
\begin{enumerate}
\item $\begin{aligned}[t]
\displaystyle\sum_{k=0}^{n-1}\omega_k & = 1 + \e^{\frac{2\i\pi}{n}} + \e^{\frac{4\i\pi}{n}}+ \e^{\frac{6\i\pi}{n}} + \cdots + \e^{\frac{2\i(n-1)\pi}{n}}\\
& = \dfrac{1-\left(\e^{\frac{2\i\pi}{n}}\right)^n}{1-\e^{\frac{2\i\pi}{n}}}\\
& = \dfrac{1-\e^{2\i\pi}}{1-\e^{\frac{2\i\pi}{n}}}\\
& = \dfrac{1-1}{1-\e^{\frac{2\i\pi}{n}}}\\
& = 0.
\end{aligned}$

\item On souhaite calculer :
\begin{align*}
\displaystyle\sum_{k=0}^{n-1}\omega^p_k & = 1 + \e^{\frac{2p\i\pi}{n}} + \e^{\frac{4p\i\pi}{n}}+ \e^{\frac{6p\i\pi}{n}} + \cdots + \e^{\frac{2p\i(n-1)\pi}{n}}\\
& = \dfrac{1-\left(\e^{\frac{2p\i\pi}{n}}\right)^n}{1-\e^{\frac{2\i\pi}{n}}}\\
& = \dfrac{1-\e^{2p\i\pi}}{1-\e^{\frac{2\i\pi}{n}}}\\
&  = \dfrac{1-\left(\e^{2\i\pi}\right)^p}{1-\e^{\frac{2\i\pi}{n}}}\\
&  = \dfrac{1-1^p}{1-\e^{\frac{2\i\pi}{n}}}\\
& = 0.
\end{align*}

\end{enumerate}
