Sketching loci in the complex plane

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jj364
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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)ei2π/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+e2πi/3 - e2πi/3/(1+e2πi/3) = a(1+e4πi/3/(1+e2πi/3))

But to be honest I really don't know where I am going with this!
Thanks in advance!
 
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I think you're missing an "e" in your second "relevant equation" and that you mean
a − z2 = (z2 − z1) ei2π/3
instead. If that's right, you might try to interpret geometrically what multiplying a complex number by ei2π/3 means.
 
e2πi/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
 
jj364 said:
e2πi/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) = ze2πi/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.
 
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?
 
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.