Tìm tất cả số phức z thỏa \({z^2} = {\left| z \right|^2} + \overline z \)
A. \(z = 0,\,\,z = - \frac{1}{2} + \frac{1}{2}i,\,\,z = - \frac{1}{2} - \frac{1}{2}i\).
B. \(z = 0,\,\,z = - \frac{1}{2} + \frac{1}{2}i,\,\,z = \frac{1}{2} - \frac{1}{2}i\).
C. \(z = 0,\,\,z = - 1 - \frac{1}{2}i,\,\,z = - 1 + \frac{1}{2}i\).
D. \(z = 0,\,\,z = - \frac{1}{4} + \frac{1}{4}i,\,\,z = - \frac{1}{4} - \frac{1}{4}i\).
A. \(z = 0,\,\,z = - \frac{1}{2} + \frac{1}{2}i,\,\,z = - \frac{1}{2} - \frac{1}{2}i\).
B. \(z = 0,\,\,z = - \frac{1}{2} + \frac{1}{2}i,\,\,z = \frac{1}{2} - \frac{1}{2}i\).
C. \(z = 0,\,\,z = - 1 - \frac{1}{2}i,\,\,z = - 1 + \frac{1}{2}i\).
D. \(z = 0,\,\,z = - \frac{1}{4} + \frac{1}{4}i,\,\,z = - \frac{1}{4} - \frac{1}{4}i\).
Đặt \(z = x + yi,\,\,x,y \in \mathbb{R}\)\( \to \overline z = x - yi\)
Ta có: \({z^2} = {\left| z \right|^2} + \overline z \Leftrightarrow 2{y^2} + x - (2xy + y)i = 0 \Leftrightarrow \left\{ \begin{array}{l}2{y^2} + x = 0\\2xy + y = 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x = 0\\y = 0\end{array} \right. \vee \left\{ \begin{array}{l}x = - \frac{1}{2}\\y = \frac{1}{2}\end{array} \right. \vee \left\{ \begin{array}{l}x = - \frac{1}{2}\\y = - \frac{1}{2}\end{array} \right.\)
\( \Rightarrow \)\(z = 0,\,\,z = - \frac{1}{2} + \frac{1}{2}i,\,\,z = - \frac{1}{2} - \frac{1}{2}i\)
Vậy chọn đáp án A.
Ta có: \({z^2} = {\left| z \right|^2} + \overline z \Leftrightarrow 2{y^2} + x - (2xy + y)i = 0 \Leftrightarrow \left\{ \begin{array}{l}2{y^2} + x = 0\\2xy + y = 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x = 0\\y = 0\end{array} \right. \vee \left\{ \begin{array}{l}x = - \frac{1}{2}\\y = \frac{1}{2}\end{array} \right. \vee \left\{ \begin{array}{l}x = - \frac{1}{2}\\y = - \frac{1}{2}\end{array} \right.\)
\( \Rightarrow \)\(z = 0,\,\,z = - \frac{1}{2} + \frac{1}{2}i,\,\,z = - \frac{1}{2} - \frac{1}{2}i\)
Vậy chọn đáp án A.