De Moivre's theorem for Individual sine functions. Need help

[joshnfsmw]

De Moivre's theorem for Individual sine functions. Need urgent help!

Hey guys I need some help with this question! I've only been able to do the first part by factorising out (1-z) for the first 12 terms and then multiplying (1+z)/(1+z) to the whole equation. For part b), I am totally lost. Your help is much appreciated! Thanks

[PLAIN]http://img651.imageshack.us/img651/6011/mathso.jpg

 

[Hootenanny]

Welcome to Physics Forums, John.

What does DeMoivre's theorem state? How could you represent some complex variable z so that you can apply the theorem?

 

[joshnfsmw]

De Moivre's Theorem: [cos(x)+jsin(x)]^n= cos(nx)+jsin(nx), where j is the imaginary unit

This is all that the question gave. It asks to prove it, so you don't need any complex variable. Like for part a),

[(1-z)(z+z^3+z^5+z^7+z^9+z^11)+z^13]*[(1+z)/(1+z)] = (z+z^14)/(1+z) (shown)

 

[Hootenanny]

De Moivre's Theorem: [cos(x)+jsin(x)]^n= cos(nx)+sin(nx).

This is all that the question gave. It asks to prove it, so you don't need any complex variable. Like for part a),

[(1-z)(z+z^3+z^5+z^7+z^9+z^11)+z^13]*[(1+z)/(1+z)] = (z+z^14)/(1+z) (shown)

By using the variable z, the question implied we are dealing with complex variables. Just as when you see x or y, you assume them to be real.

So, what is [cos(x)+jsin(x)]^n in terms of z?

 

[joshnfsmw]

By using the variable z, the question implied we are dealing with complex variables. Just as when you see x or y, you assume them to be real.

So, what is [cos(x)+jsin(x)]^n in terms of z?

That's the puzzling part. So we let z=(cosx+jsinx)^n ?

 

[Hootenanny]

That's the puzzling part. So we let z=(cosx+jsinx)^n ?

Very close. You let z=\cos x+j\sin x and then use DeMoivre's formula for consecutive powers.

 

[joshnfsmw]

ok then how to we get rid of the extra cosines that were introduced in the formula?

 

[Hootenanny]

ok then how to we get rid of the extra cosines that were introduced in the formula?

If you have two complex numbers that are equal to each other, what do you know about their real and imaginary parts?

 

[Quinzio]

If you have two complex numbers that are equal to each other, what do you know about their real and imaginary parts?

Hi, I just got involved in this exercise but I'm unable to solve it and I would like to get what I'm missing.
I solved the last equations by using a trig. identity, that is:

\sin\alpha\sin\beta=\frac{1}{2}\left[\sin(\alpha+\beta)+\sin(\alpha\beta)\right]

Using this identity and taking \cos(\frac{\theta}{2}) on the left it is easy to prove it.
But I cannot understand how De Moivre can help solve the problem. I understand you can decompose the single terms using the imaginary part of the De Moivre sum, but you end up with a huge sum of terms which I cannot simplify or take advantage of.
How is De Moivre useful in this ?

 

[Hootenanny]

Hi, I just got involved in this exercise but I'm unable to solve it and I would like to get what I'm missing.
I solved the last equations by using a trig. identity, that is:

\sin\alpha\sin\beta=\frac{1}{2}\left[\sin(\alpha+\beta)+\sin(\alpha\beta)\right]

Using this identity and taking \cos(\frac{\theta}{2}) on the left it is easy to prove it.
But I cannot understand how De Moivre can help solve the problem. I understand you can decompose the single terms using the imaginary part of the De Moivre sum, but you end up with a huge sum of terms which I cannot simplify or take advantage of.
How is De Moivre useful in this ?

So on the left hand side we have,

\sum_{n=1}^{13} (-1)^{n+1}z^n = \sum_{n=1}^{13} (-1)^{n+1}[\cos\theta + i\sin\theta]^n

Using De Moivre we can write,

\sum_{n=1}^{13} (-1)^{n+1}z^n = \sum_{n=1}^{13} (-1)^{n+1}[\cos(n\theta) + i\sin(n\theta)]

And on the right hand side we have.

\frac{z+z^{14}}{1+z} = \frac{\cos\theta+i\sin\theta + \cos(14\theta) + i\sin(14\theta)}{1+\cos\theta + i\sin\theta}

After a bit of tedious algebra, you will arrive at the result.

