Bài 4.18 trang 202 SBT giải tích 12

Đề bài

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. ${z_1}{z_2} - \overline {{z_1}} \overline {{z_2}}  \in \mathbb{R}$

Lời giải chi tiết

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.