= Solution
Dividing Newton's equation by the mass gives the stated model: $\ddot z$ is the acceleration, $\gamma\dot z$ represents viscous damping, $(k/m)z$ is the spring's restoring acceleration, and the right side is the applied driving force per unit mass. The complex notation is a convenience; the physical displacement is the real part. Put $\omega_0^2=k/m$. The homogeneous characteristic roots are
$$
s=-\frac\gamma2\pm\sqrt{\frac{\gamma^2}4-\omega_0^2}.
$$
For $m,k>0$ and damping $\gamma\ge0$, oscillations occur when $\gamma^2<4\omega_0^2$, the underdamped case of the <damped harmonic oscillator>.
For a steady response $z=Ze^{i\omega t}$, substitution gives
$$
Z=\frac{F_0/m}{\omega_0^2-\omega^2+i\gamma\omega}=re^{-i\phi}.
$$
Taking the <modulus> and phase gives
$$
\boxed{r=\frac{F_0/m}{\sqrt{(\omega_0^2-\omega^2)^2+\gamma^2\omega^2}},\qquad\phi=\operatorname{atan2}(\gamma\omega,\omega_0^2-\omega^2).}
$$
The phase is between zero and $\pi$ for positive damping and frequency; an unqualified arctangent loses the correct quadrant above resonance.
With the real drive $F_0\cos\omega t$ and real response $r\cos(\omega t-\phi)$, the average driving <power> equals the dissipated viscous power:
$$
W_{\mathrm{av}}=m\gamma\langle\dot z^2\rangle=\frac12m\gamma r^2\omega^2=\frac{F_0^2\gamma\omega^2}{2m[(\omega_0^2-\omega^2)^2+\gamma^2\omega^2]}.
$$
For $\gamma>0$, write this as
$$
\frac{W_{\mathrm{av}}}{F_0^2/(2m\gamma)}=\frac1{1+[(\omega_0^2-\omega^2)/(\gamma\omega)]^2}.
$$
It is largest when the squared term vanishes, giving
$$
\boxed{\omega=\omega_0,\qquad W_{\mathrm{av,max}}=\frac{F_0^2}{2m\gamma}.}
$$
This power peak is at $\omega_0$ even when the displacement-amplitude peak is shifted. The <frequency bandwidth of viscous oscillator power absorption> follows from the half-power condition $|\omega_0^2-\omega^2|=\gamma\omega$:
$$
\boxed{\omega_1=\frac{\sqrt{4\omega_0^2+\gamma^2}-\gamma}{2},\quad\omega_2=\frac{\sqrt{4\omega_0^2+\gamma^2}+\gamma}{2},\quad\omega_2-\omega_1=\gamma.}
$$
In fact the bandwidth identity is exact in this model. Sharp resonance additionally gives $\omega_{1,2}\simeq\omega_0\mp\gamma/2$. With zero damping at exact resonance, there is no bounded steady harmonic response to which these finite-power formulas apply.
\Image[/past-exam-of-the-mathematical-methods-i-course-of-the-university-of-oxford/2015-resit-oscillator-power.png]
{title=Driven-oscillator power peaks at the natural frequency; the exact separation of the half-power frequencies is the damping coefficient}
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