 

[Quinzio]

It is right the "tedious algebra" the part that doesn't convince me.
I had taken the way you are suggesting but that implies writing down 13(13-1)/2 term and then simplifying them, plus getting the binomials.
I don't think that the problem, giving that it should be solved in less than 25 minutes is to be solved in that way.

I have spent an hour on it, and I am convinced De Moivre is of little help here.

 

[Hootenanny]

It is right the "tedious algebra" the part that doesn't convince me.
I had taken the way you are suggesting but that implies writing down 13(13-1)/2 term and then simplifying them, plus getting the binomials.
I don't think that the problem, giving that it should be solved in less than 25 minutes is to be solved in that way.

I'm not quite following you there - I see no need for binomials. If you weren't to solve it this way, how would you do it?

 

[Quinzio]

I'm not quite following you there - I see no need for binomials. If you weren't to solve it this way, how would you do it?

OK, forgive the binomials, my mistake.
I still fail to see how

\Im\left[ \frac{\cos\theta+i\sin\theta + \cos(14\theta) + i\sin(14\theta)}{1+\cos\theta + i\sin\theta}\right] = \frac{\sin{7\theta} \cos \frac {13 \theta}{2}}{\cos \frac{\theta}{2} }

Because, if I am not wrong, this is where we have to go.

The left side of the equation is trivial, that is:

\Im \left[ \sum_{n=1}^{13} (-1)^{n+1}[\cos(n\theta) + i\sin(n\theta)]\right] = \sum_{n=1}^{13} (-1)^{n+1} \sin({n\theta})

You get the sine sum just by extracting the imaginary part, which is trivial for the left side, but not for the right one.

 

[Hootenanny]

OK, forgive the binomials, my mistake.
I still fail to see how

\Im\left[ \frac{\cos\theta+i\sin\theta + \cos(14\theta) + i\sin(14\theta)}{1+\cos\theta + i\sin\theta}\right] = \frac{\sin{7\theta} \cos \frac {13 \theta}{2}}{\cos \frac{\theta}{2} }

Have you actually tried to compute the imaginary part of the RHS?

 

[Quinzio]

Have you actually tried to compute the imaginary part of the RHS?

The best I could get is

\frac{z+z^{14}}{1+z} = \frac{z(1+z^{13})}{1+z}
then if I write the factors in terms of modulus and argument, I can write

for the moduli
\left|\frac{z(1+z^{13})}{1+z}\right|= \frac{|z||1+z^{13}|}{|1+z|}

for the angles
\angle{\frac{z(1+z^{13})}{1+z}}= \angle z +\angle (1+z^{13}) - \angle (1+z)

\angle z = \theta
\angle (1+z^{13}) = \frac{13\theta}{2}
\angle (1+z )= \frac{\theta}{2}

For what I can understand so far it must be:
\frac{\sin{7\theta} \cos \frac {13 \theta}{2}}{\cos \frac{\theta}{2}} = \frac{|z||1+z^{13}|}{|1+z|} \angle(\theta+ \frac{13\theta}{2}-\frac{\theta}{2})=\frac{|1+z^{13}|}{|1+z|} \angle({7\theta})
as |z|=1

And more:
|1+z^n|^2= (1+cos (n\theta))^2+sin^2(n\theta) = 1+2cos(n\theta)+cos^2(n\theta)+sin^2(n\theta)=2(1+cos(n\theta))= 2(cos^2(\frac{n\theta}{2}))
|1+z^n| = \sqrt 2 (cos(\frac{n\theta}{2}))

So:
|1+z| = \sqrt 2 (cos(\frac{\theta}{2}))
|1+z^{13}| = \sqrt 2 (cos(\frac{13\theta}{2}))

OK, I think we are done:
\Im \left[\frac{|z||1+z^{13}|}{|1+z|}\angle({7\theta})\right] = \frac{sin (7\theta) cos \frac{13\theta}{2}}{cos \frac{\theta}{2}}

[PLAIN]http://free-animated-screensavers.org/wp-content/uploads/2007/04/spectacular-fireworks-screensaver.jpg

Ahhhh... ok... finally this was the way to go.
It was hard but finally...
ok show me someone how does it in 20 minutes...

 

[Hootenanny]

Congratulations!

It's only 15 lines, surely you can do that in twenty minutes ...