12.3. Let the complex number $z$, for which $z^{2020}|z|+\bar{z} i=0$. Determine the set of values of the expression

$$
E=z^{2020}-z^{2019} i+\cdots+(-1)^{k} z^{2020-k} i^{k}+\cdots+z i+1
$$
Solution 1. $z^{2020}|z|+\bar{z} i=0 \Leftrightarrow z^{2021}|z|+z \bar{z} i=0 \Leftrightarrow z^{2021}|z|+|z|^{2} i=0 \Leftrightarrow$

$$
\Leftrightarrow z^{2021}|z|=-|z|^{2} i \Rightarrow|z|^{2021}=|z| \Leftrightarrow\left[\begin{array}{l}
|z|=0 \\
|z|=1
\end{array}\right.
$$

For $|z|=0, E=z^{2020}-z^{2019} i+\cdots+(-1)^{k} z^{2020-k} i^{k}+\cdots+z i+1=1$.

For $|z|=1$, from the equality $z^{2021}|z|+|z|^{2} i=0$ we get $z^{2021}=-i$.

In this case for $z i \neq 1$

$$
\begin{gathered}
E=z^{2020}\left(1+\frac{1}{z i}+\left(\frac{1}{z i}\right)^{2}+\cdots+\left(\frac{1}{z i}\right)^{k}+\left(\frac{1}{z i}\right)^{2020}\right)=z^{2020} \frac{1-\left(\frac{1}{z i}\right)^{2021}}{1-\frac{1}{z i}}= \\
=z^{2020} \frac{(z i)^{2021}-1}{(z i)^{2021}} \cdot \frac{z i}{z i-1}=\frac{z^{2021} i-1}{z i-1}=\frac{-i^{2}-1}{z i-1}=0
\end{gathered}
$$

However, if $z i=1 \Leftrightarrow z=-i$, then $E=(-i)^{2020}+(-i)^{2020}+\cdots+(-i)^{2020}+1=2021$.

In conclusion, $E=z^{2020}-z^{2019} i+\cdots+(-1)^{k} z^{2020-k} i^{k}+\cdots+z i+1=\left\{\begin{array}{c}1, z=0 \\ 0, z \neq-i, \quad z \neq 0 \\ 2021, z=-i\end{array}\right.$