# De Moivre's theorem

1. Aug 30, 2007

### rock.freak667

Using De Moivre's theorem to prove the sum of a series

1. The problem statement, all variables and given/known data
Write down an expression in terms of z and N for the sum of the series:
$$\sum_{n=1}^N 2^{-n} z^n$$

Use De Moivre's theorem to deduce that

$$\sum_{n=1}^{10} 2^{-n} \sin(\frac{1}{10}n\pi)$$ = $$\frac{1025\sin(\frac{1}{10}\pi)}{2560-2048\cos(\frac{1}{10}\pi)}$$

2. Relevant equations

$$e^{in\theta}=(\cos{\theta}+i\sin{\theta})^n = \cos{n\theta}+i\sin{n\theta}$$

3. The attempt at a solution
To find a sum for the series it is a GP with first term,$$a=2^{-1}z$$ common ration,$$r=2^{-1}z$$
then $$S_N = \frac{2^{-1}z(1-(2^{-1}z)^{N})}{1-2^{-1}z}$$

For the second part I was thinking to just replace $$z^n$$ with $$\sin(\frac{1}{10}n\pi)$$ would that work?

(NOTE:Also, even though I think I typed the LATEX thing correctly it doesn't display what i actually typed when i previewed the post, so if something looks weird please check if I typed it correctly,such as
\frac{1025\sin(\frac{1}{10}\pi)}{2560-2048\cos(\frac{1}{10}\pi) appears as $$a^3<-9b-3c-3$$

Last edited: Aug 30, 2007
2. Aug 30, 2007

### dextercioby

You're missing a } at the end. EDIT:Not anymore.

Suggestion. $z=e^{i\pi/10}$.

LATER EDIT: Yes, it's already visible in post #1.

Last edited: Aug 30, 2007
3. Aug 30, 2007

### rock.freak667

ah....strange things appear in my browser...$z=e^{i\pi/10}$...so simple...shall try it now

4. Apr 29, 2011

### ima_shah2002

Did u get the answer...can u post the solution...

5. Apr 29, 2011

### eumyang

Even if the OP found the answer (almost 4 years ago!), if you need help with the same question, he can't just post the solution. That would be against forum rules. YOU need to post your attempt FIRST, and then maybe we can help.