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\begin{document}
\thispagestyle{empty}
\noindent\begin{tabularx}{\textwidth}{@{} lCr @{}}
    Lycee Taghzirt\textbf{/}Prof MOSAID &
    2023-2024\textbf{/}Devoir Libre 1 S02&
    TCSF-3\textbf{/}2h\\
    \bottomrule
\end{tabularx}
\mylink \hfill \mylink\\
\exe{4}\\
     Résoudre les équation complexe suivantes :
    \( z^2 + 2z + 5 = 0, \quad z^2+z+1=0, \quad z^4+1=0\) \\
    \hspace*{1cm}\( z^2 + 2z + 5 = 0, \quad 2z^2-4z+7=0 \) \\

\begin{center}
\textcolor{red}{\textbf{\underline{Solution:}}}\\
\end{center}

\mylabel[green]{1.} Pour \( z^2 + 2z + 5 = 0 \) :

\( \Delta = b^2 - 4ac = 2^2 - 4 \times 1 \times 5 = 4 - 20 = -16 \)

Puisque \( \Delta \) est négatif, nous pouvons directement utiliser la formule :
\( z = \frac{{-b \pm i\sqrt{{-\Delta}}}}{{2a}} \)\\

Donc \(z_1= \frac{{-2 - i\sqrt{{16}}}}{{2 \times 1}} = \frac{{-2 - 4i}}{2} = -1 - 2i \)
et
\( z_2 = \frac{{-2 + i\sqrt{{16}}}}{{2 \times 1}} = \frac{{-2 + 4i}}{2} = -1 + 2i \)\\

\mylabel[green]{2.} Pour \( z^2 + z + 1 = 0 \) :

\( \Delta = b^2 - 4ac = 1^2 - 4 \times 1 \times 1 = 1 - 4 = -3 \)

Encore une fois, \( \Delta \) est négatif, donc :

\( z = \frac{{-1 - i\sqrt{{3}}}}{{2 \times 1}} = \frac{{-1 - i\sqrt{{3}}}}{2} \)
et
\( z = \frac{{-1 + i\sqrt{{3}}}}{{2 \times 1}} = \frac{{-1 + i\sqrt{{3}}}}{2} \)\\

\mylabel[green]{3.} Pour \( z^4 + 1 = 0 \) :\\

On a
\begin{align*}
    z^4+1&=0\\
    z^4-(-1)&=0\\
    (z^2)^2-i^2&=0\\
    (z^2-i)(z^2+i)&=0\\
    z^2-i=0 \qquad\text{ou} &\qquad z^2+i=0\\
    z^2=i \qquad\text{ou} &\qquad z^2=-i\\
    z^2=e^{i\frac{\pi}{2}} \qquad\text{ou} &\qquad z^2=e^{-i\frac{\pi}{2}}
    \qquad\text{remarquer que \(i = 0+1\times i = \cos \frac{\pi}{2} +\sin\frac{\pi}{2}\times i\)}\\
    z = \pm(e^{i\frac{\pi}{2}})^{\frac{1}{2}} \qquad\text{ou}
    &\qquad z = \pm(e^{-i\frac{\pi}{2}})^{\frac{1}{2}}\\
    z = \pm e^{i\frac{\pi}{4}} \qquad\text{ou}
    &\qquad z = \pm e^{-i\frac{\pi}{4}}\\
    z = \pm(\cos\frac{\pi}{4} + i \sin\frac{\pi}{4}) \quad\text{ou}&
    \quad z = \pm(\cos\frac{-\pi}{4} + i \sin\frac{-\pi}{4}) \\
\end{align*}
Alors
\begin{align*}
    \quad z = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}i \quad\text{ou}&
    \quad z = -\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}i \quad\text{ou}&
    \quad z = \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}i \quad\text{ou}&
    \quad z =  -\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}i \\
\end{align*}



\mylabel[green]{4.} Pour \( 2z^2 - 4z + 7 = 0 \) :

\( \Delta = b^2 - 4ac = (-4)^2 - 4 \times 2 \times 7 = 16 - 56 = -40 \)

Donc :

\( z_1 = \frac{{4 - i\sqrt{{40}}}}{{2 \times 2}} = \frac{{4 - 2i\sqrt{{10}}}}{4} = 1 -i\frac{{\sqrt{{10}}}}{2} \)
et
\( z_2 = \frac{{4 + i\sqrt{{40}}}}{{2 \times 2}} = \frac{{4 + 2i\sqrt{{10}}}}{4} = 1 + i\frac{{\sqrt{{10}}}}{2} \)
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