Sum of Complex series

Question

Let $\theta\in\mathbb{R}$ and $\theta \neq k\pi$ for $k\in\Bbb Z$.

By summing a geometric progression show that

$$1 + e^{2i\theta} + e^{4i\theta}+e^{6i\theta} + e^{8i\theta}= \frac{e^{9i\theta}-e^{-i\theta}}{2i\sin\theta}.$$

Hence show that $$\sin (2\theta)+\sin(4\theta)+\sin (6\theta)+ \sin(8\theta)= \frac{\cos(\theta)-\cos(9\theta)}{2\sin\theta}.$$

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Could we treat the progression as

(e$^{2i\theta})^{0}+$(e$^{2i\theta})$$^{1}+$(e$^{2i\theta})^{2}+$(e$^{2i\theta})^{3}+(e^{2i\theta})^{4}$

Which would give us the terms for the sum in the geometric progression with parameters

$$\begin{cases}a=1 \\ r= e^{2i\theta} \\ m=5\end{cases}.$$

giving the sum $$1\cdot\left(\frac{1-e^{10i\theta}}{1-e^{2i\theta}}\right)$$

Am I on the right track?

Answer 1

You are indeed almost there.

Next write

$$\sin\theta = {e^{i\theta}-e^{-i\theta}\over 2i}$$

and conclude

$${1-e^{10i\theta}\over 1-e^{2i\theta}}={e^{10i\theta}-1\over e^{i\theta}{2i\over 2i}(e^{i\theta}-e^{-i\theta})}$$ $$={e^{-i\theta}(e^{10i\theta}-1)\over 2i\sin\theta} = {e^{9i\theta}-e^{-i\theta}\over 2i\sin\theta}$$

From there you just do

$$\sin(2\theta)+\sin(4\theta)+\sin(6\theta)+\sin(8\theta)= {1\over 2i}\bigg((1+e^{2i\theta}+e^{4i\theta}+e^{6i\theta}+e^{8i\theta})-(1+e^{-2i\theta}+e^{-4i\theta}+e^{-6i\theta}+e^{-8i\theta})\bigg)$$

and from what you've already done, that's just

$${1\over 2i}\bigg({e^{9i\theta}-e^{-i\theta}\over 2i\sin\theta}-{e^{-9i\theta}-e^{i\theta}\over 2i\sin(-\theta)}\bigg)$$

since $\sin(-\theta)=-\sin\theta$ we get $$=-{1\over 2\sin\theta}\bigg({(e^{9i\theta}+e^{-9i\theta})-(e^{i\theta}+e^{-i\theta})\over 2}\bigg)$$

Finally use

$$\cos\theta = {e^{i\theta} +e^{-i\theta}\over 2}$$

to conclude your identity is

$$\sin(2\theta)+\sin(4\theta)+\sin(6\theta)+\sin(8\theta)={\cos\theta-\cos(9\theta)\over 2\sin\theta}$$

Answer 2

$$\sum_{r=0}^{n-1} (e^{2i\theta})^r=\frac{1-(e^{2i\theta})^n}{1-e^{2i\theta}}$$

$$=\frac{-e^{in\theta}(e^{in\theta}-e^{-in\theta})}{-e^{i\theta}(e^{i\theta}-e^{-i\theta})}$$

$$=e^{i(n-1)\theta}\frac{2i\sin n\theta}{2i\sin\theta}$$

$$=[\cos(n-1)\theta+i\sin(n-1)\theta]\frac{\sin n\theta}{\sin\theta}$$

As $\displaystyle\sin\phi=$ imaginary part of $e^{i\phi},$

$\displaystyle\sum_{r=0}^{n-1}\sin(2r\theta)=\frac{\sin(n-1)\theta\sin n\theta}{\sin\theta}$

Using Werner's formula, $\displaystyle2\sin(n-1)\theta\sin n\theta=\cos\theta-\cos(2n-1)\theta$

Here $n=5$