Exercise 4.3.27 (Compute $ \sum_{a=1}^n e^{2\pi i a/n}$)

Question

Let $F(n) = \sum_{a=1}^n e^{2\pi i a/n}$. Show that $F(1) = 1$ and that $F(n) = 0$ for all $n>1$.

Richard Ganaye · Accepted Answer

\begin{proof} 
\begin{itemize}
\item If $n = 1$ then $F(n) = F(1) = \sum_{a=1}^1 e^{2\pi i a} = e^{2\pi i } = 1$.

\item Suppose now that $n>1$. Then $\zeta = e^{2\pi i/n} 
e 1$. Since $\zeta^n = 1 = \zeta^0$, 
\begin{align*}
F(n) &= \sum_{a = 1}^n \zeta^a\
&= \sum_{a = 0}^{n-1} \zeta^a\
& = \frac{\zeta^n - 1}{\zeta - 1}&(	ext{since } \zeta 
e 1)\
&=0.
\end{align*}
\end{itemize}
In conclusion,
$$F(n) = \sum_{a=1}^n e^{2\pi i a/n} =
\begin{cases}
1  &	ext{if } n = 1,\
0 & 	ext{if } n>1.
\end{cases}
$$
\end{proof}
Note: If we use the hint, we can also compute the sum of the roots of the polynomial $x^n - 1$, which is $0$ if $n>1$.

Exercise 4.3.27 (Compute $ \sum_{a=1}^n e^{2\pi i a/n}$)

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Exercise 4.3.27 (Compute $ \sum_{a=1}^n e^{2\pi i a/n}$)

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