Có bao nhiêu số phức \(z\) thỏa mãn \({\left| {z - 1} \right|^2} + \left| {z - \overline z } \right|i + \left( {z + \overline z } \right){i^{2019}} = 1\,\,?\)

* Hướng dẫn giải

Gọi \(z = a + bi \Rightarrow \overline z  = a - bi\,\,\left( {a,\,b \notin \mathbb{R}} \right).\)

\(\begin{array}{l}\,\,\,\,\,{\left| {z - 1} \right|^2} + \left| {z - \overline z } \right|i + \left( {z + \overline z } \right){i^{2019}} = 1\\ \Leftrightarrow {\left| {a + bi - 1} \right|^2} + \left| {a + bi - a + bi} \right|i + \left( {a + bi + a - bi} \right)\left[ {{{\left( {{i^2}} \right)}^{1009}}.i} \right] = 1\\ \Leftrightarrow {\left( {a - 1} \right)^2} + {b^2} + \left| {bi} \right|i - 2ai = 1\\ \Leftrightarrow {\left( {a - 1} \right)^2} + {b^2} + \left( {\sqrt {{b^2}}  - 2a} \right)i = 1 \Leftrightarrow \left\{ \begin{array}{l}{\left( {a - 1} \right)^2} + {b^2} = 1\\\left| b \right| - 2a = 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}{\left( {a - 1} \right)^2} + {b^2} = 1\\\left[ \begin{array}{l}b = 2a\\b =  - 2a\end{array} \right.\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}b = 2a\\{a^2} - 2a + 1 + 4{a^2} = 1\end{array} \right.\\\left\{ \begin{array}{l}b =  - 2a\\{a^2} - 2a + 1 + 4{a^2} = 1\end{array} \right.\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}b = 2a\\\left[ \begin{array}{l}a = 0\\a = \frac{2}{5}\end{array} \right.\end{array} \right.\\\left\{ \begin{array}{l}b =  - 2a\\\left[ \begin{array}{l}a = 0\\b = \frac{2}{5}\end{array} \right.\end{array} \right. \Rightarrow \left[ \begin{array}{l}z = \frac{2}{5} + \frac{4}{5}i\\z = 0\\z = \frac{2}{5} - \frac{4}{5}i\end{array} \right..\end{array} \right.\end{array}\)

Chọn D.