Conjugate symmetry of the DFT of real-valued sequences I have read about Fourier transformation that real signals are &quot;mirrored&quot; in the real and negative halves of the Fourier transform because of the nature of the Fourier transform.  For example, if we have the real signal $x$ whose length is $N$, its $X = FFT(x)$ should be conjugate symmetric. That means the $X^*[m] = X[N-m]$.
That is ok, and I have read it clearly HERE
My question: assuming you have a real signal $x = [-1, 1, -1, 1];$ its $X = FFT(x) = [0,     0,    -4 ,    0]$
So, in that case, we cannot say that $X^*[m] = X[N-m]$.  Why ?
Is what I said above is right ? in which cases is right?
We can! In your example, $N=4$, and the DFT is real-valued, so you get $X[k]=X[4-k]$, and that's true:
$$\begin{align}k=0:\;X[0]&amp;=X[4]=0\\k=1:\;X[1]&amp;=X[3]=0\\k=2:\;X[2]&amp;=X[2]=-4\end{align}$$
Note that by definition the DFT is $N$-periodic, so $X[N]=X[0]$.