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

In summary, De Moivre's theorem states that if two complex numbers are equal to each other, their real and imaginary parts are also equal.
  • #1
joshnfsmw
4
0
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 [Broken]
 
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  • #2


Welcome to Physics Forums, John.

What does DeMoivre's theorem state? How could you represent some complex variable [itex]z[/itex] so that you can apply the theorem?
 
  • #3


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)
 
  • #4


joshnfsmw said:
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 [itex]z[/itex], the question implied we are dealing with complex variables. Just as when you see [itex]x[/itex] or [itex]y[/itex], you assume them to be real.

So, what is [itex][cos(x)+jsin(x)]^n[/itex] in terms of [itex]z[/itex]?
 
  • #5


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

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

That's the puzzling part. So we let z=(cosx+jsinx)^n ?
 
  • #6


joshnfsmw said:
That's the puzzling part. So we let z=(cosx+jsinx)^n ?
Very close. You let [itex]z=\cos x+j\sin x[/itex] and then use DeMoivre's formula for consecutive powers.
 
  • #7


ok then how to we get rid of the extra cosines that were introduced in the formula?
 
  • #8


joshnfsmw said:
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?
 
  • #9


Hootenanny said:
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:

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

Using this identity and taking [itex]\cos(\frac{\theta}{2})[/itex] 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 ?
 
  • #10


Quinzio said:
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:

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

Using this identity and taking [itex]\cos(\frac{\theta}{2})[/itex] 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,

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

Using De Moivre we can write,

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

And on the right hand side we have.

[tex]\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}[/tex]

After a bit of tedious algebra, you will arrive at the result.
 
  • #11


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 [itex]13(13-1)/2[/itex] 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.
 
  • #12


Quinzio said:
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 [itex]13(13-1)/2[/itex] 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?
 
  • #13


Hootenanny said:
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

[tex]\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} }[/tex]

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

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

[tex]\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}) [/tex]

You get the sine sum just by extracting the imaginary part, which is trivial for the left side, but not for the right one.
 
  • #14


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

[tex]\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} }[/tex]

Have you actually tried to compute the imaginary part of the RHS?
 
  • #15


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

The best I could get is

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

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

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

[tex]\angle z = \theta[/tex]
[tex]\angle (1+z^{13}) = \frac{13\theta}{2}[/tex]
[tex]\angle (1+z )= \frac{\theta}{2}[/tex]

For what I can understand so far it must be:
[tex] \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})[/tex]
as [tex]|z|=1[/tex]

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

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

OK, I think we are done:
[tex]\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}}[/tex]

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

Ahhhh... ok... finally this was the way to go. :biggrin:
It was hard but finally...
ok show me someone how does it in 20 minutes... :cry: :cry: :cry:
 
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  • #16


Congratulations! :biggrin:

It's only 15 lines, surely you can do that in twenty minutes ... :wink: :tongue2:
 

1. What is De Moivre's theorem for individual sine functions?

De Moivre's theorem for individual sine functions states that for any complex number z = r(cosθ + isinθ), the nth power of z can be expressed as z^n = r^n(cos(nθ) + isin(nθ)). This theorem is useful in simplifying complex exponential expressions involving sine functions.

2. How is De Moivre's theorem related to trigonometric identities?

De Moivre's theorem is closely related to the trigonometric identities for sine and cosine functions. It can be used to prove various trigonometric identities, such as the double angle formula for sine and cosine.

3. Can De Moivre's theorem be used for any value of n?

Yes, De Moivre's theorem can be used for any integer value of n. This includes negative values, which can be expressed using the reciprocal property of complex numbers.

4. Are there any limitations to using De Moivre's theorem for individual sine functions?

One limitation of De Moivre's theorem is that it can only be used for individual sine functions, and not for combinations of sine and cosine functions. It also only applies to complex numbers in polar form.

5. How can De Moivre's theorem be applied in real-world situations?

De Moivre's theorem has many applications in fields such as engineering, physics, and mathematics. It can be used to solve problems involving oscillations, vibrations, and alternating currents. It is also useful in solving problems related to waves and harmonic motion.

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