# Sum of two infinite complex geometric sums

let $$C=\frac{\cos\theta}{2}-\frac{\cos2\theta}{4}+\frac{\cos3\theta}{8}+...$$ $$S=\frac{\sin\theta}{2}-\frac{\sin2\theta}{4}+\frac{\sin3\theta}{8}+...$$ I want to find the sum of the series $$C+iS$$ and thus find expressions for $$C$$ and $$S$$.

so the sum of $$C+iS$$ is $$C+iS=(\frac{\cos\theta}{2}-\frac{\cos2\theta}{4}+\frac{\cos3\theta}{8}+...)+i(\frac{\sin\theta}{2}-\frac{\sin2\theta}{4}+\frac{\sin3\theta}{8}+...)$$ $$=\frac{1}{2}(cos\theta+i\sin\theta)-\frac{1}{4}(cos2\theta+i\sin2\theta)+\frac{1}{8}(cos3\theta+i\sin3\theta) + ...$$ $$=\frac{1}{2}(cos\theta+i\sin\theta)-\frac{1}{4}(cos\theta+i\sin\theta)^2+\frac{1}{8}(cos\theta+i\sin\theta)^3 + ...$$

So I know i have to now put his into a geometric series where i can find the first term and common ratio as i will want to sum to infinity, but I'm not sure where to go from here?

So I've continued and got the series to here: $$\frac{1}{2}[e^{i\theta}-\frac{1}{2}(e^{i\theta})^2+\frac{1}{4}(e^{i\theta})^3...]$$ But I'm still not sure where to continue.

• Use $\cos\theta+i\sin\theta=e^{i\theta}$ and factor out $\frac{1}{2}$. Nov 15, 2018 at 11:21

Hint:

Using Intuition behind euler's formula and the special case $$e^{i\pi}=-1$$

$$\dfrac{(\cos t+i\sin t)^n(-1)^{n-1}}{2^n}=-\dfrac{e^{int}(e^{i\pi})^n}{2^n}=-\left(\dfrac{e^{i(t+\pi)}}2\right)^n$$

Now for the common ratio$$(r)$$ of Geometric Series,

$$|r|=\left|\dfrac{e^{i(t+\pi)}}2\right|=\dfrac12<1$$

HINT

We have that

$$C+iS=\sum_{k=1}^\infty\frac{(-1)^{k+1}}{2^k}e^{\left(ik\theta\right)} =-\sum_{k=1}^\infty\left(-\frac{e^{\left(i\theta\right)}}{2}\right)^k$$

then refer to geometric series which holds also for $$r$$ complex $$|r|<1$$.

• I tried to sum to infinity using your sum with $a=0.5e^i\theta$ and $r=-0.5e^i\theta$ but i obtained $\frac{exp(i\theta)+1}{exp(i\theta)+2exp(-i\theta)+3}$which isn't the correct answer. I was told i should be $\frac{2exp(i\theta)+1}{5+4cos\theta}$ Nov 17, 2018 at 18:31
• @H.Linkhorn We have that $$-\sum_{k=1}^\infty\left(-\frac{e^{\left(i\theta\right)}}{2}\right)^k=-\frac{1}{1+\frac{e^{i\theta}}{2}}+1=\frac{e^{i\theta}}{2+e^{i\theta}}$$ Then multiply by $2+e^{-i\theta}$ denominator and numerator to obtain teh result.
– user
Nov 17, 2018 at 21:36