z² (1 − cos(z))=0
z = 0 or cos (z) = 1
z = 0 or 𝑧 = 2𝑘𝜋
When z = 0
let 𝑓(𝑧) = z², ℎ(𝑧) = 1− cos (𝑧)
then 𝑓(0) = 0, 𝑓′(0) = 2𝑧 = 0, 𝑓″(0) = 2 ≠ 0
thus order = 2
ℎ(0)=0, ℎ′(0)= 2 sin(0)=0, h″(0)= cos(𝑧)≠0
thus order = 2
Therefore, order (0) = 4
When 𝑧 = 2𝑘𝜋 (𝑘≠0)
let 𝑓(𝑧) = z² and ℎ(𝑧) = 1−cos(𝑧)
𝑓(𝑧) = z² = (2𝑘𝜋)² ≠ 0
thus order = 0
ℎ(𝑧)= 1− cos(𝑧)=0, ℎ′(𝑧)= sin(𝑧)=0, ℎ″= cos(𝑧)=0
thus order = 2
Therefore order (2𝑘𝜋) = 2, 𝑘 ≠ 0