Discrete Time Fourier Transform of a real signal

Question

I want to prove that if we have a real signal x[n] then for the DTFT it is applied that we have an even symmetry:

| X(Ω+1/10) | = | X(-(Ω+1/10)) | (I mean the magnitude)

I have tried many things but nothing really worked to prove it.

Answer 1

The DTFT of the sequence $x_n$ is

$$X(\omega)=\sum_{n=-\infty}^{\infty}x_ne^{-in\omega}$$

Its complex conjugate is (given that $x_n$ is real-valued)

$$X^*(\omega)=\sum_{n=-\infty}^{\infty}x_ne^{in\omega}=X(-\omega)$$

So for real-valued $x_n$ you have

$$X^*(\omega)=X(-\omega)\tag{1}$$

and if you take the magnitude on both sides of (1) you get

$$|X(\omega)|=|X(-\omega)|$$