2. Let $u$, $v$, $w$ be complex numbers, where $w=a+b \mathrm{i}$ $\left(a, b>3, a^{2}+b^{2}=25\right), u-w=3 v$. If $|v|=$ 1, then when the principal value of the argument of $u$ is the smallest, the value of $\frac{u}{w}$ is $\qquad$
2. $\frac{16}{25}-\frac{12}{25}$ i.

Since $|v|=1$, we have $|u-w|=3|v|=3$.
Thus, the point $P$ corresponding to $u$ lies on the circle $C$ with the point $M$ corresponding to $w$ as the center and a radius of 3.

When the principal value of the argument of $u$ is the smallest, $OP$ is tangent to the circle $C$. Given $|OM|=5, |PM|=3$, then $|OP|=4$.
Therefore, $\frac{|w|}{|u|}=\frac{5}{4}$.
The principal value of the argument of $\frac{w}{u}$ is $\theta=\angle POM$, with $\cos \theta=\frac{4}{5}$ and $\sin \theta=\frac{3}{5}$. Thus,
$$
\begin{array}{l}
\frac{w}{u}=\frac{5}{4}\left(\frac{4}{5}+\frac{3}{5} \mathrm{i}\right)=1+\frac{3}{4} \mathrm{i} . \\
\text { Hence } \frac{u}{w}=\frac{16}{25}-\frac{12}{25} \mathrm{i} .
\end{array}
$$