Question

Cum se rezolvă ex 3? ​

Answer 1

Explicație pas cu pas:

conjugatul numărului complex z = a + bi este:

$\boxed {\overline {z} = a - bi}$

a)

$2 - 4i = \overline {m + {m}^{2} i} \\ 2 - 4i = m - {m}^{2} i$

$\begin{cases}2 = m \\- 4 = - {m}^{2} \end{cases} \iff \begin{cases}m = 2 \\{m}^{2} = {2}^{2} \end{cases}$

$\begin{cases}m = 2 \\ \\ m \in \Big\{ - 2; 2\Big\} \end{cases} \implies m = 2$

b)

$\boxed {\overline {\overline {z}} = z}$

$\overline {\overline {9 + 5 i}} = \overline { {m}^{2} + 8 + ({m}^{2} - 6m) i}$

$9 + 5 i = ({m}^{2} + 8) - ({m}^{2} - 6m) i$

$\begin{cases} 9 = {m}^{2} + 8 \\ 5 = - ({m}^{2} - 6m) \end{cases} \iff \begin{cases}{m}^{2} = 9 - 8 \\{m}^{2} - 6m + 5 = 0 \end{cases}$

$\begin{cases}{m}^{2} = 1 \\(m - 5)(m - 1) = 0 \end{cases} \iff \begin{cases}m \in \Big\{ - 1; 1\Big\} \\ \\ m \in \Big\{1; 5\Big\} \end{cases}$

$\implies m = 1$

c)

$\boxed { {i}^{2} = - 1}$

$\dfrac{1 - i}{1 + i} = \dfrac{(1 - i)(1 - i)}{(1 + i)(1 - i)} = \dfrac{1 - 2i + {i}^{2} }{1 - {i}^{2} } = \dfrac{1 - 2i - 1}{1 + 1} = \dfrac{- 2i}{2} = - i$

$\overline { \dfrac{1 - i}{1 + i} } = {m}^{2} i \iff \overline {- i} = {m}^{2} i$

$\overline {0 - i} = {m}^{2} i \implies 0 + i = 0 + {m}^{2} i$

$\begin{cases} 0 = 0 \\ 1 = {m}^{2} \end{cases} \iff \begin{cases}m \in \mathbb{R} \\ \\ m \in \Big\{ - 1; 1\Big\} \end{cases}$

$\implies m \in \Big\{ - 1; 1\Big\}$

d)

$\overline { {(1 - 2i)}^{2} } = \overline {1 - 4i + 4 {i}^{2}} = \overline {1 - 4i - 4} = \overline { - 3 - 4i} = - 3 + 4i$

$\overline { {(1 - 2i)}^{2} } = m - 1 + {m}^{2} i$

$- 3 + 4i = (m - 1) + {m}^{2} i$

$\begin{cases} - 3 = m - 1 \\ 4 = {m}^{2} \end{cases} \iff \begin{cases} m = - 3 + 1 \\ {m}^{2} = {2}^{2} \end{cases}$

$\begin{cases} m = - 2 \\ {m}^{2} = {2}^{2} \end{cases} \iff \begin{cases} m = - 2 \\ \\ m \in \Big\{ - 2; 2\Big\} \end{cases}$

$\implies m = - 2$