Step 1: Set up a test-source approach. A clean way to find an input impedance is to imagine forcing a known voltage $V_{in}$ at the terminal and working out how much current $I_{in}$ the source must supply to hold it there. The ratio $V_{in}/I_{in}$ is then the impedance.
Step 2: Track where the input current can possibly go. The $V_{in}$ terminal connects only to the op-amp's non-inverting ($+$) input node. That node has exactly one other path out: resistor $R$, running to the output. Because an ideal op-amp draws zero current at its input pins, the entire current $I_{in}$ supplied by the source has nowhere to go except through $R$: \[ I_{in} = \frac{V_{in} - V_{out}}{R} \]
Step 3: Find $V_{out}$ using the feedback network on the other input. Because the op-amp keeps its two inputs at equal voltage, $V_- = V_+ = V_{in}$. At the inverting node, $R_1$ pulls current in from ground and $R_F$ pulls current in from the output, and these two currents must balance since no current enters the op-amp: \[ \frac{0-V_{in}}{R_1} = \frac{V_{in}-V_{out}}{R_F} \] Cross-multiplying: \[ -V_{in}R_F = R_1V_{in} - R_1V_{out} \] \[ R_1 V_{out} = V_{in}R_F + V_{in}R_1 \] \[ V_{out} = V_{in}\left(\frac{R_F+R_1}{R_1}\right) \] This says the circuit behaves like a standard non-inverting amplifier with gain $1+R_F/R_1$, as far as $V_{out}$ is concerned.
Step 4: Substitute back into the current expression. \[ I_{in} = \frac{V_{in} - V_{in}\left(\dfrac{R_F+R_1}{R_1}\right)}{R} = \frac{V_{in}\left(\dfrac{R_1-R_F-R_1}{R_1}\right)}{R} = \frac{-V_{in}R_F}{R_1R} \]
Step 5: Take the ratio to get $Z_{in}$. \[ Z_{in} = \frac{V_{in}}{I_{in}} = -\frac{R R_1}{R_F} \] The current the source has to deliver actually grows as $V_{in}$ rises, rather than the source having to push current in, which is the hallmark of a negative resistance seen at the input. This circuit is a well-known Negative Impedance Converter (NIC).
Step 6: Ruling out the distractors. If $R$ were absent (an open circuit at the $+$ node instead), the answer would genuinely be infinite, option (A), but here $R$ closes a real current path, so infinity cannot be right. Options (C) and (D) do not fall out of the node equations at all; they simply rearrange the resistors incorrectly or drop $R_1$ and $R_F$ from the picture.
Final Answer: \[ Z_{in} = -\frac{R \times R_1}{R_F} \] This matches option (B).
