The Fourier Transform of this function
\begin{equation} f(n)=u(n)-u(n-m) \end{equation}
(where $u$ is the unity step)
is:
\begin{equation} F(\omega)=\frac{\sin(\omega m/2)}{\sin(\omega/2)}e^{-i(m-1)/2} \end{equation}
The phase of $F$ is: \begin{equation} \phi=\angle{\frac{\sin(\omega m/2)}{\sin(\omega/2)}}+\angle{e^{-i(m-1)/2}} \end{equation}
My question is how to find the phase of: \begin{equation} \angle{\frac{\sin(\omega m/2)}{\sin(\omega/2)}} \end{equation}
As it is a real number, $0$ or $\pi$ (modulo $2k\pi$, $k\in\mathbb{Z}$), depending on its sign. Since $e^{(0+2k\pi)\imath}=1$, and $e^{(\pi+2k\pi)\imath}=-1$, every positive real number can be written as $r = |r|e^{(0+2k\pi)\imath}$, and every negative real number as $r = |r|e^{(\pi+2k\pi)\imath}$. And there is an indeterminacy for $0$, since any phase would suit.
Finally, it is conventional to choose, for each number, a "simple" phase, for instance that looks constant, or "relatively continuous". Yet to be clean, on each non-zero number the phase is defined (modulo $2k\pi$).
