Proving Im(\frac{1-e^{i(n+1)\theta}}{1-e^{i\theta}})

[Mentallic]

Homework Statement

Show that if \theta is not a multiple of 2\pi then

Im\left(\frac{1-e^{i(n+1)\theta}}{1-e^{i\theta}}\right)=\frac{sin\left(\frac{1}{2}(n+1)\theta\right)sin(\frac{1}{2}n\theta)}{sin\frac{1}{2}\theta}

Homework Equations

e^{i\theta}=cos\theta+isin\theta

The Attempt at a Solution

I noticed that \frac{1-e^{i(n+1)\theta}}{1-e^{i\theta}}\right) is a geometric summation with e^{i\theta}=r then we have:

1+e^{i\theta}+e^{i2\theta}+...+e^{in\theta}

So,

Im\left(1+e^{i\theta}+e^{i2\theta}+...+e^{in\theta}\right)=sin\theta+sin2\theta+...+sin(n\theta)

I have no idea how to show this summation is equal to what I have to show. Most likely I'm not even headed in the right direction.

 

[Dick]

If you can express your quantity in the form a+ib, then it's easy to find Im(a+ib)=b. First multiply numerator and denominator by the complex conjugate of the denominator and then split stuff into real and imaginary parts.

 

[Mentallic]

Ok then I have <br /> \frac{1-e^{i(n+1)\theta}}{1-e^{i\theta}}\right)=\frac{1-cos(n+1)\theta-isin(n+1)\theta}{1-cos\theta-isin\theta}<br />

Multiplying through by the conjugate and expanding:\frac{1-cos\theta+isin\theta-cos(n+1)\theta+isin(n+1)\theta+cos\theta cos(n+1)\theta +icos\theta sin(n+1)\theta +isin\theta cos(n+1)\theta -sin\theta sin(n+1)\theta}{2-2cos\theta}Just the imaginary part now, and simplifying since sin\theta cos(n+1)\theta +cos\theta sin(n+1)\theta=sin(n+2)\theta\frac{sin\theta+sin(n+1)\theta+sin(n+2)\theta}{2(1-cos\theta)}Seems interesting, because of the pattern. But I'm unsure what I can do from here...

 

[Mentallic]

If the expression I showed is supposed to be equal to what I'm trying to show, then I must've failed to expand correctly.
Using a simple case of n=1, \frac{sin\left(\frac{1}{2}(n+1 )\theta\right)sin(\frac{1}{2}n\theta)}{sin\frac{1} {2}\theta} is supposed to be sin\theta. I graphed my expression and it is definitely equal to sin\theta...

I'll check over my expanding again.

 

[Dick]

Use some trig identities. I can see right away that your denominator is sin(theta/2)^2.

 

[Mentallic]

Ok I did the algebra slowly and got the correct result now.

It's

\frac{sin\theta+sin n\theta-sin(n+1)\theta}{2-2cos\theta}

I have to somehow show this is equal to

<br /> \frac{sin\left(\frac{1}{2}(n+1 )\theta\right)sin(\frac{1}{2}n\theta)}{sin\frac{1} {2}\theta}<br />

 

[Mentallic]

It's been a while since I used half-angle trig identities so I don't remember them off-hand. I'll see if I can derive them in that case...

But really, this topic is on complex numbers. I don't see why it would go into such great depths to test my algebra and trigonometry skills.

 

[Dick]

It's been a while since I used half-angle trig identities so I don't remember them off-hand. I'll see if I can derive them in that case...

But really, this topic is on complex numbers. I don't see why it would go into such great depths to test my algebra and trigonometry skills.

Yes, I suppose. Might have been better to have you prove the result in the form you got first and skip the tricky trig.

 

[Mentallic]

Yes, I suppose. Might have been better to have you prove the result in the form you got first and skip the tricky trig.

Well, we've already started on this method, let's not lose the momentum thus far

So 2-2cos\theta=4sin^2(\theta/2)

and to get anywhere with the numerator, I had to cheat a little bit. Using the answer that I need to find, by the double-angle formula:

sin\left(\frac{1}{2}(n+1)\theta\right)sin\left(\frac{n\theta}{2}\right)=sin\left(\frac{\theta}{2}\right)cos\left(\frac{n\theta}{2}\right)sin\left(\frac{n\theta}{2}\right)+cos\left(\frac{\theta}{2}\right)sin^2\left(\frac{n\theta}{2}\right)

and

2sinxcosx=sin(2x)

so it can be simplified to

\frac{1}{2}sin\left(\frac{\theta}{2}\right)sin\left(n\theta\right)+cos\left(\frac{\theta}{2}\right)sin^2\left(\frac{n\theta}{2}\right)Any other identities I can exploit?

 

[Dick]

Arrrgh. Finally got it. I was going in circles for while. I'll write t for theta.
i) Expand sin(nt+t) in the numerator using the sin addition formula.
ii) Combine with the other two terms looking to get (1-cos(something*t)) factors.
iii) Change the (1-cos(something*t)) stuff into sin(something*t/2)^2.
iv) Change the remaining t stuff into t/2 stuff using sin(x)=2*sin(x/2)*cos(x/2).
v) Pull out the common factors and recognize what's left as another sin addition formula.

Let's not do that again soon. And I totally agree, not much to do with complex numbers here.

 

[vela]

One trick you can use is to note that

1-e^{ix} = e^{ix/2}(e^{-ix/2}-e^{ix/2})=-2ie^{ix/2}\sin(x/2)

 

[Dick]

One trick you can use is to note that

1-e^{ix} = e^{ix/2}(e^{-ix/2}-e^{ix/2})=-2ie^{ix/2}\sin(x/2)

Ah! That helps. I was looking for an easier way to do it once I started messing with the trig. But I didn't look back far enough. Thanks!

 

[Mentallic]

Ah! That helps. I was looking for an easier way to do it once I started messing with the trig. But I didn't look back far enough. Thanks!

Well this question is unfair for those few people that don't know that equality off by heart