There's one caveat. Strictly speaking a pure state is characterized by a normalized vector in Hilbert space modulo a phase, a socalled ray. That becomes important later when it comes, e.g., to the question of which spin quantum numbers are allowed, and since a state is determined by only a ray not by a Hilbert space vector, it's allowed that the rotation around 360 degrees multiplies the state with a phase. As the analysis of the angular-momentum commutation relations show there are indeed half-integer spins allowed by the algebra. Now a angular-momentum component in direction of an axis defines a rotation around this axis, and for a half-integer spin the rotation around 360 degrees multiplies any (allowed) state vector with a factor of -1. But that doesn't matter, because this state vector is in the ray defined by the original state vector and thus describes the same state. So for the very important fact that there are spin-1/2 particles making up the matter around us (and we are made ourselves) it is important to keep in mind that pure states are represented by rays in Hilbert space.
These rays are a bit unconvenient to work with. But as was explained in the recent threads here in the forum you can describe all states, including pure ones as statistical operators, i.e., self-adjoint positive semidefinite operators of trace 1. The pure states are then precisely those that can be written as
$$\hat{\rho}=|\psi \rangle \langle \psi|$$
with ##|\psi \rangle## being a vector normalized to 1. Note that any representant of a ray leads to the same statistical operator, because if
$$|\psi' \rangle=\exp(\mathrm{i} \varphi) |\psi \rangle, \quad \varphi \in \mathbb{R}$$
then
$$\langle \psi'|=\exp(-\mathrm{i} \varphi) \langle \psi|$$
and thus
$$\hat{\rho}'=|\psi' \rangle \langle \psi'| = |\psi \rangle \langle \psi|=\hat{\rho},$$
i.e., the freedom to choose an arbitrary phase to represent the same pure state doesn't matter if you use the statistical operator for the pure state.
