Verify (2.16). To do this, it is useful to form the so-called "step-up" and "step-down" operators
$$
J_{ \pm}=J_1 \pm i J_2 .
$$
First, show that
$$
J_{ \pm}|j m\rangle=(C-m(m \pm 1))^{1 / 2}|j, m \pm 1\rangle,
$$
that is, $J_{ \pm}$step $m$ up and down by one unit, respectively. Show that $C=j(j+1)$.
A state $|j m\rangle$ is transformed under a rotation through an angle $\theta$ about the 2-axis into a linear combination of the $2 j+1$ states $\left|j m^{\prime}\right\rangle$, with $m^{\prime}=-j,-j+$ $1, \ldots, j$ :
$$
e^{-i \theta J_2}|j m\rangle=\sum_{m^{\prime}} d_{m^{\prime} m}^j(\theta)\left|j m^{\prime}\right\rangle,
$$
where the coefficients $d_{m^{\prime} m}^j$ are written in conventional notation and are frequently called rotation matrices. From (2.19), we see the states having the same $j$ but all possible $m$ values transform among themselves under rotations. In fact, all the $2 j+1$ states are mixed by rotations. They form the basis of a $(2 j+1)$ dimensional irreducible representation of the rotation group. The set of states is called a multiplet.