Proving/Creating a conjecture on the roots of complex numbers

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SUMMARY

The discussion centers on formulating and proving a conjecture for the equations (z^3)-1=0, (z^4)-1=0, and (z^5)-1=0, focusing on the roots of complex numbers. Participants conclude that the solutions can be expressed as θ = (2π/n)k for any integer k, where n represents the degree of the polynomial. The conjecture is supported by the relationship cos(nθ) = 1, leading to the identification of specific angles that yield these roots. The final proof is established by substituting the conjecture into the original equations.

PREREQUISITES
  • Understanding of complex numbers and their roots
  • Familiarity with Euler's formula e^(iθ) = cos(θ) + i sin(θ)
  • Knowledge of trigonometric identities, particularly cos(nθ)
  • Basic algebraic manipulation skills for proof construction
NEXT STEPS
  • Study the derivation of roots of unity in complex analysis
  • Learn about the geometric interpretation of complex roots on the unit circle
  • Explore the use of Euler's formula in solving polynomial equations
  • Investigate advanced proof techniques in mathematical conjectures
USEFUL FOR

Students and educators in mathematics, particularly those focusing on complex analysis, algebra, and proof techniques. This discussion is beneficial for anyone looking to deepen their understanding of the roots of complex numbers and their applications.

  • #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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