How to find the sum of this complex series? Question says:
The convergent infinite series C and S are defined as
$$
C=1+\frac{1}{4}\cos(2\theta)+\frac{1}{16}\cos(4\theta)+\frac{1}{64}\cos(6\theta)+...
$$
$$
S=\frac{1}{4}\sin(2\theta)+\frac{1}{16}\sin(4\theta)+\frac{1}{64}\sin(6\theta)+...
$$
Show that $C+iS=\frac{k}{k-e^{2i\theta}}$ where $k$ is an integer to be determined.
(Side note: Sorry if there is a formatting problem; first time using LaTeX)

Add the corresponding terms of C and iS
You get
$$1+\frac{1}{4}\left(\cos 2\theta +i\sin 2\theta \right)+\frac{1}{16}\left(\cos 4\theta +i\sin \theta \right)...$$
use the identity $\begin{array}{l}e^{i\theta }=\cos \theta +i\sin \theta \end{array}$
$$C+iS=\left(\frac{e^{i2\theta}}{4}\right)^0+\left(\frac{e^{i2\theta}}{4}\right)^1+\left(\frac{e^{i2\theta}}{4}\right)^2...$$
Now use sum of terms in a GP formula to get
$$C+iS=\frac{4}{4-e^{2i\theta }}$$
Therefore k=4
