
\begin{rem}
Pour calculer $(a+\i b)^n$, on peut bien entendu utiliser la propriété 6 du cours, mais on peut aussi utiliser directement le binôme de Newton, ce qui est peut-être plus simple.
\end{rem}

\begin{enumerate}
\item $\begin{aligned}[t]
(2 - \i)^5 & = \sum_{k=0}^5\dbinom{5}{k}2^k \times (-\i)^{5-k}\\
& = \dbinom{5}{0}2^0(-\i)^5 + \dbinom{5}{1}2^1(-\i)^4 + \dbinom{5}{2}2^2(-\i)^3 + \dbinom{5}{3}2^3(-\i)^2 + \dbinom{5}{4}2^4(-\i)^1 + \dbinom{5}{5}2^5(-\i)^0\\
& = -\i +10 +40\i - 80 - 80\i + 32\\
\Aboxed{(2 - \i)^5 & = -38 - 41\i}
\end{aligned}$

\item $\begin{aligned}[t]
(2 + \i)^5 & = \overline{(2-\i)}^5\\
& = \overline{(2-\i)^5}\\
& = \overline{-38 - 41\i} \quad\text{d'après la question 1}\\& \Aboxed{(2 + \i)^5 & = -38 + 41\i}
\end{aligned}$

\item $\begin{aligned}[t]
(1 + 2\i)^4 & = \sum_{k=0}^4\dbinom{4}{k}1^k \times (2\i)^{4-k}\\
& = \dbinom{4}{0}1^0\times(2\i)^4
+ \dbinom{4}{1}1^1\times(2\i)^3
+ \dbinom{4}{2}1^2\times(2\i)^2
+ \dbinom{4}{3}1^3\times(2\i)^1
+ \dbinom{4}{4}1^4\times(2\i)^0 \\
& = 16 - 32\i - 24 + 8\i + 1\\
\Aboxed{(1 + 2\i)^4 & = -7 - 24\i}
\end{aligned}$

\item $\begin{aligned}[t]
(1+2\i)^8 & = \big[(1+2\i)^4\big]^2\\
& = (-7 - 24\i)^2\quad\text{d'après la question 3}\\
& = 49 + 2\times7\times24\i + (24\i)^2\\
\Aboxed{(1+2\i)^8& = -527 + 336\i}
\end{aligned}$
\end{enumerate}
