Proving/Creating a conjecture on the roots of complex numbers

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Homework Help Overview

The discussion revolves around formulating and proving a conjecture related to the roots of complex numbers, specifically for the equations (z^3)-1=0, (z^4)-1=0, and (z^5)-1=0. Participants explore the implications of the polar form of complex numbers and the conditions under which certain values for angles yield roots of unity.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss potential conjectures regarding the angles that satisfy the equations, with some suggesting values like 2π/n and 2π/n + π. There are attempts to clarify the conditions under which cos(nθ) equals 1, and questions arise about the generality of the conjectures and the notation used.

Discussion Status

The discussion is active, with various participants contributing ideas and questioning each other's reasoning. Some guidance has been offered regarding how to articulate conjectures and proofs, and there is a recognition of the need to clarify notation and assumptions. The conversation reflects a collaborative effort to refine the conjecture and explore its implications.

Contextual Notes

Participants express uncertainty about how to properly format mathematical notation and articulate their thoughts, indicating a level of stress due to impending deadlines. There is also a focus on ensuring that conjectures are clearly defined and proven within the constraints of the homework assignment.

  • #31
λ ε {2 π k/k}k=0∞
 
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  • #32
where's n ? :confused:
 
  • #33
tiny-tim said:
where's n ? :confused:


Oh woops θ ε {2nkπ/n}k=0∞
 
  • #34
now, too many n's ! :smile:
 
  • #35
AAAAAAH. I swear there's a lot wrong up there. θ ε {2nkπ/k}k=0∞
 
  • #36
erm :redface: …

too many k's ?​
 
  • #37
θ ε {2nkπ/k}k=0∞
 
  • #38
that's the same as your last one

(and 2nkπ/k = 2nπ))
 
  • #39
2kpi/n ? I think I'm lost.
 
  • #40
yes, θ = (2π/n) times k, for any integer k, are the solutions to cos(nθ) = 1

(they're also the solutions for einθ = 1 … they correspond to n equally-spaced positions on the unit circle)

is that the answer to the original question? :smile:​
 
  • #41
Yes! How would I prove it though? :/
 
  • #42
Algebraically, I already have the graph.
 
  • #43
Daaniyaal said:
Yes! How would I prove it though? :/

which part of the proof are you not clear about? :confused:
 
  • #44
I figured it out, if I were to just replace the statement with my conjecture it would be proven, thanks!
 

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