Trigonometric Expression for $1 + \cos \alpha + \cos 2\alpha + \cdots + \cos n \alpha$ using complex numbers This question is not a duplicate because I am asked here to use the fact that  $1 + \cos \alpha + \cos 2 \alpha + \cdots + \cos n \alpha = Re (1 + z + z^{2} + \cdots + z^{n})$, where the question this is suspected of being a duplicate of does not use this
I am supposed to use the fact that $1 + \cos \alpha + \cos 2 \alpha + \cdots + \cos n \alpha = Re (1 + z + z^{2} + \cdots + z^{n})$, where $z = \cos \alpha + i \sin \alpha$, ro find a trigonometric expression for $1 + \cos \alpha + \cos 2 \alpha + \cdots + \cos n \alpha$.
Now, when $z \neq 1$, we have the geometric sum $\displaystyle 1 + z + z^{2} + \cdots + z^{n} = \frac{1-z^{n+1}}{1-z}$. (When $z = 1$, $1 + z + z^{2} + \cdots + z^{n} = n + 1$, so $Re(1 + z + z^{2} + \cdots + z^{n}) = Re(n+1) = n+1$, finished.)
So, in the case where $z \neq 1$, $\displaystyle 1 + \cos \alpha + \cos 2 \alpha + \cdots + \cos n \alpha = Re(1 + z + z^{2} + \cdots + z^{n}) = Re\left(\frac{1-z^{n+1}}{1-z} \right) = Re \left(\frac{1-(\cos \alpha + i \sin \alpha)^{n+1}}{1-(\cos \alpha + i \sin \alpha)} \right) = Re \left[ \left(\frac{1-(\cos \alpha + i \sin \alpha)^{n+1}}{(1-\cos \alpha) - i \sin \alpha} \right)\cdot \frac{(1-\cos \alpha) + i \sin \alpha}{(1 - \cos \alpha)+ i \sin \alpha}\right] = Re \left( \frac{(1-\cos \alpha)+i\sin \alpha - ( 1- \cos \alpha)(\cos \alpha + i \sin \alpha)^{n+1}-i \sin \alpha (\cos \alpha + i \sin \alpha)^{n+1}}{(1-\cos \alpha)^{2} + \sin^{2} \alpha}\right)$.
Then, to make a long story short, after an application of De Moivre's Theorem to turn the $(\cos \alpha + i \sin \alpha)^{n+1}$'s into $(\cos(n \alpha + \alpha)+ i \sin (n \alpha + \alpha))$'s, usage of the sine and cosine of sum identities, collecting of like terms and cancelling some things, we obtain 
$\displaystyle Re \left(\frac{1-(\cos \alpha + i \sin \alpha)^{n+1}}{1-(\cos \alpha + i \sin \alpha)} \right) = Re \left(\frac{(1-\cos \alpha + \cos (n\alpha)) + i(\sin \alpha - \cos (n\alpha)\sin \alpha + \sin (n \alpha))}{2-2\cos \alpha} \right) = \frac{1-\cos \alpha + \cos (n \alpha)}{2-2\cos \alpha}$.
At least I think it does.
Which brings me to my question: is this correct? If not, where did I go wrong and how do I fix it?
 A: Since all your work was not shown, I cannot comment on where any problems were, but here is a simple derivation of a few formulas for the sum in the question:
$$
\begin{align}
\sum_{k=0}^n\cos(k\alpha)
&=\frac12\sum_{k=0}^n\left(e^{ik\alpha}+e^{-ik\alpha}\right)\tag{1}\\
&=\frac12\frac{e^{i(n+1)\alpha}-1}{e^{i\alpha}-1}+\frac12\frac{1-e^{-i(n+1)\alpha}}{1-e^{-i\alpha}}\tag{2}\\
&=\frac12\frac{e^{i(n+1/2)\alpha}-e^{-i\alpha/2}}{e^{i\alpha/2}-e^{-i\alpha/2}}+\frac12\frac{e^{i\alpha/2}-e^{-i(n+1/2)\alpha}}{e^{i\alpha/2}-e^{-i\alpha/2}}\tag{3}\\
&=\frac12\left(1+\frac{\sin((2n+1)\alpha/2)}{\sin(\alpha/2)}\right)\tag{4}\\
&=\frac12\left(1+\frac{\sin((n+1)\alpha)+\sin(n\alpha)}{\sin(\alpha)}\right)\tag{5}\\
&=\frac12\left(1+\frac{\cos(n\alpha)-\cos((n+1)\alpha)}{1-\cos(\alpha)}\right)\tag{6}
\end{align}
$$
Explanation:
$(1)$: $\cos(x)=\frac{e^{ix}+e^{-ix}}2$
$(2)$: sum of a geometric series
$(3)$: multiply left summand by $\frac{e^{-i\alpha/2}}{e^{-i\alpha/2}}$ and the right summand by $\frac{e^{i\alpha/2}}{e^{i\alpha/2}}$
$(4)$: $\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}$
$(5)$: multiply fraction in $(4)$ by $\frac{2\cos(\alpha/2)}{2\cos(\alpha/2)}$
$(6)$: multiply fraction in $(4)$ by $\frac{2\sin(\alpha/2)}{2\sin(\alpha/2)}$
