We have that:
\begin{align*}
\lim_{n \rightarrow \infty} \frac{1}{n} \sum_{\ell= 1}^n \dfrac{\sin \left( e^{2 \pi i \ell / n} \right)}{1 - ce^{- 2 \pi i \ell / n}} & = \frac{1}{2 \pi} \int_0^{2 \pi} \dfrac{ \sin \left( e^{i \theta} \right) }{ 1 - c e^{- i \theta} } d \theta
\nonumber \\
& = \frac{1}{2 \pi i} \oint_{|z|=1} \dfrac{\sin z}{1 - c z^{-1}} \frac{dz}{z}
\nonumber \\
& = \frac{1}{2 \pi i} \oint_{|z|=1} \dfrac{\sin z}{z - c} dz
\end{align*}
Since ##\sin z## is an entire function (having no singularities at any point in the complex plane), the function
\begin{align*}
f(z) = \dfrac{\sin z}{z - c}
\end{align*}
only has a (simple) pole at ##z_0 = c##.
Therefore, if ##|c| < 1## then
\begin{align*}
\lim_{n \rightarrow \infty} \frac{1}{n} \sum_{\ell= 1}^n \dfrac{\sin \left( e^{2 \pi i \ell / n} \right)}{1 - ce^{- 2 \pi i \ell / n}} & = \frac{1}{2 \pi i} \oint_{|z|=1} \dfrac{\sin z}{z - c} dz
\nonumber \\
& = \sin c .
\end{align*}
If ##|c| > 1## then
\begin{align*}
\lim_{n \rightarrow \infty} \frac{1}{n} \sum_{\ell= 1}^n \dfrac{\sin \left( e^{2 \pi i \ell / n} \right)}{1 - ce^{- 2 \pi i \ell / n}} & = \frac{1}{2 \pi i} \oint_{|z|=1} \dfrac{\sin z}{z - c} dz
\nonumber \\
& = 0 .
\end{align*}
Proof that ##\sin z## is an entire function: Note
\begin{align*}
\sin (x+iy) & = \dfrac{e^{i x - y} - e^{-i x + y}}{2i}
\nonumber \\
& = \dfrac{e^{i x} - e^{-i x}}{2i} \dfrac{e^y + e^{- y}}{2} + i \dfrac{e^{i x} + e^{-i x}}{2} \dfrac{e^y - e^{- y}}{2}
\nonumber \\
& = \sin x \cosh y + i \cos x \sinh y
\nonumber \\
& =u (x,y) + i v (x,y)
\end{align*}
Note
\begin{align*}
u_x & = \cos x \cosh y ,
\nonumber \\
u_y & = \sin x \sinh y
\nonumber \\
v_x & = - \sin x \sinh y ,
\nonumber \\
v_y & = \cos x \cosh y .
\end{align*}
We read off that
\begin{align*}
u_x = v_y , \qquad v_x=- u_y
\end{align*}
for all ##x,y \in \mathbb{R}##, hence the CR conditions are satisfied for ##x,y \in \mathbb{R}##, and so the function ##\sin z## is entire.
