6. Given positive real numbers $a, b$ satisfy
$$
a^{2}+b^{2}=25 \ (b>3),
$$

complex numbers $u, v, w$ satisfy
$$
w=a+b \mathrm{i}, \ u-w=3 v .
$$

If $|v|=1$, then when the principal value of the argument of $u$ is the smallest, the value of $\frac{u}{w}$ is $\qquad$
6. $\frac{16}{25}-\frac{12}{25}$ i.

Given $|v|=1$, we know $|u-w|=3|v|=3$.
Thus, in the complex plane, the point $P$ corresponding to $u$ lies on the circle $C$ with center at the point $M$ corresponding to $w$ and radius 3.

When the principal value of the argument of $u$ is the smallest, $OP$ is tangent to the circle $C$, and $|OM|=5,|PM|=3$, then $|OP|=4$.
Therefore, $\frac{|w|}{|u|}=\frac{5}{4}$.
The principal value of the argument of $\frac{w}{u}$ is $\theta=\angle POM$, and
$$
\cos \theta=\frac{4}{5}, \sin \theta=\frac{3}{5},
$$

Thus, $\frac{w}{u}=\frac{5}{4}\left(\frac{4}{5}+\frac{3}{5} \mathrm{i}\right)=1+\frac{3}{4} \mathrm{i}$.
Therefore, $\frac{u}{w}=\frac{16}{25}-\frac{12}{25} \mathrm{i}$.