High $18^{\circ}$, no real roots for the $n$-th degree equation
$$
b_{n} x^{n}+b_{n-1} x^{n-1}+\cdots+b_{1} x+b_{0}=0
$$

all roots are imaginary, $b_{k}=a_{k}+a_{n-k} \cdot i(k=0,1, \cdots, n)$ and satisfy
$$
0<a_{0}<a_{1}<a_{2}<\cdots<a_{n-1}<a_{n} .
$$

If $z$ is a complex root of the equation, find $|z|$.
Let $f(x)=a_{n} x^{n}+a_{n-1} x^{n-1}+\cdots+a_{1} x+a_{0}$, and $z_{0}$ is a root of the equation $f(x)=0$. Then $\left|z_{0}\right|<1$.
In fact, by the condition, we know that $z_{0} \neq 0,1$. Because
$$
\begin{array}{l}
0=\left(z_{0}-1\right)\left(a_{n} z_{0}^{n}+a_{n-1} z_{0}^{n-1}+\cdots+a_{1} z_{0}+a_{0}\right) \\
=a_{n} z_{0}^{n+1}+\left(a_{n-1}-a_{n}\right) z_{0}^{n}+\left(a_{n-2}-a_{n-1}\right) z_{0}^{n-1} . \\
+\cdots+\left(a_{0}-a_{1}\right) z_{0}-a_{0} \text {, so, } \\
a_{n} z_{0}^{n+1}=\left(a_{n}-a_{n-1}\right) z_{0}^{n}+\left(a_{n-1}-a_{n-2}\right) z_{0}^{n-1} \\
\quad+\cdots+\left(a_{1}-a_{0}\right) z_{0}+a_{0} .
\end{array}
$$

Assume $\left|z_{0}\right| \geqslant 1$, then
$$
\begin{array}{l}
\left|a_{n} z_{0}^{n+1}\right|=\mid\left(a_{n}-a_{n-1}\right) z_{0}^{n}+\left(a_{n-1}-a_{n-2}\right) z_{0}^{n-2} \\
\quad+\cdots+\left(a_{1}-a_{0}\right) z_{0}+a_{0} \mid \\
<\left(a_{n}-a_{n-1}\right)\left|z_{0}\right|^{n}+\left(a_{n-1}-a_{n-2}\right)\left|z_{0}\right|^{n-1} \\
+\cdots+\left(a_{1}-a_{0}\right)\left|z_{0}\right|+a_{0}
\end{array}
$$
(This is because the angles of $\left(a_{n}-a_{n-1}\right) z_{0}^{*},\left(a_{n-1}-a_{n-2}\right) z_{0}^{-1}, \cdots,\left(a_{1}-a_{0}\right) \cdot$ $x_{0}, a_{0}$ cannot be completely equal, unless $x_{0} \geqslant 0$, but in this case $x_{0}$ is not a root of $f(x)=0$)
$$
\begin{array}{l}
\leqslant\left[\left(a_{n}-a_{n-1}\right)+\left(a_{n-1}-a_{n-2}\right)+\cdots\right. \\
\left.+\left(a_{1}-a_{0}\right)+a_{0}\right]\left|z_{0}\right|^{n} \\
=a_{n}\left|z_{0}\right|^{n} .
\end{array}
$$

Therefore, $\left|a_{n}\right|\left|z_{0}\right|^{n+1}<\left|a_{n}\right|\left|z_{0}\right|^{n} \Rightarrow\left|z_{0}\right|<1$, which contradicts the assumption $\left|x_{0}\right| \geqslant 1$.
Let $z_{k}(k=1,2, \cdots, n)$ be the $n$ roots of $f(x)=0$. Then
$$
\begin{array}{l}
a_{n} x^{n}+a_{n-1} x^{n-1}+\cdots+a_{1} x+a_{0} \\
=a_{n}\left(x-z_{1}\right)\left(x-z_{2}\right) \cdots\left(x-z_{n}\right),
\end{array}
$$

