Electronic – Second order all-pass filter input impedance

I have a question about the second order all-pass filter input impedance when ω = ∞ and ω = 0 and I would appreciate if you could help me to understand the logic.

1. When ω = 0, all capacitors can be seen as open circuit, because we are at DC level. In this case I expect the Zin to be R3+R4, what would be 3 kOhms, because the voltage at opamp inputs would be 0V. From the simulation I see that the expected input impedance is 1.5 kOhms. Why?
2. When ω = ∞, capacitors can be seen as short circuits, and therefore the current will travel along C with no electrical impedance and R3, R4 are the only impedances that separates V1 from ground. In this case I do not understand why the input impedance is 3 kOhms.

At DC $$\\omega = 0\$$ (all capacitors can be seen as open circuit) we have this situation:

simulate this circuit – Schematic created using CircuitLab

And $$R_{IN_{DC}} = R_3 + R_4 = 3k\Omega$$

But at high frequency ($$\\omega = ∞\$$) when all capacitors can be seen as short circuits we have this situation:

simulate this circuit

Therefore the input resistance is now equal to:

$$R_{IN_{HF}} = \left[R_1 \times\left(1 + \frac{R_4}{R_3}\right)\right]||(R_3+R_4) = 3k\Omega||3k\Omega = 1.5k\Omega$$

Why? because now the voltage across $$\R_1\$$ resistor is no longer equal to $$\V_{IN}\$$ but to the difference between $$\V_{IN}\$$ and op-amp output voltage. And the op-amp is working as a voltage follower, meaning that the op-amp output voltage is the same as the input voltage (at non-inverting input). And the input voltage is the output voltage produced by the voltage divider build around $$\R_3, R_4\$$.