If z1 and z2 are complex numbers such that |z1|=|z2| and |z1+z2|=1 find z1/z2
- JOHNLv 72 months agoFavorite Answer
In the diagram showing a unit circle, ∠FAB = ∠GAB = π/3 → Δs FAB, GAB are equilateral. Let F = z1, G = z2, B = w . We have |z1| = |z2| = |w| = |z1 + z2| = 1. Let AFBG rotate, say anticlockwise, by θ. The geometry doesn’t change and we still have the relations |z1| = |z2| = |z1 + z2| =1. In fact from the geometry it is evident that positions of the rotated rhombus are the only possible ones satisfying these relations. Thus z1 = exp[i(π/3 + θ), z2 = exp[i(-π/3 + θ) → z1/z2 = exp[i(π/3 + θ)/ exp[i(-π/3 + θ) = [exp(iθ)exp(iπ/3)]/[exp(iθ)exp(-iπ/3)] = exp(2iπ/3) = -1/2 + i√3/2.
This, I think, vindicates a z lender’s conjecture.
- az_lenderLv 72 months ago
One possibility is that z1 = 1/2 + i*sqrt(3)/2 and z2 = 1/2 - i*sqrt(3)/2.
In that case, z1/z2 would be
(1/2 + i*sqrt(3)/2)^2 / 1 = -1/2 + i*sqrt(3)/2.
You'd have to do some more work to see if this is the only possible result.