• Support PF! Buy your school textbooks, materials and every day products Here!

Complex numbers

  • Thread starter Mentallic
  • Start date
  • #1
Mentallic
Homework Helper
3,798
94

Homework Statement


Show that if [tex]\theta[/tex] is not a multiple of [tex]2\pi[/tex] then

[tex]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}[/tex]


Homework Equations


[tex]e^{i\theta}=cos\theta+isin\theta[/tex]


The Attempt at a Solution



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

[tex]1+e^{i\theta}+e^{i2\theta}+...+e^{in\theta}[/tex]

So,

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

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.
 

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
26,258
618
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
Mentallic
Homework Helper
3,798
94
Ok then I have [tex]
\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}
[/tex]

Multiplying through by the conjugate and expanding:


[tex]\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}[/tex]


Just the imaginary part now, and simplifying since [tex]sin\theta cos(n+1)\theta +cos\theta sin(n+1)\theta=sin(n+2)\theta[/tex]


[tex]\frac{sin\theta+sin(n+1)\theta+sin(n+2)\theta}{2(1-cos\theta)}[/tex]


Seems interesting, because of the pattern. But I'm unsure what I can do from here...
 
  • #4
Mentallic
Homework Helper
3,798
94
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, [tex]\frac{sin\left(\frac{1}{2}(n+1 )\theta\right)sin(\frac{1}{2}n\theta)}{sin\frac{1} {2}\theta}[/tex] is supposed to be [tex]sin\theta[/tex]. I graphed my expression and it is definitely equal to [tex]sin\theta[/tex]...

I'll check over my expanding again.
 
  • #5
Dick
Science Advisor
Homework Helper
26,258
618
Use some trig identities. I can see right away that your denominator is sin(theta/2)^2.
 
  • #6
Mentallic
Homework Helper
3,798
94
Ok I did the algebra slowly and got the correct result now.

It's

[tex]\frac{sin\theta+sin n\theta-sin(n+1)\theta}{2-2cos\theta}[/tex]

I have to somehow show this is equal to

[tex]
\frac{sin\left(\frac{1}{2}(n+1 )\theta\right)sin(\frac{1}{2}n\theta)}{sin\frac{1} {2}\theta}
[/tex]
 
  • #7
Mentallic
Homework Helper
3,798
94
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
Dick
Science Advisor
Homework Helper
26,258
618
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.
 
  • #9
Mentallic
Homework Helper
3,798
94
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 :smile:

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

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:

[tex]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)[/tex]

and

[tex]2sinxcosx=sin(2x)[/tex]

so it can be simplified to

[tex]\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)[/tex]


Any other identities I can exploit?
 
  • #10
Dick
Science Advisor
Homework Helper
26,258
618
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
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
14,623
1,257
One trick you can use is to note that

[tex]1-e^{ix} = e^{ix/2}(e^{-ix/2}-e^{ix/2})=-2ie^{ix/2}\sin(x/2)[/tex]
 
  • #12
Dick
Science Advisor
Homework Helper
26,258
618
One trick you can use is to note that

[tex]1-e^{ix} = e^{ix/2}(e^{-ix/2}-e^{ix/2})=-2ie^{ix/2}\sin(x/2)[/tex]
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
Mentallic
Homework Helper
3,798
94
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 :biggrin:
 

Related Threads on Complex numbers

  • Last Post
Replies
5
Views
722
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
6
Views
2K
  • Last Post
Replies
6
Views
473
  • Last Post
Replies
1
Views
698
  • Last Post
Replies
10
Views
2K
  • Last Post
Replies
1
Views
872
  • Last Post
2
Replies
35
Views
3K
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
3
Views
2K
Top