Có bao nhiêu số phức $z$ thỏa mãn $\left| z \right|=1$ và $\left|...

Ta có $z.\overline{z}={{\left| z \right|}^{2}}=1\Rightarrow \overline{z}=\dfrac{1}{z}$
Khi đó $\left( * \right)\Leftrightarrow \left| {{z}^{3}}+2024z+\dfrac{1}{z} \right|-2\sqrt{3}\left| z+\dfrac{1}{z} \right|=2019$ $$
$\Leftrightarrow \left| z \right|\left| {{z}^{2}}+\frac{1}{{{z}^{2}}}+2024 \right|-2\sqrt{3}\left| z+\frac{1}{z} \right|=2019$
$\Leftrightarrow \left| {{\left( z+\frac{1}{z} \right)}^{2}}+2022 \right|-2\sqrt{3}\left| z+\frac{1}{z} \right|=2019\left( ** \right)$
Đặt $z=a+bi\left( a,b\in \mathbb{R} \right)$ $\Rightarrow z+\frac{1}{z}=z+\frac{\overline{z}}{z.\overline{z}}$ $=2a$ ( số thực) ( Vì $z.\overline{z}={{\left| z \right|}^{2}}=1$ )
Vậy, $\left( ** \right)\Leftrightarrow \left| 4{{a}^{2}}+2022 \right|-2\sqrt{3}\left| 2a \right|=2019$
$\Leftrightarrow 4{{a}^{2}}-4\sqrt{3}\left| a \right|+3=0$
$\Leftrightarrow {{\left( 2\left| a \right|-\sqrt{3} \right)}^{2}}=0\Leftrightarrow \left| a \right|=\frac{\sqrt{3}}{2}\Leftrightarrow $ $\left[ \begin{aligned}
& a=\dfrac{\sqrt{3}}{2}\Rightarrow \left[ \begin{aligned}
& b=\dfrac{1}{2} \\
& b=-\dfrac{1}{2} \\
\end{aligned} \right. \\
& a=-\dfrac{\sqrt{3}}{2}\Rightarrow \left[ \begin{aligned}
& b=\dfrac{1}{2} \\
& b=-\dfrac{1}{2} \\
\end{aligned} \right. \\
\end{aligned} \right.$
Vậy có $4$ số phức thỏa mãn.