What is the real part of the complex number $z=\frac{e^{i(n+1)\theta}-1}{e^{i\theta}-1}$? I keep getting stumped on how to find the real part of the following complex number without putting a ton of effort into converting it into rectangular coordinates, performing division, etc. Any advice on how to solved it?

$$z = \frac{e^{i(n+1)\theta}-1}{e^{i\theta}-1}$$

EDIT: For those of you giving me hints, I sincerely apologize. This is part of a larger proof of the following: 
$$1 + \cos \theta + \cos 2\theta + \ldots + \cos n\theta = .5 + \frac{\sin((n+.5)\theta)}{2\sin(\theta/2)}$$
I used both of your hints as steps to get to finding the real part of this number. I appreciate the support greatly, but I'm afraid that it is of no use here.
 A: And another Hint: $$\frac{\exp(i(n+1)\theta)-1}{\exp(i\theta)-1}=\frac{\exp(i(n+1)\theta/2)}{\exp(i\theta/2)}\frac{\exp(i(n+1)\theta/2)-\exp(-i(n+1)\theta/2)}{\exp(i\theta/2)-\exp(-i\theta/2)}$$ and use $2i\sin(x)=\exp(ix)-\exp(-ix)$.
A: Hint : $e^{i(n+1)\theta}-1 = (e^{i\theta}-1)\sum_{j=0}^ne^{j\theta}$
A: Another hint: consider $\dfrac{x^{n+1}-1}{x-1}$.
A: Using:


*

*Euler's formula, when $\text{a}\in\mathbb{C}$:
$$\text{a}=\Re[\text{a}]+\Im[\text{a}]i=|\text{a}|e^{(\arg(\text{a})+2\pi k)i}=|\text{a}|\cos(\arg(\text{a})+2\pi k)+|\text{a}|\sin(\arg(\text{a})+2\pi k)i$$
Where $|\text{a}|=\sqrt{\Re^2[\text{a}]+\Im^2[\text{a}]}$, $\arg(\text{a})$ is the complex argument of $\text{a}$ and $k\in\mathbb{Z}$.

*When $\text{b}\space\wedge\space\text{c}\in\mathbb{C}$:
$$\frac{\text{b}}{\text{c}}=\frac{\text{b}\cdot\overline{\text{c}}}{\text{c}\cdot\overline{\text{c}}}=\frac{\text{b}\cdot\overline{\text{c}}}{|\text{c}|^2}=\frac{\left(\Re[\text{b}]+\Im[\text{b}]i\right)\cdot\left(\Re[\text{c}]-\Im[\text{c}]i\right)}{\Re^2[\text{c}]+\Im^2[\text{c}]}$$



So, we get (assuming that $n\space\wedge\space\theta\in\mathbb{R}$):
$$\text{R}=\Re\left[\text{z}\right]=\Re\left[\frac{e^{i(n+1)\theta}-1}{e^{i\theta}-1}\right]=$$
$$\Re\left[\frac{\left(e^{i(n+1)\theta}-1\right)\cdot\overline{e^{i\theta}-1}}{\left(e^{i\theta}-1\right)\cdot\overline{e^{i\theta}-1}}\right]=\Re\left[\frac{\left(e^{i(n+1)\theta}-1\right)\cdot\overline{e^{i\theta}-1}}{\left|e^{i\theta}-1\right|^2}\right]$$
So, we get:


*

*$$\left|e^{i\theta}-1\right|^2=\left(\cos(\theta)-1\right)^2+\sin^2(\theta)=2-2\cos(\theta)$$

*$$\left(e^{i(n+1)\theta}-1\right)\cdot\overline{e^{i\theta}-1}=$$
$$\left(\cos(\theta(1+n))-1+\sin(\theta(1+n))i\right)\cdot\left(\cos(\theta)-1-\sin(\theta)i\right)$$


Sowe get:
$$\Re\left[\left(e^{i(n+1)\theta}-1\right)\cdot\overline{e^{i\theta}-1}\right]=1-\cos(\theta)+\cos(n\theta)-\cos(\theta(1+n))$$
Now, we know:
$$\text{R}=\frac{1-\cos(\theta)+\cos(n\theta)-\cos(\theta(1+n))}{2-2\cos(\theta)}$$
