Geometric Series with a Complex Number The assignment: Consider the series of complex numbers
$\sum^{\infty}_{k=1} r^k \big(cos(v)+i\space sin(v)\big)^k\space with\space r \in [0,1[$
(a) Show that the series is a convergent geometric series and calculate its sum
(b) Show that the series can be expressed as $\sum^{\infty}_{k=1} r^k \big(cos(kv)+i\space sin(kv)\big)$
(c) Find expressions for the two series of real numbers,
$\sum^{\infty}_{k=1} r^k \big(cos(kv)\big)$ & $\sum^{\infty}_{k=1} r^k \big(sin(kv)\big)$

My (attempt at) solution:
(a) Showing that it's a geometric series is easy enough, simply recognizing that $cos(v)+i\space sin(v)=e^{iv}$ and thus $\sum^{\infty}_{k=1}r^k \big( e^{iv} \big)^k=\sum^{\infty}_{k=1}(re^{iv})^k=\sum^{\infty}_{k=1}z^k$. 
That $|r|<1$ is given, and $|cos(v)+i\space sin(v)|<1$ is also simple to show. Thus the series converges to $\frac{re^{iv}}{1-re^{iv}}$.
However; the assignment also asks me to reduce this to the neat formula $a+ib$. And I'm not sure how to reduce it, when I tried, I simply ended up making it even wierder. I would very much like some help on rewriting this complex number "$\frac{re^{iv}}{1-re^{iv}}$".
(b) That $e^{ikv}=cos(kv)+isin(kv)$ follows from de Moivre's formula.
(c) I'm a little unsure what the assignment wants here. Is it simply that
$\sum_{k=1}^{\infty}r^k\space cos(kv)=\sum_{k=1}^{\infty} r^k \space \frac{1}{2}\big(e^{ikv}+e^{-ikv}\big)$
and similarly for sine?
Thanks for reading

We can write directly, as you already said:
$$\sum_{k=1}^\infty r^k(\cos v+i\sin v)^k=\sum_{k=1}^\infty z^k\;,\;\;\text{with}\;\;z=r(\cos v+i\sin v)=re^{iv}$$
Thus, since $\;|z|=r<1\;$ , by the well known formula we get
$$\sum_{k=1}^\infty z^k=\frac z{1-z}=\frac{re^{iv}}{1-re^{iv}}=\frac{re^{iv}}{1-r\cos v-ir\sin v}=\frac{r(\cos v+i\sin v)(1-r\cos v+ir\sin v)}{1-2r\cos v+r^2}=$$
$$=\frac{r(\cos v-r\cos^2v-r\sin^2v+(r\cos v\sin v+\sin v-r\cos v\sin v)i}{1-2r\cos v+r^2}=\frac{r(\cos v-r+i\sin v)}{1-2r\cos v+r^2}$$
and thus the sum of the series, in cartesian form, is
$$r\frac{\cos v-r}{1-2r\cos v+r^2}+r\frac{\sin v}{1-2r\cos v+r^2}\,i$$
Part (c) follows at once from the above.
