= Solution
The original circuit has $R_1$ and $C$ in series between the input and the inverting node, with $R_2$ providing feedback and the noninverting node grounded. An ideal <operational amplifier> in stable linear <negative feedback> draws no input <electric current> and holds the inverting node at <virtual ground>. For <phasors> proportional to $e^{i\omega t}$, the input <complex impedance> is $Z_1=R_1+1/(i\omega C)$. <Kirchhoff's current law> gives $\widetilde V_{\rm in}/Z_1=-\widetilde V_{\rm out}/R_2$, hence
$$
\boxed{H(\omega)=\frac{\widetilde V_{\rm out}}{\widetilde V_{\rm in}}
=-\frac{i\omega R_2C}{1+i\omega R_1C},\qquad
|H|=\frac{\omega R_2C}{\sqrt{1+\omega^2R_1^2C^2}}.}
$$
This is an inverting <high-pass filter>. Its <phase> can be chosen continuously as
$$
\boxed{\arg H=-\frac\pi2-\tan^{-1}(\omega R_1C).}
$$
Thus $V_{\rm out}(t)=V_0|H|\cos(\omega t+\arg H)$. The ratio above is a <phasor> ratio, not an instantaneous ratio at input zero crossings. Setting $R_1=R_2$ does not remove the <capacitor>'s <phase>: the output lags by $90^\circ+\tan^{-1}(\omega R_1C)$, between $90^\circ$ and $180^\circ$. At $\omega R_1C=1$ the lag is $135^\circ$; only for $\omega R_1C\gg1$ does it approach $180^\circ$. Being within the amplifier's own <bandwidth> does not specify the external RC corner <frequency>.
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