Analysis of Negative Input Resistance Op Amp
The circuit is not correct: you can't apply a voltage on that node. 

Instead, you shold apply a current, let's say Iin, and then measure the voltage of the inverting input, and after that, you can calculate the Zin as Vinv/Iin (Vinv = inverting input voltage). Iin is the input current, with passive convention sign (positive if it goes rightward).

If you look at your paper, the second equation tells exactly that (With different notations). The first equation says that Iin is $$Iin=\frac{Vinv-Vout}{R1}.$$
Then the paper assumes that negative feedback works, therefore the voltage of the inverting input is equal to the voltage of the non inverting input. As a result, Vinv (which is the same as Vninv) is the classic impedance divider formula (the paper assumes a zero input current on the OA).

As a result (equation 5):
$$Vinv=Vninv=Vout\frac{K \cdot R2}{R2 + \frac{1}{j\omega C}}.$$
Where K is the fraction of R2, selected by the potentiometer wiper.

Substituting Vinv in Iin and calculating the input impedance (Vinv/Iin), one should get exactly equation 6 on your paper.

EDIT:
It turns out it doesn't! Even after some possible approximations. A result which is similar to eq. 6 (i.e. a negative resistance + negative capacitance) is achieved by swapping the inverting input with the non-inverting input.

EDIT2: After some algebra, you obtain something like:
$$Zin = \frac {R1 \cdot K} {(k-1)^2+\frac{1}{(\omega CR2)^2}}(k-1+\frac{1}{j\omega C \cdot R2})$$ 

Under the assumptions (not stated in the paper, though) of k&lt;&lt;1, and "high enough" frequency, the first denominator becomes 1, and the term in the parethesis becomes $$-1+\frac{1}{j\omega C \cdot R2}$$.

Then you get eq. 6.