# How to show that $\hat{e}_m(k) = e^{−2πimk/N}$ for all $k$? How to show that $\hat{F}_m = e_m$?

Let $\{e_0, e_1, \dots, e_{N-1} \}$ be the Euclidean basis for $l^2(Z_N)$, and let $\{F_0, F_1, \dots, F_{N−1} \}$ be the Fourier basis( where $F_m(n)= \frac{1}{N} e^{2 \pi i m n/N}$ and $\hat{z}(m) = \sum_{n=0}^{N-1} z(n) e^{-2 \pi i m n/N}$).

1. How to show that $\hat{e}_m(k) = e^{−2πimk/N}$ for all $k$? Notice that $\hat{e}_m$ is very nearly (up to a reflection and a normalization) an element of the Fourier basis.
2. How to show that $\hat{F}_m = e_m$?
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But how do they follow immediately from the definition of the Fourier transform? Could you explain a little bit, because I have got following, but I'm not satisfied with this solution of mine. $\hat{e}_m(k) = \sum_{n=0}^{N-1} e_m(k) e^{-2 \pi i m k/N} = e^{-2 \pi i m k/N}$ ii. How to show that $\hat{F}_m = e_m$? (my solution: $\hat{F}_m = \sum_{n=0}^{N-1} \frac{1}{N}e^{2 \pi i m n/N} e^{-2 \pi i m n/N} = \sum_{n=0}^{N-1} \frac{1}{N} = ?$ Any comments on my solutions. – alvoutila Jan 10 '11 at 10:12