How to plot solutions to equations on a unit circle in complex plane

I have the following code to solve the equations, and wanted to plot them on unit circle on the complex plane, one by one. For example, this code gives me $$6$$ solutions, thus I would like to plot six circles, and on each, plot one of solutions as a unit vector on it.

sol = Solve[
Sin[a - b] + Sin[a] == 0 && Sin[b - a] + Sin[b] == 0 &&
Sin[-a] + Sin[-b] == 0, {a, b}]


Any idea? Thanks a lot.

• It is not clear to me,what exactly you want. a and b are only defined up to a multiple of Pi. Therefore it makes sense to plot them on the unit cycle. That would give 2 points, one for a and one for b. However, what do you mean by "as a vector"? Feb 6, 2022 at 16:14
• @DanielHuber For example, the first solution gives $a=2\pi c_1,b=-2\pi c_2$, therefore I need to plot two vectors on the unit circle: $(\cos(2\pi c_1),\sin(2\pi c_1)$ and $(\cos(-2\pi c_2),\sin(-2\pi c_2))$
– M.K
Feb 6, 2022 at 16:33
• Since c is required to be an integer, both {Cos[2 Pi c], Sin[2 Pi c]} and {Cos[-2 Pi c], Sin[-2 Pi c]} are the point {1, 0} Feb 6, 2022 at 16:38

ContourPlot[{Sin[a - b] + Sin[a] == 0, Sin[b - a] + Sin[b] == 0,
Sin[-a] + Sin[-b] == 0}, {a, -10, 10}, {b, -10, 10},
PlotPoints -> 50]


From above plot,we can find that all of {a,b} is the intersection of three contours,it means that only two independent equations in the original three equations. We can verified this by

Simplify[Sin[-a] + Sin[-b] == 0,
Sin[a - b] + Sin[a] == 0 && Sin[b - a] + Sin[b] == 0]


True

And we can plot all the points by MeshFunction.

ContourPlot[{Sin[a - b] + Sin[a] == 0} , {a, -10, 10}, {b, -10, 10},
MeshFunctions -> {Function[{a, b}, Sin[a - b] + Sin[a]],
Function[{a, b}, Sin[b - a] + Sin[b]]}, Mesh -> {{0}},
MeshStyle -> Directive[Opacity[1], Red], ContourStyle -> Opacity[.1],
PlotPoints -> 50]


The same as

Graphics[{Red, Point[{a, b}]} /.
Table[Solve[
Sin[a - b] + Sin[a] == 0 && Sin[b - a] + Sin[b] == 0 &&
Sin[-a] + Sin[-b] == 0, {a, b}] /. {C[1] -> c1,
C[2] -> c2}, {c1, -2, 2}, {c2, -2, 2}], Axes -> True]


OP

For a or b,AngleVector[a] or AngleVector[b] have only four points.

Graphics[{Red, Point[AngleVector[a]], Point[AngleVector[b]]} /.
Table[Solve[
Sin[a - b] + Sin[a] == 0 && Sin[b - a] + Sin[b] == 0 &&
Sin[-a] + Sin[-b] == 0, {a, b}] /. {C[1] -> c1,
C[2] -> c2}, {c1, -10, 10}, {c2, -10, 10}], Axes -> True]

• Thank you so much!! Could you explain why 'we can find that all of {a,b} is the intersection of three contours,it means that only two independent equations in the original three equations.' ?
– M.K
Feb 7, 2022 at 11:16