SUM OF COMPLEX EXPONENTIALS
Consider the geometric series
$$\begin{eqnarray}
D(u) &=& \sum_{n=-k}^{k} e^{-i u n} \\
&=& e^{-i u k}+e^{-i u (k-1)}+\ldots+e^{-i u (k-1)}+ e^{-i u k}\\
&=& e^{-i u k} \left(1+e^{i u }+e^{i 2u}+\ldots+e^{i (2k-1)u}+ e^{i 2ku}\right)\\
&=&e^{-i ku} \frac{1-e^{(2k+1)ui }}{1-e^{i u}}
\end{eqnarray} $$
To ease the notation, let us define $r=e^{i u}$. Then, the above expression can be simplified as
$$\begin{eqnarray}
r^{-k}\frac{1-r^{2k+1}}{1-r}
&=& r^{-k} \frac{r^\frac{{2k+1}}{2}\left[r^{-\frac{{2k+1}}{2}}-r^\frac{{2k+1}}{2}\right]}{r^{\frac{1}{2}}\left[ r^{-\frac{1}{2}}-r^{\frac{1}{2}}\right]}\\
&=&\frac{r^{-\frac{{2k+1}}{2}}-r^\frac{{2k+1}}{2}}{ r^{-\frac{1}{2}}-r^{\frac{1}{2}}}
\end{eqnarray} $$
Substituting back $r=e^{-i u}$,
$$\begin{eqnarray}
&=&\frac{e^{\frac{2k+1}{2}i u}-e^{-\frac{{2k+1}}{2}i u}}{ e^{\frac{i u}{2}}-e^{-\frac{i u}{2}}}\\\\
&=&\frac{\frac{e^{\frac{2k+1}{2}i u}-e^{-\frac{{2k+1}}{2}i u}}{2i}}{ \frac{e^{\frac{i u}{2}}-e^{-\frac{i u}{2}}}{2i}}\\\\
&=& \frac{\sin\left(\frac{2k+1}{2} u \right)}{\sin\left(\frac{ u}{2}\right)}
\end{eqnarray} $$
Or, in summary, we have proved that
$$\begin{eqnarray}
D(u) &=& \sum_{n=-k}^{k} e^{-i u n} &=& \frac{\sin\left(\frac{2k+1}{2} u\right)}{\sin\left(\frac{ u}{2}\right)}
\end{eqnarray} $$
The denominator becomes zero at $\frac{u}{2}=m\pi$ or $u = 2 m \pi$ for any integer $m$. By using L’Hospital’s rule
$$\begin{eqnarray}
D(u=2 m \pi) = 2k+1
\end{eqnarray} $$