= Solution
\b[The conventional relaxation-loss peak near $\omega\tau=1$ applies to fixed <strain> amplitude, not to fixed <stress> amplitude.] This distinction matters because the question explicitly describes oscillating <stress>.
The full <complex modulus> of the series Maxwell element is
$$
E^*=\frac{ki\omega\tau}{1+i\omega\tau},\qquad
E''=\frac{k\omega\tau}{1+\omega^2\tau^2}.
$$
For imposed <strain> $\varepsilon=\varepsilon_0\cos\omega t$, the energy dissipated per volume per cycle is
$$
W_\varepsilon=\pi E''\varepsilon_0^2.
$$
Since $x/(1+x^2)$ has its maximum at $x=1$, \b[fixed-strain loss per cycle peaks at $\omega^2=\tau^{-2}$]. At that scale the imposed cycle time is comparable with the <relaxation time>, and substantial viscous flow occurs while <stress> remains appreciable.
For the literally stated comparison with fixed <stress> amplitude, use the compliance
$$
J^*=\frac1k+\frac1{i\omega\eta}.
$$
The <dashpot> <strain> rate is $\dot\varepsilon_d=\sigma/\eta$. Its dissipation integrated over a period gives
$$
\boxed{W_\sigma=\int_0^{2\pi/\omega}\frac{\sigma_0^2\cos^2\omega t}{\eta}\,dt
=\frac{\pi\sigma_0^2}{\eta\omega}.}
$$
This decreases monotonically with frequency; \b[there is no finite-frequency maximum at fixed <stress> amplitude in the ideal <Maxwell model>]. Its mean dissipated <power> is $\sigma_0^2/(2\eta)$, independent of frequency. The formal low-frequency divergence in energy per cycle accompanies very large <strain> excursions and cannot be extrapolated beyond the linear model's range. The graph identifies both results rather than assigning the fixed-strain peak to the wrong loading condition.
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