7. If the complex numbers $a, b, c$ satisfy
$$
\begin{aligned}
|a|=|b|=|c|=1=a^{2}+b^{2}+c^{2}, \\
\text { then }\left|a^{2020}+b^{2020}+c^{2020}\right|=
\end{aligned}
$$
$$
\begin{array}{c}
\Rightarrow \cos 2 x+\cos 2 y+\cos 2 z=1, \\
\quad \sin 2 x+\sin 2 y+\sin 2 z=0 .
\end{array}
$$

The above two equations, when transformed using sum-to-product identities, yield
$$
\begin{array}{l}
2 \cos (x+y) \cdot \cos (x-y)=2 \sin ^{2} z, \\
2 \sin (x+y) \cdot \cos (x-y)=-2 \sin z \cdot \cos z .
\end{array}
$$

If $\cos (x-y)=0$, then
$$
\sin z=0, a= \pm \mathrm{i} b, c=1 \text {. }
$$

If $\cos (x-y) \neq 0$, then $\sin z \neq 0$.
(1) divided by (2) gives
$$
\begin{array}{l}
\cos (x+y) \cdot \cos z+\sin (x+y) \cdot \sin z=0 \\
\Rightarrow \cos (x+y-z)=0 \\
\Rightarrow \cos (x+y)= \pm \sin z \\
\Rightarrow \cos (x+y)=\cos (x-y) \\
\Rightarrow \sin x \cdot \sin y=0 .
\end{array}
$$

Assume without loss of generality that $\sin x=0$.
Substituting this in, we get $a=1, b= \pm \mathrm{i} c$, which is consistent with the previous result.
Therefore, $\left|a^{2020}+b^{2020}+c^{2020}\right|=1$.
7.1.

Let $a=\cos x+\mathrm{i} \sin x$,
$b=\cos y+\mathrm{i} \sin y$,
$c=\cos z+\mathrm{i} \sin z$.
Then $a^{2}+b^{2}+c^{2}=1$