Rotation formula Complex numbers

[erisedk]

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.

 

[SammyS]

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 ?

 

[erisedk]

1 + ω + ω^2 = 0
$$ω = e^\frac{i2\pi}{3}$$
How does that help?

 

[SammyS]

1 + ω + ω^2 = 0
$$ω = e^\frac{i2\pi}{3}$$
How does that help?

I was thinking in terms of the complex conjugate.

 

[erisedk]

ω is the conjugate of ω^2.

 

[Dick]

You have that (z-ω)=k(z-ω^2) where k is real and positive. Take the real part of both sides.

 

[haruspex]

Your target is Re(z). How else can you write that?