Rotation formula Complex numbers

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Homework Help Overview

The discussion revolves around proving that if the argument of the ratio of two complex expressions involving non-real cube roots of unity is zero, then the real part of a complex variable must equal -1/2. The problem is situated within the context of complex numbers and their geometric interpretations.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the implications of the equality of arguments, with one suggesting a relationship between the expressions involving the cube roots of unity. There is a discussion about the potential use of the rotation formula and how to analytically prove the result. Questions arise regarding the nature of the cube roots of unity and their relationships, as well as how to express the real part of the complex variable.

Discussion Status

The discussion is active, with participants sharing insights and attempting to clarify the relationships between the cube roots of unity. Some guidance has been offered regarding taking the real part of the expressions, but no consensus or resolution has been reached yet.

Contextual Notes

Participants are working under the constraints of the problem statement and the properties of cube roots of unity, with some uncertainty about how to proceed with the proof analytically.

erisedk
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Homework Statement


If arg(\frac{z-ω}{z-ω^2}) = 0, \ then\ prove \ that\ Re(z) = -1/2

Homework Equations


ω and ω^2 are non-real cube roots of unity.

The Attempt at a Solution


arg(z-ω) = arg(z-ω^2)
So, z-ω = k(z-w^2)
Beyond that, I'm not sure how to proceed. Using the rotation formula may also be required, but I'm not sure how to use it here.
Furthermore, I can sort of intuitively visualise the answer, but I'm not sure how to prove it analytically.
 
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erisedk said:

Homework Statement


\text{If}\ \arg \left( \frac{z-ω}{z-ω^2} \right) = 0, \ \text{then prove that }\ Re(z) = -1/2

Homework Equations


ω and ω^2 are non-real cube roots of unity.

The Attempt at a Solution


arg(z-ω) = arg(z-ω^2)
So, z-ω = k(z-w^2)
Beyond that, I'm not sure how to proceed. Using the rotation formula may also be required, but I'm not sure how to use it here.
Furthermore, I can sort of intuitively visualise the answer, but I'm not sure how to prove it analytically.
If ω is one of the non-real cube roots of unity, then what is ω2, and how is it related to ω ?
 
1 + ω + ω^2 = 0
ω = e^\frac{i2\pi}{3}
How does that help?
 
erisedk said:
1 + ω + ω^2 = 0
ω = e^\frac{i2\pi}{3}
How does that help?
I was thinking in terms of the complex conjugate.
 
ω is the conjugate of ω^2.
 
You have that (z-ω)=k(z-ω^2) where k is real and positive. Take the real part of both sides.
 
Your target is Re(z). How else can you write that?
 

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