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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2 Answers

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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|}\,.