# Starting to prove by De Moivre's Theorem

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

## Homework Statement

Use de Moivre's theorem to show that

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

## Homework Equations

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

## 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?

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Hurkyl
Staff Emeritus
Gold Member
Just need a little help on how to start this question. Where would I get the $\frac{sinn\theta}{2^n}$ from?
As the imaginary part of $(cos\theta + isin\theta)^n$, presumably. Oh, I suppose that doesn't work literally.

tiny-tim
Homework Helper
As the imaginary part of $(cos\theta + isin\theta)^n$, presumably. Oh, I suppose that doesn't work literally.
(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. Homework Helper
But if I expand $(cos\theta +isin\theta)^n$ how do I know when to stop since n varies?

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

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

$$(cos\theta +isin\theta)^n=(isin\theta)^n + ^nC_1(isin\theta)^{n-1}cos\theta +^nC_2(isin\theta)^{n-2}cos^2\theta +...$$
Don't expand it! Use deMoivre!

Homework Helper
Don't expand it! Use deMoivre!
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.

Hurkyl
Staff Emeritus
Gold Member
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.
Not even after plugging it into the sum and simplifying?

Homework Helper
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 ?

Dick
Homework Helper
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 ?
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.

Homework Helper
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.
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?

Homework Helper
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.

Dick
Homework Helper
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?
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.

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

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!