Complex Trigonometric Solutions

Click For Summary

Homework Help Overview

The discussion revolves around solving the polynomial equation x7 = 1, specifically focusing on finding all solutions, including complex solutions, using trigonometric methods. Participants are also tasked with plotting these solutions on the unit circle.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the steps taken to express the equation in terms of trigonometric identities, particularly using the Rcis notation. Questions arise regarding the elimination of the R variable and the implications of the cis notation. There is also a focus on solving for θ and the conditions under which the solutions are derived.

Discussion Status

The discussion is active, with participants sharing their reasoning and clarifying steps. Some guidance has been offered regarding the interpretation of the roots and the plotting of solutions on the unit circle. Multiple interpretations of the steps are being explored, particularly concerning the use of trigonometric identities.

Contextual Notes

Participants note the importance of understanding the relationship between the real and imaginary components in the context of complex numbers. There are mentions of typos and clarifications regarding notation, which may affect the clarity of the discussion.

Bogrune
Messages
60
Reaction score
0

Homework Statement


2.) Solve
x7=1
I am to solve for all the solutions to this polynomial, including the complex solutions using trigonometry. I'm supposed to plot my solutions on the unit circle, with the x-axis being the real axis, and the y-axis being the imaginary (or complex) axis.

Homework Equations


I'm supposed to use the following Identities to solve the equation: Rcis, meaning: R (real) times cosine + i(complex) sine.
cisn(θ)=cis(nθ)

The Attempt at a Solution


I've numbered my steps taken in order:

x7=1
1.)(Rcisθ)7=1

2.)Rcis(θ)7=1

3.)cis7(θ)=1

4.)cis(7θ)=1

5.)cos(7θ) + isin(7θ)=1 + i0

5a.) cos(7θ)=1
7θ= cos-1(1)
7θ= 0 + 2πn

5b.) sin(7θ)=0
7θ= sin-1(0)


And that's where I got stuck.
 
Last edited:
Physics news on Phys.org
Bogrune said:

I've numbered my steps taken in order:

x7=1
1.)(Rcisθ)7=1

2.)Rcis(θ)7=1

3.)cis7(θ)=1



Do you have a "relevant equation" that explains why you can get rid of the R. (I don't use the cis notation, myself, so I'm not sure what you are supposed to say)

4.)cis(7θ)=1

5.)cos(7θ) + isin(7θ)=1 + i0

5a.) cos(5θ)=1

it should say cos(7θ) , of course

7θ= cos-1(1)
7θ= 0 + 2πn

5b.) sin(7θ)=0
7θ= sin-1(θ)


It should say sin-1(0). (The solution won't add any new information because it will also imply that 7θ= 0 + 2πn


And that's where I got stuck.

Solve 7θ= 0 + 2πn for θ

Let n = 0,1,2... and you'll come back to where you started when n/7 = 1
 
I simply got rid of the "R" because it's not really a variable. I forgot to mention that "R" simply stands for "real", because of the fact that cosθ lies on the x-axis (or the real axis, in this case) and isinθ lies on the y (imaginary)-axis.
And sorry about the typos, I make those pretty often! ^^;
Solve 7θ= 0 + 2πn for θ

Let n = 0,1,2... and you'll come back to where you started when n/7 = 1
Oh, I think I see it now. So I plot those on the unit circle as θ= (0 +2πn)/7, and I get my final solution, right?

Though, thanks for the help!
 
Last edited:
Yes, that's right. You are plotting what are called "the seventh roots of unity". In the complex numbers, 1 has 2 square roots, 3 cube roots, 4 fourth roots, etc.
 

Similar threads

  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
Replies
13
Views
2K
  • · Replies 7 ·
Replies
7
Views
9K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K
Replies
39
Views
6K
  • · Replies 11 ·
Replies
11
Views
3K