Bài 4.18 trang 202 SBT giải tích 12

Cho \({z_1},{z_2} \in \mathbb{C}\). Khẳng định nào sau đây sai?

A. \({z_1}\overline {{z_2}}  + \overline {{z_1}} {z_2} \in \mathbb{R}\)

B. \({z_1}{z_2} + \overline {{z_1}} \overline {{z_2}}  \in \mathbb{R}\)

C. \({z_1}\overline {{z_2}} \overline {{z_1}} {z_2} \in \mathbb{R}\)

D. \({z_1}{z_2} - \overline {{z_1}} \overline {{z_2}}  \in \mathbb{R}\)

Đáp án A:

Đặt \(z = {z_1}\overline {{z_2}}  + \overline {{z_1}} {z_2}\) ta có: \(\overline z  = \overline {{z_1}\overline {{z_2}}  + \overline {{z_1}} {z_2}}  = \overline {{z_1}\overline {{z_2}} }  + \overline {\overline {{z_1}} {z_2}} \)\( = \overline {{z_1}} .\overline {\overline {{z_2}} }  + \overline {\overline {{z_1}} } .\overline {{z_2}}  = \overline {{z_1}} {z_2} + {z_1}\overline {{z_2}}  = z\).

Do đó \({z_1}\overline {{z_2}}  + \overline {{z_1}} {z_2} \in \mathbb{R}\).

Đáp án B:

Đặt \(z = {z_1}{z_2} + \overline {{z_1}} \overline {{z_2}} \) ta có: \(\overline z  = \overline {{z_1}{z_2} + \overline {{z_1}} \overline {{z_2}} }  = \overline {{z_1}{z_2}}  + \overline {\overline {{z_1}} \overline {{z_2}} } \) \( = \overline {{z_1}} .\overline {{z_2}}  + {z_1}{z_2} = z\) nên \(z \in \mathbb{R}\).

Đáp án C:

Đặt \(z = {z_1}\overline {{z_2}} \overline {{z_1}} {z_2}\) ta có: \(\overline z  = \overline {{z_1}\overline {{z_2}} \overline {{z_1}} {z_2}}  = \overline {{z_1}} \overline {\overline {{z_2}} } \overline {\overline {{z_1}} } .\overline {{z_2}} \) \( = \overline {{z_1}} .{z_2}.{z_1}.\overline {{z_2}}  = z\) nên \(z \in \mathbb{R}\).

Đáp án D:

Đặt \(z = {z_1}{z_2} - \overline {{z_1}} \overline {{z_2}} \) ta có: \(\overline z  = \overline {{z_1}{z_2} - \overline {{z_1}} \overline {{z_2}} } \) \( = \overline {{z_1}{z_2}}  - \overline {\overline {{z_1}} .\overline {{z_2}} }  = \overline {{z_1}} .\overline {{z_2}}  - {z_1}{z_2} \ne z\) nên \(z \notin \mathbb{R}\).

Chọn D.