>>62
z := x + y*i (x, y∈R)
e^z = -1
⇔ exp((x + y*i)(p.v.Log|e| + i*arg(e))) = -1
⇔ exp(x - 2nπy)*exp((2nπx + y)i) = -1
⇔ x = 2nπy and 2nπx + y = (2m + 1)π
⇔ x = 2nπy and y = (2m + 1)π/((2nπ)^2 + 1)
⇔ z = (2m + 1)π/(2nπ - i)
⇒ |z| = |2m + 1|π/sqrt(4n^2π^2 + 1) (m, n∈N)
だからS = {(2n-1)π}はありえないよ