# How to show that $\hat{e}_m(k) = e^{-2 \pi mk/N}$ for all k?

Let $\{e_0, e_1, \dots, e_{N-1} \}$ be the Euclidian basis for $l^2(Z_N)$, and let $\{F_0, F_1, \dots, F_{N-1} \}$ be the fourier basis i. How to show that $\hat{e}_m(k) = e^{-2 \pi mk/N}$ for all k? Notice that $\hat{e}_m$ is very nearly ( up to a reflection and a normalization ) an element of the Fourier basis. ii. How to show that $\hat{F}_m = e_m$?

First some definitions $F_m=N^{-1/2} E_m = \frac{1}{N}e^{2 \pi i m n/N}$ and $\hat{z}(m) = \sum_{n=0}^{N-1} z(n) e^{-2 \pi i m n/N}$ i.How to show that $\hat{e}_m(k) = e^{-2 \pi mk/N}$? (my) Solution: $\hat{e}_m(k) = \sum_{n=0}^{N-1} e_m(k) e^{-2 \pi i m k/N} = e^{-2 \pi i m k/N}$ 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.

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Are you missing an $i$ in the title? – KennyTM Nov 25 '10 at 20:01