# Homework Help: Starting to prove by De Moivre's Theorem

1. May 18, 2008

### rock.freak667

[SOLVED] Starting to prove by De Moivre's Theorem

1. The problem statement, all variables and given/known data

Use de Moivre's theorem to show that

$$\sum_{n=1} ^{\infty} \frac{sinn\theta}{2^n}=\frac{2^{N+1}sin\theta+sinN\theta-2sin(N+1)\theta}{2^N(5-4cos\theta)}$$

2. Relevant equations

$$(cos\theta + isin\theta)^n=cosn\theta + isinn\theta$$

3. The attempt at a solution

Just need a little help on how to start this question. Where would I get the $\frac{sinn\theta}{2^n}$ from?

2. May 18, 2008

### Hurkyl

Staff Emeritus
As the imaginary part of $(cos\theta + isin\theta)^n$, presumably. Oh, I suppose that doesn't work literally.

3. May 19, 2008

### tiny-tim

(btw, the ∑ is to N, not to ∞)

Yes it does … sum $\left(\frac{1}{2}(cos\theta + isin\theta)\right)^n$ from 1 to N, get rid of the complex denominator in the usual way, and take the imaginary part.

4. May 19, 2008

### rock.freak667

But if I expand $(cos\theta +isin\theta)^n$ how do I know when to stop since n varies?

$$(cos\theta +isin\theta)^n=(isin\theta)^n + ^nC_1(isin\theta)^{n-1}cos\theta +^nC_2(isin\theta)^{n-2}cos^2\theta +...$$

5. May 19, 2008

### Dick

Don't expand it! Use deMoivre!

6. May 19, 2008

### rock.freak667

But then that would just give me that

$sin n\theta = Im[(cos\theta +i sin\theta)^n$

the left has what I need, the right doesn't seem to be giving me any ideas.

7. May 19, 2008

### Hurkyl

Staff Emeritus
Not even after plugging it into the sum and simplifying?

8. May 19, 2008

### rock.freak667

Forgive me if I am slow with this, I am very bad with these problems...

But back to it. $sin n\theta = Im[(cos\theta +i sin\theta)^n]$

putting that into

$$\sum_{n=1} ^{N} \frac{sinn\theta}{2^n}$$

I get

$$\sum_{n=1} ^{N} \frac{Im[(cos\theta +i sin\theta)^n]}{2^n} =\sum_{n=1} ^{N} (\frac{Im[(cos\theta +i sin\theta)}{2})^n$$

$$\sum_{n=1} ^{N} (\frac{sin\theta}{2})^n$$

Which is a GP with first term,a=$\frac{sin\theta}{2}$ and common ratio,r=$\frac{sin\theta}{2}$ and |r|<1 ?

9. May 19, 2008

### Dick

You can't factor the nth power outside the Im, and you don't want to. The left side is a geometric progression with common ratio exp(i*theta)/2. Sum it first, then take the imaginary part of both sides.

10. May 19, 2008

### rock.freak667

Not too clear here.

I know that $e^{i\theta}=cos\theta +i sin\theta$ so the imaginary part would just be the sine. But if I can't factor the n outside of the Im, how do I get it as a GP?

11. May 19, 2008

### rock.freak667

ah I now got what you meant.

I was to consider the sum from 1 to N of [exp(i*theta)/2]^n and then take the Im part. Well I got everything in the Denominator and two of the terms in the numerator. I am just missing the $2^{N+1}sin\theta$, I have $2^Nsin\theta$. Shall recheck my algebra in a while.

12. May 19, 2008

### Dick

Forget the original series. Sum (exp(i*theta)/2)^n. Then take the imaginary part. The original series will appear on the left side. It will be a surprise.

13. Sep 8, 2010

### usman94

Re: [SOLVED] Starting to prove by De Moivre's Theorem

Can some1 post the whole solution? How to get the 2^N[5-4Cos(theta)]

How do we get cos???? If we are taking the imaginry part???? Please solve it and post it... thanx in advance

14. Sep 9, 2010

### usman94

Re: [SOLVED] Starting to prove by De Moivre's Theorem

We want to show that
Σ(n=1 to N) sin(nθ) / 2^n = [2^(N+1) sin θ + sin (Nθ) - 2 sin(N+1)θ] / [2^N (5 - 4 cos θ)].
-------------------------
Note that e^(iθ) = cos θ + i sin θ.
So, De Moivre's Theorem yields for any positive integer n
e^(inθ) = (cos θ + i sin θ)^n = cos(nθ) + i sin(nθ).

Next, note that
Σ(n=1 to N) sin(nθ) / 2^n
= Im [Σ(n=1 to N) e^(inθ) / 2^n], by the first observation
= Im {Σ(n=1 to N) [e^(iθ)/2]^n}.

Thus, we need to compute the imaginary part of the finite geometric series
Σ(n=1 to N) [e^(iθ)/2]^n
= (e^(iθ)/2) Σ(k=0 to N-1) [e^(iθ)/2]^k, by re-indexing the sum
= (e^(iθ)/2) (1 - (e^(iθ)/2)^N) / (1 - e^(iθ)/2)
= e^(iθ) (1 - (e^(iθ)/2)^N) / (2 - e^(iθ))
= e^(iθ) (2^N - e^(iNθ)) / [2^N (2 - e^(iθ))]

Multiply numerator and denominator by the conjugate (2 - e^(-iθ)):
==> e^(iθ)(2 - e^(-iθ)) (2^N - e^(iNθ)) / [2^N (2 - e^(iθ))(2 - e^(-iθ))]
= (2e^(iθ) - 1) (2^N - e^(iNθ)) / [2^N (5 - 2 (e^(iθ) + e^(-iθ))]
= (2e^(iθ) - 1) (2^N - e^(iNθ)) / [2^N (5 - 2 (2 cos θ))]
= (2e^(iθ) - 1) (2^N - e^(iNθ)) / [2^N (5 - 4 cos θ)].

Now, it's a matter of simplifying the numerator.
(2e^(iθ) - 1) (2^N - e^(iNθ))
= 2^(N+1) e^(iθ) - 2^N - 2e^(i(N+1)θ) + e^(iNθ)
= 2^(N+1) (cos θ + i sin θ) - 2^N - 2 (cos(N+1)θ + i sin(N+1)θ) + (cos(Nθ) + i sin(Nθ))
= [2^(N+1) cos θ - 2^N - 2 cos(N+1)θ + cos(Nθ)] + i [2^(N+1) sin θ - 2 sin(N+1)θ) + sin(Nθ)]

Its imaginary part is 2^(N+1) sin θ - 2 sin(N+1)θ) + sin(Nθ).

Therefore, we have obtained
Σ(n=1 to N) sin(nθ) / 2^n = [2^(N+1) sin θ + sin (Nθ) - 2 sin(N+1)θ] / [2^N (5 - 4 cos θ)]
as required.

I hope this helps!