6. Let $z \in \mathbf{C}, f(z)=a z^{2}+b z+c$ be a polynomial with complex coefficients, and suppose that when $|z| \leqslant 1$, $|f(z)| \leqslant 1$. Then, when $|z| \leqslant 1$, the range of $|a z+b|$ is $\qquad$
$$
\text { 6. } 0 \leqslant|a z+b| \leqslant 2 \text {. }
$$

When $|z| \leqslant 1$, because
$$
\left|\frac{z+1}{2}\right| \leqslant 1, \left|\frac{z-1}{2}\right| \leqslant 1 \text {, }
$$

so, $\left|f\left(\frac{z+1}{2}\right)\right| \leqslant 1, \left|f\left(\frac{z-1}{2}\right)\right| \leqslant 1$.
Therefore, $|a z+b|$
$$
\begin{array}{l}
=\left| a\left(\frac{z+1}{2}\right)^{2}+b\left(\frac{z+1}{2}\right)+c-\right. \\
{\left.\left[a\left(\frac{z-1}{2}\right)^{2}+b\left(\frac{z-1}{2}\right)+c\right] \right|} \\
=\left|f\left(\frac{z+1}{2}\right)-f\left(\frac{z-1}{2}\right)\right| \\
\leqslant\left|f\left(\frac{z+1}{2}\right)\right|+\left|f\left(\frac{z-1}{2}\right)\right| \\
\leqslant 1+1=2 \text {. } \\
\end{array}
$$

Thus, $0 \leqslant|a z+b| \leqslant 2$.