be real. What must be real is the probability density that is "carried" by $\psi(x)$. In some loose and imprecise intuitive way, you may think about a TV image carried by electromagnetic waves. The signal that travels is not itself the image, but it carries it, and you can recover the image by decoding the signal properly.
Somewhat similarly, the complex wave function that is found by solving Schrödinger equation carries the information of "where the particle is likely to be", but in an indirect manner. The information on the probability density $P(x)$ of finding the particle is recovered from $\psi(x)$ simply by multiplying it times its complex conjugate:
$$\psi(x)^*\psi(x) = P(x)$$
that gives a real function as a result. Note that it is a density: what you compute eventually is the probability of finding the particle between $x=a$ and $x=b$ as $\int_{a}^{b} P(x) dx$
As you know, when you multiply a complex number(/function) times its complex conjugate, the information on the phase is lost:
$$\rho e^{i \theta}\rho e^{-i \theta}=\rho^{2}$$
For that reason, in some places one can (not quite correctly) read that the phase has no physical meaning (see footnote), and then you may wonder "if I eventually get real numbers, why did not they invent a theory that directly handles real functions?".
The answer is that, among other reasons, complex wave functions make life interesting because, since the Schrödinger equation is linear, the superposition principle holds for its solutions. Wave functions add, and it is in that addition where the relative phases play the most important role.
The archetypical case happens in the double slit experiment. If $\psi_{1}$ and $\psi_{2}$ are the wave functions that represent the particle coming from the hole number $1$ and $2$ respectively, the final wave function is
$$\psi_{1}+\psi_{2}$$
and thus the probability density of finding the particle after it has crossed the screen with two holes is found from
$$P_{1+2}= (\psi_{1}+\psi_{2})^{*}(\psi_{1}+\psi_{2}) $$
That is, you shall first add the wave functions representing the individual holes to have the combined complex wave function, and then compute the probability density. In that addition, the phase informations carried by $\psi_{1}$ and $\psi_{2}$ play the most important role, since they give rise to interference patterns.
Comment: Feynman is quoted to have said "One of the miseries of life is that everybody names things a little bit wrong, and so it makes everything a little harder to understand in the world than it would be if it were named differently." It is quite similar here. Every book says that the phase of the wave function has no physical meaning. That is not 100% correct, as you see.
Your solution is valid. It has zero kinetic energy. It doesn't necessarily have zero energy. It can have any potential energy you'd like. Just because your particle is "freely moving," that doesn't mean the potential is zero. You could have $V(x)=k$ for any constant $k$. The value of $k$ is not observable and has no physical significance.
In general there is no special significance to having zero energy in a solution to the Schrodinger equation. Any solution can be defined to have zero energy simply by changing the potential appropriately like $V\rightarrow V+c$, where $c$ is some constant.
A realistic example involving zero kinetic energy and a constant wavefunction would be some particle-rotor models of nuclei, in which the deformed (prolate) nucleus (rotor) has some orientation in space, specified by one or two angular coordinates. If the odd particle has some component $K$ of its angular momentum along the symmetry axis, you get a rotational band with energies proportional to $J(J+1)$, starting with a ground state at $J=K$. In the ground state for the $K=0$ case, the rotor has zero kinetic energy, and its wavefunction is a constant as a function of the angular coordinates.
Best Answer
A charged particle in external magnetic field has the following Hamiltonian:
$$\hat H=\frac1{2m}\left(\hat{\vec p}-q\vec A\right)^2+qV,$$
where $\vec A$ is vector potential, $V$ is scalar potential and $\hat{\vec p}=-i\hbar\nabla$ is momentum operator.
If you set $\vec A\not=0$, you'll get non-trivially complex wavefunction, and there will be no degeneracy due to time inversion symmetry as is the case for usual running wave, because magnetic field breaks time-inversion symmetry.