Exercise 6.10 Frequency dependencies in a unity gain configuration

a)

An answer:

$$\begin{array}{rcll}& r_{\mathit{out},\mathit{circuit}}& =\frac{v_o}{i_o}& \\ & & i_o=\frac{(v_o-\frac{A_o{v}_x}{1+\mathit{j\omega }{\tau }_1})}{r_{\mathit{out}}}& \\ & & v_x=-v_o& \\ & & i_o=\frac{v_o}{r_{\mathit{out}}}+\frac{A_o{v}_o}{r_{\mathit{out}}(1+\mathit{j\omega }{\tau }_1)}& \\ & & =\frac{v_o}{r_{\mathit{out}}}(1+\frac{A_o}{1+\mathit{j\omega }{\tau }_1})& \\ & r_{\mathit{out},\mathit{circuit}}& =\frac{r_{\mathit{out}}}{1+\frac{A_o}{1+\mathit{j\omega }{\tau }_1}}& \\ & & =\frac{r_{\mathit{out}}(1+\mathit{j\omega }{\tau }_1)}{1+\mathit{j\omega }{\tau }_1+A_o}& \\ & & =\frac{r_{\mathit{out}}}{1+A_o}\cdot \frac{1+\mathit{j\omega }{\tau }_1}{1+\frac{\mathit{j\omega }{\tau }_1}{1+A_o}}& \end{array}$$

Assuming for simplicity reasons that $1+A_0=100$, the Bode plot would be similar to: