\documentclass[12pt,a4paper]{article}
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        {\textcolor{darkpurple}{\textbf{Exercise #1:}}};
    \end{tikzpicture}
}

% Define header and footer
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\fancyhf{} % Clear header and footer
\fancyhead[L]{Lycee Taghzirt\textbf{/}Prof MOSAID}
\fancyhead[C]{\hspace*{2.5cm}2023-2024\textbf{/}Correction Série: Complexes}
\fancyhead[R]{2BAC-PC/SVT}
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\begin{document}
\mylink \hfill \mylink\\
%============================
% add exercices here
%============================
\noindent
\begin{minipage}[t]{0.48\textwidth}
\exe{1}\\
\\
\hspace*{0.1cm}\mylabel[green]{1}
\vspace{-2\baselineskip}
\begin{align*}
    (2+3i)+(5-2i)&= 2+5+(3-2)i \\
    &=7+i
\end{align*}
\vspace{-2\baselineskip}
\mylabel[green]{2}\begin{align*}
    (4-2i)-(3+6i)&= 4-3-2i-6i \\
    &= 1-8i
\end{align*}
\mylabel[green]{3}\begin{align*}
    (2+i)(3-2i)&=2\times3-2\times2i+i\times3-2\times i^2  \\
    &=6-4i+3i+2  \\
    &=8-i  \\
\end{align*}
\vspace{-2\baselineskip}
\mylabel[green]{4}\begin{align*}
    (3-4i)^2 &= 9-24i-16 \\
    %\intertext{}
    &= -7-24i\\
\end{align*}
\vspace{-2\baselineskip}
\mylabel[green]{5}
\begin{align*}
    (2+i)^3 &= 8+3\times4i+3\times2\times i^2 +i^3 \\
    %\intertext{}
    &= 8+12i-6-i\\
    &= 2+11i
\end{align*}
\end{minipage}
\vline
\begin{minipage}[t]{0.51\textwidth}
\mylabel[green]{6}
\begin{align*}
    \frac{3-4i}{1+i}\times \frac{1-i}{2+3i} &= \frac{(3-4i)\textcolor{red}{(1-i)}}{(1+i)\textcolor{red}{(1-i)}}\times \frac{(1-i)\textcolor{red}{(2-3i)}}{(2+3i)\textcolor{red}{(2-3i)}} \\
    %\intertext{}
    &=\frac{3-3i-4i+4i^2}{1-i^2}\times \frac{2-3i-2i+3i^2}{4-9i^2} \\
    &= \frac{3-7i-4}{1+1}\times \frac{2-5i-3}{4+9}\\
    &=\frac{-1-7i}{2}\times \frac{-1-5i}{13}\\
    &=\frac{1+5i+7i+35i^2}{26}\\
    &=\frac{-34+12i}{26}\\
    &=\frac{-34}{26}+\frac{12}{26}i\\
    &=\frac{-17}{13}+\frac{6}{13}i
\end{align*}\\
\mylabel[green]{7}
\vspace{-2\baselineskip}
\begin{align*}
    \frac{2+3i}{1-i} &= \frac{(2+3i)(1+i)}{(1-i)(1+i)} \\
    %\intertext{}
    &= \frac{2+2i+3i-3}{1+1}\\
    &= \frac{-1+5i}{2}\\
    &= \frac{-1}{2}+\frac{5}{2}i
\end{align*}

\noindent

\end{minipage}
\noindent
\exe{2}\hfill \mylink\\
\begin{tabularx}{\textwidth}{L{0.10\textwidth}L{0.10\textwidth}L{0.20\textwidth}L{0.65\textwidth}}
\hline
\text{nombre} & \text{conjugué}&\text{module} & \text{forme trigonométrique.} \\
a + ib&a-ib&r=\sqrt{a^2 + b^2} & \theta = \tan^{-1}\left(\frac{\text{Im}(z)}{\text{Re}(z)}\right) =  \tan^{-1}\left(\frac{b}{a}\right),\quad z = r(\cos \theta + i \sin \theta)\\
\hline
4+2i&4-2i&\sqrt{4^2+2^2} = \sqrt{20} &\theta = \tan^{-1}\left(\frac{2}{4}\right)\\
&&& z=\sqrt{20}\bigg(\cos\left(\tan^{-1}(\frac{1}{2})\right) + i\sin\left(\tan^{-1}(\frac{1}{2})\right)\bigg)\\
\hline
1-\sqrt{2}i & 1+\sqrt{2}i& \sqrt{1^2+(-\sqrt{2})^2}=\sqrt{3} & \theta = \tan^{-1}\left(-\sqrt{2}\right), \quad z = \sqrt{3}(\frac{1}{\sqrt{3}}+\frac{-\sqrt{2}}{\sqrt{3}}i)\\
&&& z=\sqrt{3}\bigg(\cos\left(\tan^{-1}(-\sqrt{2})\right) + i\sin\left(\tan^{-1}(-\sqrt{2})\right)\bigg)\\
\hline
(2+2i)^5& (2-2i)^5& (\sqrt{2^2+2^2})^5& z^n=[r(\cos\theta+i\sin\theta)]^n=r^n(\cos n\theta+i\sin n\theta)\\
    &&=(2\sqrt{2})^5& z = 2\sqrt{2}(\frac{\sqrt2}{2}+i\frac{\sqrt2}{2})=2\sqrt2(\cos \frac{\pi}{4}+i\sin \frac{\pi}{4})\\
    &&& z^n = (2\sqrt{2})^5(\cos \frac{5\pi}{4}+i\sin \frac{5\pi}{4})\\
    \hline
1+i\sqrt3&1-i\sqrt2&\sqrt{1^2+\sqrt3^2}=2& z=2(\frac{1}{2}+\frac{\sqrt3}{2}i)=2(\cos\frac{\pi}{3}+i\sin\frac{\pi}{3})\\
    \hline
    &&&\\
    \frac{1+i\sqrt3}{(2+2i)^5}&&\frac{2}{(2\sqrt2)^5}&z=\frac{[r,\theta]}{[r',\theta']}=[\frac{r}{r'},\theta-\theta']=\frac{[2,\frac{\pi}{3}]}{[(2\sqrt2)^5,\frac{5\pi}{4}]}=[(2\sqrt2)^5,\frac{\pi}{3}-\frac{5\pi}{4}]\\
    &&&=[(2\sqrt2)^5,\frac{-11\pi}{12}]\\
\end{tabularx}
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\textcolor{white}{.}\hfill \underline{MOSAID le \today}\\
\textcolor{white}{.}\hfill \mylink
\end{document}