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
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