$$
\begin{array}{l}
\text { 4. If } a=1+\mathrm{i}, b=2+\mathrm{i}, c=3+\mathrm{i}, \\
x=-\frac{1}{2}+\frac{\sqrt{3}}{2} \mathrm{i},
\end{array}
$$

then $\left|a+b x+c x^{2}\right|=$
4. $\sqrt{3}$.

Notice that $x$ satisfies $x^{2}+x+1=0$.
Thus, $x^{3}=1,|x|=1, x \bar{x}=1$.
Also, the imaginary parts of $a, b, c$ are equal, and combining this with $x^{2}+x+1=0$, we only need to compute for $a=1, b=2, c=3$.
$$
\begin{array}{l} 
\text { Hence }\left|a+b x+c x^{2}\right|^{2}=\left(a+b x+c x^{2}\right)\left(a+b \bar{x}+c \bar{x}^{2}\right) \\
= a^{2}+b^{2}+c^{2}+a b(x+\bar{x})+b c\left(x \bar{x}^{2}+\bar{x} x^{2}\right)+ \\
a c\left(x^{2}+\bar{x}^{2}\right) \\
= a^{2}+b^{2}+c^{2}-a b-b c-c a .
\end{array}
$$

Substituting $a=1, b=2, c=3$ into the above equation, we get
$$
\left|a+b x+c x^{2}\right|^{2}=3 \text {. }
$$

Thus, $\left|a+b x+c x^{2}\right|=\sqrt{3}$.