# Angle between two segments described using complex numbers

Assume we have two segments, ACAC and BCBC. We can represent points AA, BB and CC on the complex plane with three complex numbers, respectively a,ba,b and c∈Cc\in\mathbb{C}. My question is: is there a nice formula for the angle between these two segments?

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arg(b−ca−c)\arg(\frac{b-c}{a-c})?
– A.G.
2 days ago

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You can do a translation of the points:

A′=A−CB′=B−CC′=C−C=0
A’ = A – C \\
B’ = B – C \\
C’ = C – C = 0

Now C′C’ is the origin, you can get the arguments of A′A’ and B′B’ and subtract them:

α=arg(B′)−arg(A′)\alpha = \arg(B’) – \arg(A’)

So α\alpha is the angle between A′C′A’C’ and B′C′B’C’. But since translations preserve angles, α\alpha is also the angle between ACAC and BCBC. To sum up,

α=arg(B−C)−arg(A−C)\alpha = \arg(B – C) – \arg(A – C)

This is so far the best answer, as it gives directly the angle.
– hetajr
2 days ago

Assuming that the order of the points is A−B−CA-B-C going counter-clockwise around the triangle, we first of all put CC in the origin (so c=0c = 0). Then aa is a complex number with absolute |a||a| and argument α\alpha, and bb is a complex number with absolute |b||b| and argument β\beta. If you multiply aa with a number zz whose absolute is |b|/|a||b|/|a| and argument is ϕ=β−α\phi = \beta – \alpha, you get a number with absolute |b||b| and argument β\beta (hence, this is bb). This number is given by

b=az,z=|b||a|eiϕ.b = az\,, \quad z = \frac{|b|}{|a|}e^{i\phi}\,.

If you now replace aa by a−ca – c and bb by b−cb – c in this derivation, you shift CC back to an arbitrary place and the formula becomes

(b−c)=(a−c)|b−c||a−c|eiϕ,(b – c) = (a – c)\frac{|b – c|}{|a – c|}e^{i\phi}\,,

where (\phi) is the angle between these two segments. This gives you

eiϕ=b−ca−c|a−c||b−c|.e^{i\phi} = \frac{b – c}{a – c}\frac{|a – c|}{|b – c|}\,.