# Thread: Series and complex number

1. ## Series and complex number

This series
$\displaystyle 1 + \frac{cos(\theta)}{sin(\theta)} + \frac{cos(2\theta)}{sin^2(\theta)} + ... + \frac{cos(n\theta)}{sin^n(\theta)}$
I think that the summation notation would be $\displaystyle \sum_{i=0}^{n} \frac{cos(i\theta)}{sin^i(\theta)}$.
But anyway, I'm having trouble with the top one. I have tried to transform it so that I can use complex numbers to find the answer. I have not succeed. Does someone know the way (or any method) to do this problem?

2. Originally Posted by vincisonfire
This series
$\displaystyle 1 + \frac{cos(\theta)}{sin(\theta)} + \frac{cos(2\theta)}{sin^2(\theta)} + ... + \frac{cos(n\theta)}{sin^n(\theta)}$
I think that the summation notation would be $\displaystyle \sum_{i=0}^{n} \frac{cos(i\theta)}{sin^i(\theta)}$.
But anyway, I'm having trouble with the top one. I have tried to transform it so that I can use complex numbers to find the answer. I have not succeed. Does someone know the way (or any method) to do this problem?
What is your goal :S? What is it you're trying to get to?

PS: Confusing to use i as your index if you're also planning on using complex numbers in the solution!

3. You're right, bad choice of index.
I try to find a "nice" general formula for $\displaystyle S_n$
$\displaystyle 1 + \frac{cos(\theta)}{sin(\theta)} + \frac{cos(2\theta)}{sin^2(\theta)} + ... + \frac{cos(n\theta)}{sin^n(\theta)} = S_n$
$\displaystyle S_n = \sum_{k=0}^{n} \frac{cos(k\theta)}{sin^k(\theta)}$

4. Hi

Your idea seems to be good

Let $\displaystyle T_n = \sum_{k=0}^{n} \frac{sin(k\theta)}{sin^k(\theta)}$

Then

$\displaystyle S_n + i T_n = \sum_{k=0}^{n} \frac{cos(k\theta)+isin(k\theta)}{sin^k(\theta)}$

$\displaystyle S_n + i T_n = \sum_{k=0}^{n} \frac{e^{ik\theta}}{sin^k(\theta)}$

$\displaystyle S_n + i T_n = \sum_{k=0}^{n} \frac{({e^{i\theta}})^k}{sin^k(\theta)}$

$\displaystyle S_n + i T_n = \sum_{k=0}^{n} \left({\frac{e^{i\theta}}{sin(\theta)}}\right)^k$

5. Originally Posted by running-gag
Hi

Your idea seems to be good

Let $\displaystyle T_n = \sum_{k=0}^{n} \frac{sin(k\theta)}{sin^k(\theta)}$

Then

$\displaystyle S_n + i T_n = \sum_{k=0}^{n} \frac{cos(k\theta)+isin(k\theta)}{sin^k(\theta)}$

$\displaystyle S_n + i T_n = \sum_{k=0}^{n} \frac{e^{ik\theta}}{sin^k(\theta)}$

$\displaystyle S_n + i T_n = \sum_{k=0}^{n} \frac{({e^{i\theta}})^k}{sin^k(\theta)}$

$\displaystyle S_n + i T_n = \sum_{k=0}^{n} \left({\frac{e^{i\theta}}{sin(\theta)}}\right)^k$
Also

$\displaystyle S_n - i T_n = \sum_{k=0}^{n} \frac{cos(k\theta)-isin(k\theta)}{sin^k(\theta)}$

$\displaystyle S_n - i T_n = \sum_{k=0}^{n} \frac{e^{-ik\theta}}{sin^k(\theta)}$

$\displaystyle S_n - i T_n = \sum_{k=0}^{n} \frac{({e^{-i\theta}})^k}{sin^k(\theta)}$

$\displaystyle S_n - i T_n = \sum_{k=0}^{n} \left({\frac{e^{-i\theta}}{sin(\theta)}}\right)^k$

Two geometric series!

6. Originally Posted by running-gag
$\displaystyle S_n + i T_n = \sum_{k=0}^{n} \left({\frac{e^{i\theta}}{sin(\theta)}}\right)^k$
I would use $\displaystyle \sum_{k=0}^{n} x^i = \frac{1-x^{n+1}}{1-x}$ and then take $\displaystyle S_n = Re (\sum_{k=0}^{n} \left({\frac{e^{i\theta}}{sin(\theta)}}\right)^k)$

7. Originally Posted by vincisonfire
I would use $\displaystyle \sum_{k=0}^{n} x^i = \frac{1-x^{n+1}}{1-x}$ and then take $\displaystyle S_n = Re (\sum_{k=0}^{n} \left({\frac{e^{i\theta}}{sin(\theta)}}\right)^k)$
Yes
You can also use the hint given by danny arrigo

$\displaystyle S_n = \frac{(S_n + i T_n) + (S_n - i T_n)}{2}$

I did not perform the job, so I don't know which is the easiest way !

8. Answer is $\displaystyle \frac{sin^{n+2}(x)-cos(x)sin^{n+1}(x)-cos((n+1)x)sin(x)+cos(nx)}{sin^n(x)(\frac{3}{2}-\frac{sin(2x+arctan(\frac{1}{2}))}{2sin(arctan(\fr ac{1}{2}))})}$ and both ways of doing it are as long.
To who it may interest. This is the form with the less "sin" and "cos" in it although it looks as complicated as the raw version.
I replaced $\displaystyle *sin^2(x)-2cos(x)sin(x)+1$ by $\displaystyle \frac{3}{2}-\frac{sin(2x+arctan(\frac{1}{2}))}{2sin(arctan(\fr ac{1}{2}))}$