Shunt feedback input impedance
This is an inverting Transimpedance Amplifier.
The current will see a very small input impedance and here is how to prove the statement:
$$I_{in} = \frac{V_{in}}{R_i} + \frac{V_{in}-V_o}{R_f}$$
$$V_o = -AV_{in}$$
$$I_{in} = \frac{V_{in}}{R_i} + \frac{V_{in}+AV_{in}}{R_f}$$
$$\frac{I_{in}}{V_{in}} = \frac{1}{R_i} + \frac{1+A}{R_f}$$
$$Z_{in} = \frac{V_{in}}{I_{in}} = R_i || \frac{R_f}{1+A}$$