What is the real part of $\frac{e^{(n+1)i\theta}-1}{e^{i\theta}-1}$? 
What is the real part of this complex number?
$$\frac{e^{(n+1)i\theta}-1}{e^{i\theta}-1}$$

I am trying to times conjugate of denominator which will be $$\frac{e^{(n+1)i\theta}-1}{e^{i\theta}-1}\cdot\frac{e^{-i\theta}-1}{e^{-i\theta}-1}$$
but it makes my fractions very complicated and I don't know how do I go on.
Thanks for any help.
 A: $$
\begin{aligned}
\frac{e^{(n+1)i\theta}-1}{e^{i\theta}-1}& =\frac{e^{-\frac{(n+1) i \theta}{2}}\left(e^{\frac{(n+1) i \theta}{2}}-e^{-\frac{(n+1) i \theta}{2}}\right)}{e^{-\frac{i \theta}{2}}\left(e^{\frac{i \theta}{2}}-e^{-\frac{i \theta}{2}}\right)} \\
& =e^{-\frac{n i \theta}{2}} \cdot \frac{2 i \sin \frac{(n+1) \theta}{2}}{2 i \sin \frac{ \theta}{2}} \\
& =\frac{\sin \frac{(n+1) \theta}{2}}{\sin \frac{\theta}{2}}\left(\cos \frac{n \theta}{2}-i \sin \frac{n\theta}{2}\right)
\end{aligned}
$$
Therefore the real part of the complex number is $\csc \frac{\theta}{2} \sin \frac{(n+1) \theta}{2} \cos \frac{n \theta}{2}.$
Wish it helps!
A: The real part of $\large\frac{a+bi}{c+di}$ is $\large\frac{ac+bd}{c^2+d^2}$ so the real part of $\large\frac{e^{(n+1)i\theta}-1}{e^{i\theta}-1}=\frac{\cos(n+1)\theta-1+i\sin(n+1)\theta}{\cos\theta-1+i\sin\theta}$ is
$$\frac{(\cos(n+1)\theta-1)(\cos\theta-1)+\sin(n+1)\theta\sin\theta}{(\cos\theta-1)^2+(\sin\theta)^2}\\
=\frac{(-2)\sin^2\frac{(n+1)\theta}{2}(-2)\sin^2\frac{\theta}{2}+2\sin\frac{(n+1)\theta}{2}\cos\frac{(n+1)\theta}{2}.2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}{2-2\cos\theta}\\
=\frac{4\sin\frac{(n+1)\theta}{2}\sin\frac{\theta}{2}\left(\sin\frac{(n+1)\theta}{2}\sin\frac{\theta}{2}+\cos\frac{(n+1)\theta}{2}\cos\frac{\theta}{2}\right)}{4\sin^2\frac{\theta}{2}}\\
=\frac{\sin\frac{(n+1)\theta}{2}\cos\frac{n\theta}{2}}{\sin\frac{\theta}{2}}\\
=\sin\frac{(n+1)\theta}{2}\cos\frac{n\theta}{2}\csc\frac{\theta}{2}$$
I don't recommend this way. It took my life. I will not check again.
