9. Given $a \in \mathbf{R}, b \in \mathbf{R}_{+}$, $\mathrm{e}$ is the base of the natural logarithm, then the minimum value of $\left(\mathrm{e}^{a}-\ln b\right)^{2}+(a-b)^{2}$ is
The problem is to find the minimum value of the square of the distance between points $A\left(a, \mathrm{e}^{a}\right)$ and $B(b, \ln b)$. Since $y=\mathrm{e}^{x}$ and $f(x)=\ln x$ are inverse functions, the minimum value of $|A B|$ is twice the minimum distance from point $B$ to the line $y=x$.
Considering the tangent line with a slope of 1 on the graph of the function $f(x)=\ln x$, we have $f^{\prime}\left(x_{0}\right)=\frac{1}{x_{0}}=1 \Rightarrow x_{0}=1$, thus the coordinates of the tangent point are $M(1,0)$, and the distance from $M$ to the line $y=x$ is $d=\frac{\sqrt{2}}{2}$.
Therefore, $|A B|_{\min }=\sqrt{2} \Rightarrow\left(\mathrm{e}^{a}-\ln b\right)^{2}+(a-b)^{2}$ has a minimum value of 2.