= Solution
Take $u>0$ to mean that the mirror recedes along the incident <plane electromagnetic wave>. Put $\beta=u/c$, with $|\beta|<1$. The <moving-mirror Doppler shift> can be obtained directly by matching <wave phase> on the moving surface. Write the incident <wave phase> as $k_i z-\omega_i t$ and the reflected <wave phase> as $-k_r z-\omega_r t$, where $k_i=\omega_i/c$ and $k_r=\omega_r/c$ are positive <wavenumbers>. At the mirror, $z=z_0+ut$, the two phases must have the same time dependence; their constant offset incorporates the reflection <phase shift>. Thus
$$
k_i u-\omega_i=-k_r u-\omega_r,\qquad
\omega_i(1-\beta)=\omega_r(1+\beta).
$$
Consequently the observed reflected <angular frequency> and ordinary <frequency> are
$$
\boxed{\omega_r=\omega_i\frac{1-u/c}{1+u/c},\qquad
f_r=f_i\frac{1-u/c}{1+u/c}.}
$$
This is a redshift for a receding mirror and a blueshift for an approaching mirror. At small <velocity>, $\omega_r\simeq\omega_i(1-2u/c)$.
Equivalently, the <relativistic Doppler effect> gives $\omega_i'=\gamma(1-\beta)\omega_i$ in the mirror rest <inertial frame>. Reflection there reverses propagation and preserves <frequency>. Transforming the backward-going reflected wave to the laboratory gives another factor $\gamma(1-\beta)$, so $\omega_r=\gamma^2(1-\beta)^2\omega_i=\omega_i(1-\beta)/(1+\beta)$. A single Doppler factor would miss the effect of the return path.
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