The instantaneous phasor can be expressed
$$\tilde{f}_{as}(t)=f_{qs}^e(t)-jf_{ds}^e(t)~~~~~~~~~~\text{(13)}$$
and
$$\begin{split}f_{as}(t) & =\text{Re}\left[\tilde{f}_{as}(t)e^{j\omega_et}\right]\\ & \text{Re}\left\{\left[f_{qs}^e(t)-jf_{ds}^e(t)\right]e^{j\omega_et}\right\}\end{split}~~~~~~~~~~\text{(14)}$$
The q and d axis voltage equations in the synchronously rotating reference frame for a symmetrical two-phase stator with a round rotor are
$$v_{qs}^e=r_si_{qs}^e+\omega_eL_{ss}i_{ds}^e+L_{ss}pi_{qs}^e~~~~~~~~~~\text{(15)}$$
$$v_{ds}^e=r_si_{ds}^e-\omega_eL_{ss}i_{qs}^e+L_{ss}pi_{ds}^e~~~~~~~~~~\text{(16)}$$
where
$$\lambda_{qs}=L_{ss}i_{qs}^e~~~~~~~~~~\text{(17)}$$
$$\lambda_{ds}=L_{ss}i_{ds}^e~~~~~~~~~~\text{(18)}$$
and
$$L_{ss}=L_{ls}+L_{ms}~~~~~~~~~~\text{(19)}$$
substituting (15) and (16) into (12) yields
$$v_{qs}^e-jv_{ds}^e=r_s\left(i_{qs}^e-ji_{ds}^e\right)+\omega_eL_{ss}\left(i_{ds}^e+ji_{qs}^e\right)+L_{ss}\left(pi_{qs}^e-jpi_{ds}^e\right)~~~~~~~~~~\text{(20)}$$
This can be written as
$$\tilde{v}_{as}=r_s\tilde{i}_{as}+j\omega_eL_{ss}\tilde{i}_{as}+L_{ss}p\tilde{i}_{as}~~~~~~~~~~\text{(21)}$$
In the figure below, which is Figure 5.7-1 in the text, Eqn. (21) is animated. Each of the instantaneous phasors in (21) is shown, and the equation is visualized as the sum of those phasors. As the system reaches steady state, the $L_{ss}p\tilde{i}_{as}$ term decays to zero, and this is illustrated by the red arrow in the animation. The trajectories of $\tilde{i}_{as}$ and $\left(r_s+j\omega_eL_{ss}\right)\tilde{i}_{as}$ are traced out. The latter of these is the steady state value of $\tilde{v}_{as}$, which can be seen as its trajectory decays to $\tilde{v}_{as}$.
To operate the animation, click the "Play" button. Three play speeds can be selected from among 0.1, 1, and 10 times the nominal speed by clicking the button to the right of "Play". You can also move to a specific moment in the animation by dragging the slider left or right. To pause the animation, simply click "Pause".