Your approach is correct. But what isn’t shown in the figure explicitly, are the non-inverting and the inverting input terminals.

This case, you can safely assume it’s a negative feedback. Hence, the feedback resistor Rf, is connected to the inverting input terminal.

Therefore, the differential input voltage V+ - V- = -v

Hence, after applying the loop equations as you have done, and assuming the output resistance getting getting swamped by negative feedback, you get the same expression as given in the book.

Your approach is correct. But what isn’t shown in the figure explicitly, are the non-inverting and the inverting input terminals.

This case, you can safely assume it’s a negative feedback. Hence, the feedback resistor \$R_{f}\$, is connected to the inverting input terminal.

Therefore, the differential input voltage \$V_{+} - V_{-} = -v_{i}\$

After applying the loop equations as you have done, $$ I_{f} = \frac{v_{i} - A(-v_{i})}{R_{f} + R_{o}} $$ $$ I_{i} = \frac{v_{i}}{R_{i}} $$ $$ I = I_{f} + I_{i} $$

and assuming the output resistance \$R_{o}\$ getting getting swamped by negative feedback, you get the same expression for \$Z_{in} = \frac{v_{i}}{I}\$ as given in the book.

Reference: Operational Amplifiers (Page 9)