Có bao nhiêu số phức z thỏa mãn \(2\left| {z – i} \right| = \left| {z – \overline z + 2i} \right|\) và \(\left| {{z^2} – {{\left( {\overline z } \right)}^2}} \right| = 16\)?

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

Theo giả thiết, ta có hệ:

\(\left\{ \begin{array}{l}2\left| {z – i} \right| = \left| {z – \overline z + 2i} \right|\\\left| {{z^2} – {{\left( {\overline z } \right)}^2}} \right| = 16\end{array} \right.\)

\( \Leftrightarrow \left\{ \begin{array}{l}2\left| {a + bi – i} \right| = \left| {a + bi – a + bi + 2i} \right|\\\left| {{{\left( {a + bi} \right)}^2} – {{\left( {a – bi} \right)}^2}} \right| = 16\end{array} \right.\)

\( \Leftrightarrow \left\{ \begin{array}{l}2\left| {a + bi – i} \right| = \left| {\left( {2b + 2} \right)i} \right|\\\left| {\left( {{a^2} + 2abi – {b^2}} \right) – \left( {{a^2} – 2abi – {b^2}} \right)} \right| = 16\end{array} \right.\)

\( \Leftrightarrow \left\{ \begin{array}{l}2\sqrt {{a^2} + {{\left( {b – 1} \right)}^2}} = \sqrt {{{\left( {2b + 2} \right)}^2}} \\\left| {4abi} \right| = 16\end{array} \right.\)

\( \Leftrightarrow \left\{ \begin{array}{l}4\left[ {{a^2} + {{\left( {b – 1} \right)}^2}} \right] = 4{\left( {b + 1} \right)^2}\\\left| {abi} \right| = 4\end{array} \right.\)

\( \Leftrightarrow \left\{ \begin{array}{l}{a^2} + {b^2} – 2b + 1 = {b^2} + 2b + 1\\{a^2}{b^2} = 16\end{array} \right.\)

\( \Leftrightarrow \left\{ \begin{array}{l}{a^2} = 4b\\\left[ \begin{array}{l}ab = 4\\ab = – 4\end{array} \right.\end{array} \right.\)

\( \Leftrightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}b = \frac{{{a^2}}}{4}\\a.\frac{{{a^2}}}{4} = 4\end{array} \right.\\\left\{ \begin{array}{l}b = \frac{{{a^2}}}{4}\\a.\frac{{{a^2}}}{4} = – 4\end{array} \right.\end{array} \right.\)

\( \Leftrightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}b = \frac{{{a^2}}}{4}\\{a^3} = 16\end{array} \right.\\\left\{ \begin{array}{l}b = \frac{{{a^2}}}{4}\\{a^3} = – 16\end{array} \right.\end{array} \right.\)

\( \Leftrightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}b = \sqrt[3]{4}\\a = 2\sqrt[3]{2}\end{array} \right.\\\left\{ \begin{array}{l}b = \sqrt[3]{4}\\a = – 2\sqrt[3]{2}\end{array} \right.\end{array} \right.\)

Vậy có 2 số phức z thỏa đề: \(z = 2\sqrt[3]{2} + \sqrt[3]{4}i\) và \(z = – 2\sqrt[3]{2} + \sqrt[3]{4}i\)