= Solution
For $\omega\tau\ll1$,
$$
G'(\omega)\simeq\frac{\eta_M^2}{G}\omega^2,\qquad
G''(\omega)\simeq(\eta_M+\eta_s)\omega,
$$
and
$$
\eta^*(\omega)=\frac{G^*}{i\omega}
=\eta_s+\frac{\eta_M}{1+i\omega\tau}
\longrightarrow\boxed{\eta_0=\eta_M+\eta_s}.
$$
Thus \b[the material flows at sufficiently low <frequency>]: its static <shear modulus> vanishes and its terminal response is viscous. The <dashpot> in the Maxwell branch allows its <spring> <stress> to relax, so there is no permanent elastic path through the parallel network.
The fact that both $G'$ and $G''$ tend to zero is not evidence of a force-free material: <stress> at fixed <shear rate> tends to $\eta_0\dot\gamma$. At higher <frequency> the <spring> stores <energy>, but a finite-frequency elastic response does not make the long-time material a solid.
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