One, (25 points) Let $f(z)=z^{2}+a z+b$ be a quadratic trinomial in the complex variable $z$ with complex coefficients $a, b$, and for all $z(|z|=1)$, $|f(z)|=1$. Find the values of $a$ and $b$.

保留源文本的换行和格式，直接输出翻译结果。
Let $a=a_{1}+i a_{2}, b=b_{1}+i b_{2}, a_{1}, a_{2}, b_{1}, b_{2}$
all be real numbers. Since $|z|=1$ implies $|f(z)|=1$, let $z=1$, we get
$\left(1+a_{2}+b_{1}\right)^{2}+\left(a_{2}+b_{2}\right)^{2}=1$.
Let $z=-1$, we get $\left(1-a_{1}+b_{1}\right)^{2}+\left(-a_{2}+b_{2}\right)^{2}=1$.
(1) - (2) gives $a_{1}+a_{1} b_{1}+a_{2} b_{2}=0$.
Similarly, let $z=i$ and $z=-i$, we get
$$
\begin{array}{l}
\left(-1-a_{2}+b_{1}\right)^{2}+\left(a_{1}+b_{2}\right)^{2}=1, \\
\left(-1+a_{2}+b_{1}\right)^{2}+\left(-a_{1}+b_{2}\right)^{2}=1 .
\end{array}
$$
(4) - (5) gives $a_{2}-a_{2} b_{1}+a_{1} b_{2}=0$.

In general, let $z=\cos \theta+i \sin \theta(0 \leqslant \theta<2 \pi)$, then
$$
\begin{array}{l}
a_{1}^{2}+a_{2}^{2}+b_{1}^{2}+b_{2}^{2}+2\left(a_{1}+a_{1} b_{1}+a_{2} b_{2}\right) \cos \theta \\
+2\left(a_{2}-a_{2} b_{1}+a_{1} b_{2}\right) \sin \theta+2 b_{1} \cos 2 \theta+2 b_{2} \sin 2 \theta \\
=0
\end{array}
$$

From (3), (6) we get
$$
\begin{array}{l}
a_{1}^{2}+a_{2}^{2}+b_{1}^{2}+b_{2}^{2}+2 b_{1} \cos 2 \theta+2 b_{2} \sin 2 \theta \\
=0,
\end{array}
$$

This equation holds for all $\theta(0 \leqslant \theta<2 \pi)$.
Let $\theta=0, \frac{\pi}{2}$, we get
$$
\begin{array}{l}
a_{1}^{2}+a_{2}^{2}+b_{1}^{2}+b_{2}^{2}+2 b_{1}=0, \\
a_{1}^{2}+a_{2}^{2}+b_{1}^{2}+b_{2}^{2}-2 b_{1}=0,
\end{array}
$$

Thus, $b_{1}=0$.
Let $\theta=\frac{\pi}{4}, \frac{3 \pi}{4}$, we get
$$
\begin{array}{c}
a_{1}^{2}+a_{2}^{2}+b_{1}^{2}+b_{2}^{2}+2 b_{2}=0, \\
a_{1}^{2}+a_{2}^{2}+b_{1}^{2}+b_{2}^{2}-2 b_{2}=0 .
\end{array}
$$

Thus, $b_{2}=0$. Substituting $b_{1}=0, b_{2}=0$ into (3), we get $a_{1}=a_{2}=0$.
In summary, $a=b=0$.