\begin{enumerate}
\item $\begin{aligned}[t]
(z+2)(z^2-2z+4)=0 & \iff z+2=0 \quad \text{ou}\quad z^2-2z+4=0\\
& \iff z=-2 \text{ ou } z=\dfrac{-(-2)\pm\sqrt\Delta}{2}\text{ avec }\Delta=(-2)^2-4\times1\times4=-12\\
& \iff z=-2 \text{ ou } z=1-\i\sqrt3 \text{ ou } z=1+\i\sqrt3.
\end{aligned}$

\medskip

\uline{L'ensemble solution de (E) est donc : $\{-2~;~1-\i\sqrt3~;~1+\i\sqrt3\}$.}

\medskip


$a=-2$, $b=1+\i\sqrt3$ et $c=1-\i\sqrt3$.

\item
\begin{enumerate}
\item $|b|=\sqrt{1^2+\big(\sqrt3\big)^2}=\sqrt{4}=2$ et donc :
\[
b=2\left(\dfrac{1}{2}+\i\dfrac{\sqrt3}{2}\right)=2\left(\cos\dfrac{\pi}{3}+\i\sin\dfrac{\pi}{3}\right)
\]
donc :
\[
\boxed{b=2\e^{\i\frac{\pi}{3}}}
\]
De plus, $c=\overline{b}$ donc $\boxed{c=2\e^{-\i\frac{\pi}{3}}}$.

\item $\begin{aligned}[t]
|a-b| & =|-2-1-\i\sqrt3|\\
& = |-3-\i\sqrt3|\\
& = \sqrt{9+3}\\
\Aboxed{|a-b| & = 2\sqrt3}
\end{aligned}$

De plus, $\begin{aligned}[t]
|a-c| & = |\overline{a}-\overline{b}|\\
& = |\overline{a-b}|\\
& = |a-b|
\Aboxed{|a-c| & = 2\sqrt3}
\end{aligned}$

\item $|a-b|=|a-c|$ donc ABC est au moins isoc\`ele en A (car $\rm AB=AC$).

Calculons CB : $\begin{aligned}[t]
|b-c| & = |1+\i\sqrt3 -1+\i\sqrt3|\\
& = |2\i\sqrt3|\\
\Aboxed{|b-c| & = 2\sqrt3}
\end{aligned}$

\medskip

Ainsi, $\rm BC = AB = AC$ donc \uline{ABC est \'equilat\'eral.}
\end{enumerate}
\newpage
\item
\begin{enumerate}
\item 
\begin{multicols}{3}
$\begin{aligned}[t]
f(a) & = f(-2)\\
& = 1-2\e^{\i\frac{\pi}{6}}\\
& = 1-2\left(\cos\dfrac{\pi}{6}+\i\sin\dfrac{\pi}{6}\right)\\
& = 1-2\left(\dfrac{\sqrt3}{2}+\dfrac{1}{2}\i\right)\\
\Aboxed{f(a) & = 1-\sqrt3-\i}
\end{aligned}$

\columnbreak

$\begin{aligned}[t]
f(b) & = f\left(2\e^{\i\frac{\pi}{3}}\right)\\
& = 1+2\e^{\i\frac{\pi}{3}}\times\e^{\i\frac{\pi}{6}}\\
& = 1+2\e^{\i\left(\frac{\pi}{3}+\frac{\pi}{6}\right)}\\
& = 1+2\e^{\i\frac{\pi}{2}}\\
\Aboxed{f(b) & = 1+2\i}
\end{aligned}$

\columnbreak

$\begin{aligned}[t]
f(c) & = f\left(2\e^{-\i\frac{\pi}{3}}\right)\\
& = 1+2\e^{-\i\frac{\pi}{3}}\times\e^{\i\frac{\pi}{6}}\\
& = 1+2\e^{\i\left(-\frac{\pi}{3}+\frac{\pi}{6}\right)}\\
& = 1+2\e^{-\i\frac{\pi}{6}}\\
& = 1+2\left(\dfrac{\sqrt3}{2}-\dfrac{1}{2}\i\right)\\
\Aboxed{f(c) & = 1+\sqrt3-\i}
\end{aligned}$
\end{multicols}

\item $\begin{aligned}[t]
|f(a)-f(b)|^2 & = | 1-\sqrt3-\i - 1-2\i |^2\\
& = |-\sqrt3-3\i|^2\\
& = 12.
\end{aligned}$

\medskip


$\begin{aligned}[t]
|f(c)-f(b)|^2 & = | 1+\sqrt3-\i - 1-2\i |^2\\
& = |\sqrt3-3\i|^2\\
& = 12.
\end{aligned}$

\medskip

$\begin{aligned}[t]
|f(a)-f(c)|^2 & = | 1-\sqrt3-\i -1-\sqrt3 +\i|^2\\
& = |-2\sqrt3|^2\\
& = 12.
\end{aligned}$

\medskip

\uline{Ainsi, l'image de ABC par $f$ est aussi \'équilat\'eral.}
\end{enumerate}
\end{enumerate}