80 Given that $n$ is a positive integer, $A$ is a complex number, $i^{2}=-1$, try to find the necessary and sufficient condition for all roots of the equation $\left(\frac{1+i \cdot x}{1-i \cdot x}\right)^{n}=A$ to be distinct real numbers (express the condition in terms of $A$).
80 From $\left(\frac{1+i x}{1-i x}\right)^{n}=A$, taking the conjugate gives $\left(\frac{1-i x}{1+i x}\right)^{n}=\bar{A}$. Multiplying the two equations yields $\bar{A} \cdot A=1$, i.e.,
$$
|A|^{2}=1,|A|=1 .
$$

To prove the sufficiency: Given $|A|=1$, we can set $A=\cos \theta+i \cdot \sin \theta(0 \leqslant \theta<2 \pi)$. Thus, the set of roots of the equation $\left(\frac{1+i \cdot x}{1-i \cdot x}\right)^{n}=A$ is the union of the sets of roots of the following $n$ equations:
$$
\begin{array}{c}
\frac{1+i \cdot x}{1-i \cdot x}=\cos \frac{\theta+2 k \pi}{n}+i \cdot \sin \frac{\theta+2 k \pi}{n}=\mathrm{e}^{\frac{\theta+2 k \pi}{n} \cdot i} \\
(k=0,1,2, \cdots, n-1) \\
x=\frac{1-\mathrm{e}^{\frac{\theta+2 k \pi}{n} \cdot i}}{1+\mathrm{e}^{\frac{\theta+2 k \pi}{n} \cdot i}} \cdot i
\end{array}
$$

Solving this, let $\beta_{k}=\frac{1}{2} \cdot \frac{\theta+2 k \pi}{n}$, then
$$
\begin{aligned}
x & =\frac{1-\cos 2 \beta_{k}-i \cdot \sin 2 \beta_{k}}{1+\cos 2 \beta_{k}+i \cdot \sin 2 \beta_{k}} \cdot i=\frac{2 \cdot \sin ^{2} \beta_{k}-2 \cdot \sin \beta_{k} \cdot \cos \beta_{k} \cdot i}{2 \cdot \cos ^{2} \beta_{k}+2 \cdot \sin \beta_{k} \cdot \cos \beta_{k} \cdot i} \cdot i \\
& =\tan \beta_{k}
\end{aligned}
$$

Therefore, the roots of the equation $\left(\frac{1+i \cdot x}{1-i \cdot x}\right)^{n}=A(|A|=1)$ are $\tan \beta_{k}(k=0,1,2, \cdots, n-1)$, and all roots are distinct real numbers. When $n$ is even and $\theta=0$ (i.e., $A=1$), $k=\frac{n}{2}$ should be removed. When $n$ is odd and $\theta=\pi$ (i.e., $A=-1$), $k=\frac{n-1}{2}$ should be removed.