3. 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
3. $[2,+\infty)$.

It is easy to see that when $|z| \leqslant 1$,
$$
\left|\frac{z+1}{2}\right| \leqslant 1, \left|\frac{z-1}{2}\right| \leqslant 1 \text{. }
$$

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