Bài tập 4.38 trang 208 SBT Toán 12

Gọi số phức \(z = a + bi,a,b \in R\)

\(\begin{array}{l}
\overline z  = {z^3}\\
 \Leftrightarrow a - bi = {(a + bi)^3}\\
 \Leftrightarrow a - bi = {a^3} + 3{a^2}bi + 3a{b^2}{i^2} + {b^3}{i^3}\\
 \Leftrightarrow a - bi = {a^3} - 3a{b^2} + (3{a^2}b - {b^3})i\\
 \Leftrightarrow \left\{ \begin{array}{l}
a = {a^3} - 3a{b^2}\\
 - b = 3{a^2}b - {b^3}
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
a = 0\\
 - b =  - {b^3}
\end{array} \right.\\
\left\{ \begin{array}{l}
1 = {a^2} - 3{b^2}\\
 - b = 3{a^2}b - {b^3}
\end{array} \right.
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
a = 0\\
\left[ \begin{array}{l}
b = 0\\
b =  \pm 1
\end{array} \right.
\end{array} \right.\\
\left\{ \begin{array}{l}
{a^2} = 1 + 3{b^2}\\
 - b = 3\left( {1 + 3{b^2}} \right)b - {b^3}
\end{array} \right.
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
a = 0\\
\left[ \begin{array}{l}
b = 0\\
b =  \pm 1
\end{array} \right.
\end{array} \right.\\
\left\{ \begin{array}{l}
{a^2} = 1 + 3{b^2}\\
8{b^3} + 4b = 0
\end{array} \right.
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
a = 0\\
\left[ \begin{array}{l}
b = 0\\
b =  \pm 1
\end{array} \right.
\end{array} \right.\\
\left\{ \begin{array}{l}
a =  \pm 1\\
b = 0
\end{array} \right.
\end{array} \right.
\end{array}\)

Vậy \(z \in \left\{ {0;1; - 1;i; - i} \right\}\)

b) 

\(\begin{array}{l}
|z| + z = 3 + 4i\\
 \Leftrightarrow \sqrt {{a^2} + {b^2}}  + a + bi = 3 + 4i\\
 \Leftrightarrow \left\{ \begin{array}{l}
\sqrt {{a^2} + {b^2}}  + a = 3\\
b = 4
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
\sqrt {{a^2} + 16}  = 3 - a\\
b = 4
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
a < 3\\
{a^2} + 16 = 9 - 6a + {a^2}\\
b = 4
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
a =  - \frac{7}{6}\\
b = 4
\end{array} \right.
\end{array}\)

Vậy \(z =  - \frac{7}{6} + 4i\)