fourier transform - why do we use $X(e^{jomega})$ instead of $X(jomega)$ in Discrete Time FT

I am studying DT-FT. But I cannot figure out why we use $X(e^{j\omega})$ instead of $X(j\omega)$ in DT FT

Thanks in advance..

Answer

The argument $e^{j\omega}$ emphasizes the $2\pi$-periodicity of the discrete-time Fourier transform (DTFT) of a sequence. Furthermore, if the $\mathcal{Z}$-transform $X(z)$ of the sequence $x[n]$ exists and if the unit circle $|z|=1$ is inside the region of convergence, then the DTFT of the sequence is simply obtained by evaluating $X(z)$ on the unit circle $z=e^{j\omega}$.

That's the same thing as with the continuous Fourier transform and the Laplace transform, referred to in your previous question.