Real part of $\frac{1-e^{(n+1)i\theta}}{1-e^{i\theta}}$ How can I compute the real part of
\begin{equation*}
\frac{1-e^{(n+1)i\theta}}{1-e^{i\theta}}, \quad \text{where}\ \theta \in \mathbb{R}?
\end{equation*}
Maybe it's a silly question, but I'm feeling rusty with complex numbers...

So the solution I've found is: 
$$ \frac{e^{(n+1)i\theta}-1}{e^{i\theta}-1} = $$
$$ \frac{\cos{(n+1)\theta} -1 +i\sin{(n+1)\theta}}{(\cos\theta - 1)^2 + \sin^2 \theta}(\cos \theta -1 -i\sin\theta)= $$
$$ \frac{\cos{n\theta} - \cos(n+1)\theta - \cos\theta + 1}{2-2\cos\theta} -i \frac{\sin{n\theta} - \sin(n+1)\theta + \sin\theta}{2-2\cos\theta}= $$
$$ \frac{\sin{(n+\frac{1}{2})\theta} + \sin\frac{\theta}{2}}{2\sin\frac{\theta}{2}} +i \frac{\sin(n+1)\frac{\theta}{2} \sin(n\frac{\theta}{2})}{\sin\frac{\theta}{2}}$$
Now it's easy to distinguish real and imaginary parts.
