Summation of $\sum_{n=1}^{10}2^{-n}\sin\left(\frac{1}{10}\pi n\right)$ using complex numbers

Question

Use de Moivre’s theorem to deduce that $$\sum_{n=1}^{10}{2^{-n}\sin{\left(\frac{1}{10}\pi n\right)}}=\frac{1025\sin{\left(\frac{1}{10}\pi\right)}}{2560-2048\cos{\left(\frac{1}{10}\pi\right)}}$$

By first considering the sum of $\sum_{n=1}^{10}{2^{-n}z^n}$, we obtain $$\begin{align}\sum_{n=1}^{10}{2^{-n}z^n}=\frac{z}{2}+\frac{z^2}{4}+\frac{z^3}{8}+\cdots+\frac{z^{10}}{1024}&=\frac{\frac{z}{2}\left(1-\left(\frac{z}{2}\right)^{10}\right)}{1-\frac{z}{2}}\\&=\frac{z-z\left(\frac{z}{2}\right)^{10}}{2-z}\\&=\frac{1024z-z^{11}}{2048-1024z}\end{align}$$ Recognizing the substitution $z=e^{\frac{i\pi}{10}}$, it follows that $z^{10}=-1$. $$\begin{align}\frac{1024z-z\left(z^{10}\right)}{2048-1024z}&=\frac{1025z}{2048-1024z}\\&=\frac{1025e^\frac{i\pi}{10}}{2048-1024e^\frac{i\pi}{10}}\\&=\frac{1025e^\frac{i\pi}{10}\left(2048-1024e^{-\frac{i\pi}{10}}\right)}{2048-1024e^\frac{i\pi}{10}\left(2048-1024e^{-\frac{i\pi}{10}}\right)}\\&=\frac{2099200e^\frac{i\pi}{10}-1049600}{5242880-4194304e^\frac{i\pi}{10}}\end{align}$$ I am, however, stumped after this point. I feel as if I have gotten myself into an unnecessary and long-winded solution when the answer relatively straightforward. How do I proceed from this point further or do I need to revaluate my solution?

Answer 1

An algebraic error occurred and should be fixed like eyeballfrog said. I will let $\alpha = \frac{\pi}{10}$ to make everything a little shorter.

After fixing the algebraic error, we have $$ \frac{1025 e^{i \alpha} (2048 - 1024e^{-i \alpha})}{(2048 - 1024e^{i \alpha})(2048 - 1024e^{-i \alpha})} $$ Which gives us $$ \frac{2099200e^{i \alpha} - 1049600}{5242880 - 2097152e^{-i \alpha} - 2097152e^{i \alpha}} \\ = \frac{2099200e^{i \alpha} - 1049600}{5242880 - 4194304 \cos(\alpha)} \\ = \frac{2050e^{i \alpha} - 1025}{5120 - 4096 \cos(\alpha)} \\ = \frac{2050 \cos(\alpha) + 2050 \sin(\alpha) - 1025}{5120 - 4096 \cos(\alpha)} $$ To get the final answer, we need the real part of the above expression $$ \Re(\frac{2050 \cos(\alpha) + 2050 i \sin(\alpha) - 1025}{5120 - 4096 \cos(\alpha)}) \\ = \frac{2050 \sin(\alpha)}{5120 - 4096 \cos(\alpha)} \\ = \frac{1025 \sin(\frac{\pi}{10})}{2560 - 2048 \cos(\frac{\pi}{10})} $$

The numbers in intermediate calculations were huge and could've been avoided rewriting $2048 - 1024z = 1024(2-z)$ like eyeballfrog said.

Answer 2

You have an algebra error here:

$$\begin{align}\frac{1024z-z\left(z^{10}\right)}{2048-1024z}&=\frac{1025z}{2048-1024z}\\&=\frac{1025e^\frac{i\pi}{10}}{2048-1024e^\frac{i\pi}{10}}\\&=\frac{1025e^\frac{i\pi}{10}\left(2048-1024e^{-\frac{i\pi}{10}}\right)}{2048-1024e^\frac{i\pi}{10}\left(2048-1024e^{-\frac{i\pi}{10}}\right)}\\&=\frac{2099200e^\frac{i\pi}{10}-1049600}{5242880-4194304e^\frac{i\pi}{10}}\end{align}$$

The third line should be $$ =\frac{1025e^\frac{i\pi}{10}\left(2048-1024e^{-\frac{i\pi}{10}}\right)}{\left(2048-1024e^\frac{i\pi}{10}\right)\left(2048-1024e^{-\frac{i\pi}{10}}\right)} $$ which will result in a purely real denominator. Also, as suggested in the comments, since $2048 - 1024z = 1024(2-z)$, multiplying the top and bottom by $2-\bar{z}$ instead will make your life easier.

Answer 3

Let S denote the sum; $$S=\sum_{n=1}^{10}\frac{\sin\left(\frac{\pi n}{10}\right)}{2^{n}}$$

Now we know $z=|r|e^{i\theta}=|r|(cos\theta+isin\theta)$

We can then write $sinx=\frac{e^{ix}-e^{-ix}}{2i}$, Further, using DMT; $$sin(\frac{n\pi}{10})=\frac{e^{i\frac{n\pi}{10}}-e^{-i\frac{n\pi}{10}}}{2i}$$

Substituting in S,

$S=\sum_{n=1}^{10}\frac{\frac{e^{i\frac{n\pi}{10}}-e^{-i\frac{n\pi}{10}}}{2i}}{2^{n}}$

$S=\frac{1}{2i}\sum_{n=1}^{10}\frac{e^{i\frac{n\pi}{10}}-e^{-i\frac{n\pi}{10}}}{2^{n}}$

S = $L_1$ - $L_1$ = $\frac{1}{2i} \left[\sum_{n=1}^{10} \left(\frac{e^{\frac{i\pi }{10}}}{2}\right)^n - \sum_{n=1}^{10} \left(\frac{e^{\frac{-i\pi }{10}}}{2}\right)^n \right]$

On using the formula for sum of n terms of GP, $S_n = \frac{a(1-r^n)}{(1-r)}$ where symbols have usual meaning. Further, $S_{10} = \frac{\frac{e^\frac{i\pi}{10}}{2}(1-e^\frac{in \pi}{10})}{(1-e^\frac{i\pi}{10})}$ in $L_1$ and $S_{10} = \frac{\frac{e^\frac{-i\pi}{10}}{2}(1-e^\frac{-in \pi}{10})}{(1-e^\frac{-i\pi}{10})}$ in $L_2$

S = $L_1$ - $L_1$ = $\frac{1}{2i} \left[ \frac{\frac{e^\frac{i\pi}{10}}{2}(1-e^\frac{in \pi}{10})}{(1-e^\frac{i\pi}{10})} - \frac{\frac{e^\frac{-i\pi}{10}}{2}(1-e^\frac{-in \pi}{10})}{(1-e^\frac{-i\pi}{10})}\right]$

$S \approx 0.5$