 Phase of a negatively charged driven oscillator

For $\ddot{z}+2\gamma \dot{z}+{\omega }_0^{2}z=(q{E}_0/m)\cos \omega t$, the $e^{i\omega t}$ convention gives complex displacement amplitude $\tilde{z}=(q{E}_0/m)/({\omega }_0^2-{\omega }^2+2i\gamma \omega )$. If the real amplitude is defined nonnegative, the phase is $-\delta$ for $q>0$ but $\pi -\delta$ for $q<0$, where $\delta =atan2(2\gamma \omega ,{\omega }_0^2-{\omega }^2)$. An electron's low-frequency displacement is opposite to the electric field; its high-frequency displacement is in phase. The induced electric dipole moment $qz$ instead has phase $-\delta$ for either sign of charge. A signed amplitude can absorb the extra $\pi$, but that convention must be stated.