Taking the conjugate on both sides, and noting that $\overline{a_{k}}=a_{k}(k=0,1, \cdots, n)$.
$$
\begin{array}{l}
a_{n} \bar{x}^{n}+a_{n-1} \bar{x}^{n-1}+\cdots+a_{1} \bar{x}+a_{0} \\
=a_{n}\left(\bar{x}-\bar{z}_{1}\right)\left(\bar{x}-\bar{z}_{2}\right) \cdots\left(\bar{x}-\bar{z}_{n}\right) . \\
\text { Substituting }\left(\frac{\overline{1}}{x}\right) \text { for } x \text { in the above equation, we get } \\
a_{n}\left(\frac{1}{x}\right)^{n}+a_{n-1}\left(\frac{1}{x}\right)^{n-1}+\cdots+a_{1}\left(\frac{1}{x}\right)+a_{0} \\
=a_{n}\left(\frac{1}{x}-\overline{z_{1}}\right)\left(\frac{1}{x}-\overline{z_{2}}\right) \cdots\left(\frac{1}{x}-\overline{z_{n}}\right),
\end{array}
$$

Substituting $\left(\frac{\overline{1}}{x}\right)$ for $x$ in the above equation, we get
$$
\begin{array}{l}
a_{n}\left(\frac{1}{x}\right)^{n}+a_{n-1}\left(\frac{1}{x}\right)^{n-1}+\cdots+a_{1}\left(\frac{1}{x}\right)+a_{0} \\
=a_{n}\left(\frac{1}{x}-\overline{z_{1}}\right)\left(\frac{1}{x}-\overline{z_{2}}\right) \cdots\left(\frac{1}{x}-\overline{z_{n}}\right),
\end{array}
$$

which is $a_{0} x^{n}+a_{1} x^{n-1}+\cdots+a_{n-1} x+a_{n}$
$$
=a_{n}\left(1-\overline{z_{1}} x\right)\left(1-\bar{z}_{2} x\right) \cdots\left(1-\overline{z_{n}} x\right) \text {. }
$$

Therefore, the equation can be rewritten as
$$
\begin{array}{l}
a_{n}\left(x-z_{1}\right)\left(x-z_{2}\right) \cdots\left(x-z_{n}\right) \\
=a_{n}\left(1-\overline{z_{1}} x\right)\left(1-\bar{z}_{2} x\right) \cdots\left(1-\overline{z_{n}} x\right)(-i) .
\end{array}
$$

Taking the modulus on both sides, we get
$$
\begin{array}{l}
\left|x-z_{1}\right| \cdot\left|x-z_{2}\right| \cdot \cdots \cdot\left|x-z_{n}\right| \\
=\left|1-\overline{z_{1}} x\right| \cdot\left|1-\overline{z_{2}} x\right| \cdot \cdots \cdot\left|1-\overline{z_{n}} x\right| .
\end{array}
$$

Thus,
$$
\begin{array}{l}
\left|z-z_{1}\right| \cdot\left|z-z_{2}\right| \cdot \cdots \cdot\left|z-z_{n}\right| \\
=\left|1-\overline{z_{1}} z\right| \cdot\left|1-\overline{z_{2}} z\right| \cdots\left|1-\overline{z_{n}} z\right| .
\end{array}
$$

Noting that $\left|z-z_{k}\right|^{2}-\left|1-\overline{z_{k}} z\right|^{2}$
$$
=\left(|z|^{2}-1\right)\left(1-\left|z_{k}\right|^{2}\right)
$$

and $\left|z_{k}\right|<1(k=1,2, \cdots, n)$, then
when $|z| \geqslant 1$, $\left|z-z_{n}\right| \geqslant\left|1-\bar{z}_{n} z\right|$,
when $|z| \leqslant 1$, $\left|z-z_{n}\right| \leqslant\left|1-\overline{z_{n}} z\right|$.

Therefore, equation (1) holds if and only if $|z|=1$.
Hence, $|z|=1$.