= Solution
Refer all layer <wave amplitudes> to the lower interface. Let a downward incident SH <displacement field> have unit <wave amplitude> there, an upward reflected wave have <wave amplitude> $R_h$, and the decaying half-space wave <wave amplitude> $T_h$. Interface continuity gives
$$
1+R_h=T_h,\qquad i\mu_0q(1-R_h)=-\mu pT_h.
$$
Solving these equations proves the <Love-wave interface reflection phase> and transmission <wave amplitude>:
$$
\boxed{R_h=\frac{\mu_0q-i\mu p}{\mu_0q+i\mu p}=e^{-2i\arctan(\mu p/(\mu_0q))},\quad T_h=\frac{2\mu_0q}{\mu_0q+i\mu p}.}
$$
Thus $|R_h|=1$: the <evanescent> substrate field has nonzero <wave amplitude> but no net downward mean flux. At the traction-free top, an upward incident wave and its downward reflection must have canceling <derivatives>, so their <displacement field> <reflection coefficient> is $R_0=+1$.
A layer wave acquires <wave phase> $2qh$ during a complete round trip. <Constructive interference> requires
$$
R_0R_he^{2iqh}=1,\qquad qh-\arctan\frac{\mu p}{\mu_0q}=n\pi.
$$
Taking the tangent gives again \b[$\tan(qh)=\mu p/(\mu_0q)$]. The reflection phases, rather than only geometric travel distance, are essential to the mode condition.
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