Complex series can only have two (or three) real parts This is related to a deleted question which asked about the complex series
$$S_u = \sum\limits_{n=1}^{\infty}e^{-iun}\frac{\sin(n)}{n} $$
All the terms are real when $u$ is a multiple of $\pi$, and when it is an even multiple this gives $S_0=\sum\limits_{n=1}^{\infty} \frac{\sin(n)}{n}=\frac{\pi-1}{2}$, while if it is an odd multiple it gives $S_\pi=\sum\limits_{n=1}^{\infty} (-1)^n\frac{\sin(n)}{n}=-\frac{1}{2}$, each with $O(\frac1n)$ convergence.
The surprising result is what happens for other values of $u$.  My empirical investigations suggest that:

*

*you get conditional convergence to a complex value except when $u=2k\pi \pm 1$;

*$Re(S_u) = \frac{\pi-1}{2}$ when $2k \pi -1 < u <2k \pi +1$;

*$Re(S_u) = -\frac{1}{2}$ when $2k \pi +1 < u <2(k+1) \pi -1$;

*there is no convergence when $u=2k\pi \pm 1$ since the imaginary parts of the partial sums diverge, but the real parts of the partial sums converge on $\frac{\pi-2}{4}$, halfway between the other two values.

My questions here are why the real part of $S_u$ only takes two (or three) values while the imaginary part can take any value, and why one of the real values is in a sense more common than the other real value.
The answer to why one value is in a sense more common may be that the two values are different in magnitude but their average over $[0,2\pi]$ is $0$.
To illustrate this, here is the real part of the sum plotted against $u$ with $u \in [-2\pi,2\pi]$

and then the imaginary part of the sum plotted against $u$

 A: Let $t \in [0, 2\pi)$ and note that with $z=e^{it}$ one has using the principal argument (which is in $(-\pi, \pi])$ of $$1-z=-2ie^{it/2}\sin \frac{t}{2}=2\sin \frac{t}{2}e^{ i\frac{t-\pi }{2}}$$ (which is well defined as $\Re(1-z) \ge 0$) that $\arg (1-z)=\frac{t-\pi}{2} \in [-\pi/2, \pi/2)$, while if $t$ goes up beyond (and including) $2\pi$ we subtract $\pi$ for every period, and down below $0$ we add $\pi$ for every period though here we do not include $t=0$ for the addition.
Note also $\sum_{n \ge 1} z^n/n =-\log (1-z)=-\log|1-z|-i\arg (1-z)$ holds for $|z|<1$ and extends to $|z|=1, z \ne 1$, while for $z=1$ where $t \to 0^+$ we have $\arg(1-e^{it}) = \frac{t-\pi}{2} \to \frac{-\pi}{2}$ and similarly when $t \to 2\pi^-$ we have $\arg(1-e^{it}) = \frac{t-\pi}{2} \to \frac{\pi}{2}$, though indeed $\log |1-e^{it}| \to -\infty$ in both cases, so we have divergence of the real part there, but the imaginary part converges to the average of the above (which is $0$) by the usual Fourier series results.
Now coming back to our problem let $u \in [0, 2\pi)$, so we have $$2iS_u = \sum\limits_{n=1}^{\infty}\frac{e^{i(-u+1)n}-e^{i(-u-1)n}}{n}$$ so $2iS_u = -\log(1-e^{i(-u+1)})+\log(1-e^{i(-u-1)})$ except when $u =1, 2\pi-1$ when the analysis above applies.
In particular, $2\Re S_u=\arg (1-e^{i(-u-1)})-\arg (1-e^{i(-u+1)})$ and we have three cases: $0 \le u < 1, 1 < u < 2\pi -1, 2\pi -1 < u <2\pi$ and in the first case we get $2\Re S_u=\pi+\frac{(-u-1-\pi)}{2}-\frac{(-u+1-\pi)}{2}=\pi-1$, in the second we get $-1$ as there is no extra $\pi$ from a jump, and in the third again $\pi-1$ since now we subtract $\pi$ from the second negative term, while clearly at the border we get the average as one term is continuous and the other gives the average by Fourier series classical results.
So putting it together for $u \in [0, 2\pi)$ we get $$\Re S_u=-\frac{1}{2}, 1< u < 2\pi -1$$ and $$\Re S_u=\frac{\pi-1}{2}, 0  \le u<1, 2\pi-1<u<1$$, with $$\Re S_u=\frac{\pi-2}{4}, u=1,2\pi-1$$ and of course the result extends by periodicity to any $u$ and can be written as in the OP.
As for the imaginary parts, that is $\frac{1}{2}\log |\frac{\sin \frac{1-u}{2}}{\sin \frac{1+u}{2}}|$ since when $-u \pm 1$ get out of the $[0, 2\pi)$ range the $\sin$ changes to its negative so we always have the $\log |\sin t/2|$ term for any real $t$ and $$2\Im S_u = \Re\log(1-e^{i(-u+1)})-\Re \log(1-e^{i(-u-1)})$$ as the sign changes because of the $i$ and by continuity this takes all values since it takes $\infty$ as $u \to 2\pi-1$ and takes $-\infty$ as $u \to 1$
