>>665
じゃ、z = re^iθとして、r = |z|, θ=∠zだから、
z1+z2 について、|z1+z2| = |1+√3| |1+i| = (1+√3)√2 = √2+√6
∠|z1+z2| = ∠|1+i| = π/4.
z1-z2 = (1-√3)(1-i) = (√3-1)(-1+i)も 同様に、
|z1-z2| = (√3-1)√2 = √6-√2、∠|z1-z2| = (3/4)π