# Complex numbers

1. Dec 27, 2014

### erisedk

1. The problem statement, all variables and given/known data
$$If arg(\frac{z-ω}{z-ω^2}) = 0, \ then\ prove \ that\ Re(z) = -1/2$$
2. Relevant equations
ω and ω^2 are non-real cube roots of unity.

3. 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.

Last edited: Dec 27, 2014
2. Dec 27, 2014

### SammyS

Staff Emeritus
If ω is one of the non-real cube roots of unity, then what is ω2, and how is it related to ω ?

3. Dec 27, 2014

### erisedk

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

4. Dec 27, 2014

### SammyS

Staff Emeritus
I was thinking in terms of the complex conjugate.

5. Dec 27, 2014

### erisedk

ω is the conjugate of ω^2.

6. Dec 28, 2014

### Dick

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

7. Dec 28, 2014

### haruspex

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