# Homework Help: Complex numbers

1. Mar 13, 2010

### Mentallic

1. The problem statement, all variables and given/known data
Show that if $$\theta$$ is not a multiple of $$2\pi$$ then

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

2. Relevant equations
$$e^{i\theta}=cos\theta+isin\theta$$

3. 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.

2. Mar 13, 2010

### 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.

3. Mar 13, 2010

### Mentallic

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

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...

4. Mar 13, 2010

### 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.

5. Mar 13, 2010

### Dick

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

6. Mar 13, 2010

### 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

$$\frac{sin\left(\frac{1}{2}(n+1 )\theta\right)sin(\frac{1}{2}n\theta)}{sin\frac{1} {2}\theta}$$

7. Mar 13, 2010

### 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.

8. Mar 13, 2010

### Dick

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

9. Mar 13, 2010

### Mentallic

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?

10. Mar 13, 2010

### 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.

11. Mar 13, 2010

### vela

Staff Emeritus
One trick you can use is to note that

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

12. Mar 13, 2010

### Dick

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!

13. Mar 14, 2010

### Mentallic

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

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