Solution:

$ \begin{aligned}\\ &\text { Given } y(n)=a y(n-1)+x(n) \\\\ &Y(z)=a z^{-1} Y(z)+X(z) \\\\ &Y(z)-a z^{-1} Y(z)=X(z) \\\\ &Y(z)\left[1-a z^{-1}\right]=X(z) \\\\ &H(z)=\frac{Y(z)}{X(z)}=\frac{1}{1-a z^{-1}}=\frac{z}{z-a} \\\\ &H\left(z^{-1}\right)=\frac{z^{-1}}{z^{-1}-a}\\ \end{aligned}\\ $

$ \begin{aligned}\\ \sigma_{\varepsilon}^2 &=\sigma_e^2 \frac{1}{2 \pi j} \oint_C A(z) H\left(z^{-1}\right) z^{-1} d z \\\\ &=\sigma_e^2 \frac{1}{2 \pi j} \oint_c \frac{z}{z-a} \frac{z^{-1}}{z^{-1}-a} z^{-1} d z \\\\ &=\sigma_e^2 \frac{1}{2 \pi j} \oint_c \frac{z^{-1}}{(z-a)\left(z^{-1}-a\right)} d z\\ \end{aligned}\\ $

$ \begin{aligned}\\ &\sigma_{\varepsilon}^2=\sigma_e^2\left[\text { residue of } \frac{z^{-1}}{(z-a)\left(z^{-1}-a\right)}\right. \text { at }\\\\ &=\left.\sigma_e^2\left[(z-a) \frac{z^{-1}}{(z-a)\left(z^{-1}-a\right)}\right]\right|_{z=a}\\\\ &=\sigma_e^2\left[\frac{1 / a}{(1 / a-a)}\right]\\\\ &=\sigma_e^2\left[\frac{1}{1-a^2}\right] \text {. }\\ \end{aligned}\\ $