A question on the intuition behind the fourier transform. I cite the answer to this mathoverflow question. The subject of the question is that the operations of differentiation and integration are reduced to multiplication and division after being transformed by an integral transform. The accepted answer states:

It might help you to think about a discrete model: consider complex valued functions on $Z/n$. The discrete Fourier transform takes $f(k)$ to 
  $g(j) :=\sum_{k=1}^n f(k) \zeta^{jk}$ where $\zeta=e^{2 \pi i/n}$. It is pretty easy to see that, if we change $f(k)$ to $f(k+1)$, we change $g(j)$ to $g(j)*\zeta^j$.
Similarly, changing $f(k)$ to $f(k+1)-f(k)$ changes $g(j)$ to $g(j)*(\zeta^j-1)$. So, in this discrete model, taking a difference becomes multiplication by $(\zeta^j-1)$. In a similar way, in the continuous setting, taking a derivative becomes multiplication by $x$.

My question is would we not get the same result if we had chosen $\zeta$ as an arbitrary real number, why did it have to be $\zeta=e^{2 \pi i/n}$, it seems to me that the calculations done follow for any $\zeta$ ?
 A: Suppose you have a linear operator $L$ with eigenvectors $e_{\lambda}$ and corresponding eigenvalues $\lambda$ so that $Le_{\lambda}=\lambda e_{\lambda}$, and suppose you can expand a vector $f$ in the space as an integral "sum" of eigenvectors with coefficient function $f_\lambda$ in the form
$$
           f  = \int f_{\lambda}e_{\lambda}d\mu(\lambda).
$$
Then the action of $L$ on $f$ becomes multiplication of the coefficient function $f_{\lambda}$ by $\lambda$:
$$
          Lf = \int f_{\lambda}\lambda e_{\lambda}d\mu(\lambda).
$$
That is, if $f\sim f_{\lambda}$, then $Lf \sim \lambda f_{\lambda}$. This is the standard framework for Fourier Analysis that arose out of Fourier's separation of variables technique. Later this was generalized to eigenvector analysis of matrices (Fourier Analysis of this type came first, and motivated the techniques for diagonalizing normal matrices.)
For example, Fourier Analysis is well-suited to the study of linear Math-Physics equations where separation of variables applies, and this Fourier framework naturally leads to representations of various Math-Physics operators as multiplication operators on the vector coefficient functions obtained through eigenfunction expansions. Integral transforms give, in this setting, the vector coefficient functions, which is why the transforms turn the operators into multiplication. Transforms such as
$$
             f \sim f_{\lambda}=\int_{-\infty}^{\infty}f(t)e^{-i\lambda t}dt
$$
are really vector coefficient functions associated with the eigenfunctions $e_{\lambda}=e^{i\lambda t}$ of the differentiation operator $\frac{1}{i}\frac{d}{dt}$. So the Fourier transform turns differentiation of $f$ into multiplication on the coefficient function use to expand $f$ in the continuous basis of eigenfunctions $e_{\lambda} = e^{-i\lambda t}$.
What you're asking about amounts to reparameterizing the eigenvalue parameter, which just throws some scale factors into the mix.
