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\fancyhead[L]{Lycee Taghzirt\textbf{/}Prof MOSAID}
\fancyhead[C]{\hspace*{2.5cm}2023-2024\textbf{/}Correction Série: Complexes}
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\begin{document}
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\noindent
\begin{minipage}[t]{0.48\textwidth}
\exe{3}\\
 Considérez les points \( A \), \( B \), et \( C \) dans le plan complexe avec
les affixes suivantes :\\
\hspace*{1cm} \( a = 2 + 3i, \quad b = -1 - i, \quad c = 4 + 2i \) \\
\mylabel[lightblue]{1}Calculez les distances  \( AB \) et \( BC \).\\
 \mylabel[lightblue]{2}Trouvez les affixes des vecteurs \( \overrightarrow{AB} \) et \( \overrightarrow{AC} \).\\
\mylabel[lightblue]{3}Calculez la mesure de l'angle \( \widehat{(\overrightarrow{AB},\overrightarrow{AC}  )} \)\\
\hspace*{1cm}en radians.\\
\\
\Centering \myhighlight{yellow}{\large Solution}\\
\mylabel[green]{1} Pour calculer la distance entre deux points dans le plan complexe, nous utilisons la formule de la distance entre deux points \(z_1\) et \(z_2\) dans le plan complexe, qui est \(|z_2 - z_1|\).\\

\begin{align*}
  AB &= |b - a| \\
  &= |-1 - i - (2 + 3i)| \\
  &= |-3 - 4i| \\
  &= \sqrt{(-3)^2 + (-4)^2} \\
  &= \sqrt{9 + 16} \\
  &= \sqrt{25}\\
  &= 5
\end{align*}

\begin{align*}
   BC &=|c - b| \\
   &= |(4 + 2i) - (-1 - i)|\\
   &= |5 + 3i| \\
   &= \sqrt{5^2 + 3^2} \\
   &= \sqrt{25 + 9} \\
   &= \sqrt{34}
\end{align*}


\mylabel[green]{2} Les affixes des vecteurs \(\overrightarrow{AB}\) et \(\overrightarrow{AC}\) sont :


\begin{align*}
\overrightarrow{AB} &( b - a)\\
&( (-1 - i) - (2 + 3i))\\
&( -1 - i - 2 - 3i )\\
&( -3 - 4i )
\end{align*}

\begin{align*}
\overrightarrow{AC} &( c - a)\\
&( (4 + 2i) - (2 + 3i) )\\
&( 4 + 2i - 2 - 3i)\\
&( 2 - i )
\end{align*}
\end{minipage}
\hspace*{0.1cm}
\vline
\hspace*{0.1cm}
\begin{minipage}[t]{0.47\textwidth}
\mylabel[green]{3}Calculez la mesure de l'angle \( \widehat{(\overrightarrow{AB},\overrightarrow{AC}  )} \)
en radians.
\[\overline{(\overrightarrow{AB},\overrightarrow{AC})} \equiv Arg\Bigg(\frac{c-a}{b-a} \Bigg) [2\pi]\]


Calculons \(\frac{c-a}{b-a}\):
   \[
\begin{aligned}
\frac{c-a}{b-a} &= \frac{(4+2i)-(2+3i)}{(-1-i)-(2+3i)} \\
&= \frac{2-i}{-3-4i}\\
&= \frac{(2-i)(-3+4i)}{(-3-4i)(-3+4i)} \\
&= \frac{-6+8i+3i+4}{9+16} \\
&= \frac{-2+11i}{25}
\end{aligned}
\]

Donc, \(\frac{c-a}{b-a} = \frac{-2-11i}{25}=\frac{-2}{25}+i\frac{-11}{25}\). \\
Donc \(\theta = \tan^-1\bigg(\frac{-11}{-2}\bigg) = 79.69^\circ\)\\
puisque les deux valeurs de \(\sin \theta\) et \(\cos \theta\) sont négatives
alors \(\theta\) est dans le \(3^\text{ème}\) quard du cercle \\
alors \(\theta = 180 + 79.69 = 259.69^\circ = 1.443\pi rad\)

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