Example 14 Let the complex-coefficient polynomial be
$$
f(z)=a z^{2}+b z+c
$$

satisfy: for any $|z| \leqslant 1$, we have $|f(z)| \leqslant 1$. Try to find the maximum value of $|a|+|b|$.
【Analysis】(1) Let
$$
\begin{array}{l}
f_{1}(z)=\mathrm{e}^{\mathrm{i} \alpha} f\left(\mathrm{e}^{\mathrm{i} \beta} z\right) \\
=\mathrm{e}^{\mathrm{i}(\alpha+2 \beta)} a z^{2}+\mathrm{e}^{\mathrm{i}(\alpha+\beta)} b z+c \mathrm{e}^{\mathrm{i} \alpha} \\
=a_{1} z^{2}+b_{1} z+c_{1} . \\
\text { Let }\left\{\begin{array}{l}
\alpha+\beta=-\arg b, \\
\alpha+2 \beta=-\arg a .
\end{array}\right.
\end{array}
$$

Then $a_{1} \geqslant 0, b_{1} \geqslant 0$, and for any $|z| \leqslant 1$, we still have
$$
\left|f_{1}(z)\right|=\left|\mathrm{e}^{\mathrm{i} \alpha} f\left(\mathrm{e}^{\mathrm{i} \beta} z\right)\right| \leqslant 1 \text {. }
$$

Therefore, we can assume without loss of generality that $a \geqslant 0, b \geqslant 0$.
(2) Let $z=\mathrm{e}^{\mathrm{i} \theta}$. Then
$$
\begin{array}{l}
|\operatorname{Im} f(z)| \\
=|a \sin 2 \theta+b \sin \theta+\operatorname{Im} c| \\
\leqslant|f(z)| \leqslant 1 .
\end{array}
$$

To maximize $|a|+|b|=a+b$, we can choose $\theta$ such that $\sin 2 \theta=\sin \theta$, which gives $\theta= \pm \frac{\pi}{3}$. Thus, we have
$$
\begin{array}{c}
\left\{\begin{array}{l}
\left|\frac{\sqrt{3}}{2}(a+b)+\operatorname{Im} c\right| \leqslant 1, \\
\left|-\frac{\sqrt{3}}{2}(a+b)+\operatorname{Im} c\right| \leqslant 1
\end{array}\right. \\
\Rightarrow 2 \geqslant\left|\frac{\sqrt{3}}{2}(a+b)+\operatorname{Im} c\right|+ \\
\quad\left|\frac{\sqrt{3}}{2}(a+b)-\operatorname{Im} c\right| \\
\geqslant \sqrt{3}(a+b) \\
\Rightarrow|a|+|b| \leqslant \frac{2}{\sqrt{3}},
\end{array}
$$

Equality holds if and only if
$$
\left\{\begin{array}{l}
a+b=\frac{2}{\sqrt{3}}, \\
c \in \mathbf{R}, \\
\frac{1}{2}(b-a)+c=0 .
\end{array}\right.
$$

Let $b=2 a$, solving gives $f(z)=\frac{2 z^{2}+4 z-1}{3 \sqrt{3}}$.
At this point, for any $|z| \leqslant 1$, we can assume
$$
z=\cos \theta+\mathrm{i} \sin \theta \text {, }
$$

then $|f(z)| \leqslant 1$
$$
\begin{array}{l}
\Leftrightarrow\left|2 z^{2}+4 z-1\right| \leqslant 3 \sqrt{3} \\
\Leftrightarrow-8 \cos ^{2} \theta+8 \cos \theta+25 \leqslant 27 \\
\Leftrightarrow 8\left(\cos \theta-\frac{1}{2}\right)^{2} \geqslant 0 .
\end{array}
$$

In summary, the maximum value of $|a|+|b|$ is $\frac{2 \sqrt{3}}{3}$.
【Comment】Converting the conditions into an estimate of $|\operatorname{Im} f(z)|$ is a bold and important step. In the equality condition discussion, taking $b=2 a$ is because when $\theta=\frac{\pi}{3}$, $|\operatorname{Im} f(z)|$ reaches its maximum value of 1, so its derivative with respect to $\theta$ at $\theta=\frac{\pi}{3}$ is 0, which can be derived.