Discrete Fourier transform symmetry I was reading the chapter on discrete Fourier transforms in Lyons' book -- Understanding Digital Signal Processing -- and could not understand the last paragraph about symmetry.


  There’s an additional symmetry property of the DFT that deserves mention at this point. In practice, we’re occasionally required to determine the DFT of real input functions where the input index $n$ is defined over both positive and negative values. If that real input function is even, then $X(m)$ is always real and even; that is, if the real $x(n) = x(−n)$, then, $X_{\textrm{real}}(m)$ is in general nonzero and $X_{\textrm{imag}}(m)$ is zero. Conversely, if the real input function is odd, $x(n) = −x(−n)$, then $X_{\textrm{real}}(m)$ is always zero and $X_{\textrm{imag}}(m)$ is, in general, nonzero.


Note: $X(m) = X_{\textrm{real}}(m) + jX_{\textrm{imag}}(m)$


Firstly, what is meant by "odd" and "even"? I suspect it's the number of samples in the input signal, but that leads me to my second question,
Why is $X_{\textrm{imag}}(m)$ zero with real input functions that are even, and why, with real input functions that are odd, is $X_{\textrm{real}}(m)$ zero and $X_{\textrm{imag}}(m)$ generally non-zero?
Even &amp; odd refer to the symmetry around $n = 0$.

Even means $x[n] = x[-n]$; you can get the part for $n &lt; 0$ by simply mirroring the part for $n &gt; 0$ at the $n=0$ line.

Odd means $x[n] = -x[-n]$; you can get the part for $n &lt; 0$ by simply mirroring the part for $n &gt; 0$ at the $n=0$ line and multiplying it by $-1$.

A cosine wave is even, sine wave is odd. 

These are all just special cases of the general symmetry that says 


  if it's real in one domain, it's conjugate symmetric in the other.


Conjugate symmetric means that the real part is even and the imaginary part is odd. Most people know that a real time domain signal as a conjugate symmetric spectrum, but it also goers the other way around: a conjugate symmetric time domain signal has a real valued spectrum.