Sketching loci in the complex plane

[jj364]

Homework Statement

Make a sketch of the complex plane showing a typical pair of complex numbers
z₁ and z₂


Describe the geometrical figure whose vertices
are z₁, z₂ and z₀ = a + i0.

Homework Equations

z₂ − z₁ = (z₁ − a)e^i2π/3

a − z₂ = (z₂ − z₁)^i2π/3

where a is a real positive constant.

The Attempt at a Solution

I really am not sure what to do on this question, my initial thoughts were that the solution would look like 3 lines in the complex plane all 2π/3 apart so that it would look like the solution to a roots of unity question.

I tried to rearrange to give z₂ in terms of a which yielded

z₂(1+e^2πi/3 - e^2πi/3/(1+e^2πi/3) = a(1+e^4πi/3/(1+e^2πi/3))

But to be honest I really don't know where I am going with this!
Thanks in advance!

 

[Michael Redei]

I think you're missing an "e" in your second "relevant equation" and that you mean

a − z₂ = (z₂ − z₁) e^i2π/3​

instead. If that's right, you might try to interpret geometrically what multiplying a complex number by e^i2π/3 means.

 

[jj364]

e^2πi/3=-1/2 +i√3/2 so multiplying by it would change the real and imaginary components accordingly.

So would it be best to split into real and imaginary components so z1=x1=iy1 and z2=x2+iy2, then substitute these into the equations?
Which i think gives

z₂=1/2(x₁+a) + i3y₁√3 /2

 

[Michael Redei]

e^2πi/3=-1/2 +i√3/2 so multiplying by it would change the real and imaginary components accordingly.

That doesn't really say a lot. What does such a multiplication f(z) = ze^2πi/3 look like geometrically? If you sketch 1 and f(1) in the complex plane, how could you describe the geometrical operation that takes you from 1 to f(1)? How about 1/2 and f(1/2)? How about 1+i and f(1+i)? If you can discover some similarity, you can apply this knowledge to your original problem.

 

[jj364]

Ok, so does it rotate them by 2π/3 keeping the same magnitude?

But I'm still struggling to work out my problem from this. Do I need to just think about it or can I actually solve the problem using the equations, because I've tried eliminating to no avail?

 

[jj364]

Actually I think I might have worked it out. I think it is just the solutions to z^3=1 so

1, e^{[itex]\frac{2\pi i}{3}[/itex]}, e^{[itex]\frac{4\pi i}{3}[/itex]}

I tried it for the equations and it worked, is this right? They are all rotations of 2pi/3 of each other so it does make sense.